Bruck’s vision of regular spreads or what is the use of a baer superspace

Abh. Math. Sem. Univ. Hamburg 63 (1993), 37-54

Bruck's Vision of Regular Spreads or

What is the Use of a Baer Superspace?

By A. BEUTELSPACHER and J. UEBERBERG

Dedicated to WALTER BENZ on the occasion of his 60 th birthday

I Introduction

In [8] BRUCK established a relation between the theory of regular spreads and the theory of Baer subspaces of a projective space. This relation is a powerful tool in the study of spreads. In order to present his ideas we first introduce the notions of regular spreads and of Baer subspaces.

Let P = PG(3, F) be a 3-dimensional pappian projective space. A regulus of P is a set ~ of disjoint lines of P satisfying the following two conditions:

�9 If a line l of P meets three lines of ~ , then l intersects every line of ~.

�9 If a line l of P meets three lines of ~ , then every point of l is incident with (exactly) one line of ~.

Let us recall the following property of 3-dimensional pappian projective spaces: If Ii, 12 and 13 a re three disjoint lines, then there is exactly one regulus containing ll, 12 and 13.

A spread of P is a set 6 e of skew lines of P such that each point of P is incident with (exactly) one line of 5 a. A spread 6: is called regular if for any three lines of 6 e the regulus determined by these three lines is contained in 6 e (for more details about regular spreads cf. section 3). More generally, if P is a (2t + 1)-dimensional projective space, then a spread of P is a set 5 e of skew lines of P such that each point of P is incident with (exactly) one line of P. A spread 6: is called regular if the following property is fulfilled: If T is a 3-dimensional subspace of P containing at least two lines of 6e, then the lines of 6: contained in T form a regular spread of T.

Let P* = PG(d, F*) be a d-dimensional pappian projective space such that the field F* admits a subfield F of index 2. The set P of points of P* with homogenous coordinates in F is a projective subspace of P" isomorphic to PG(d, F). The projective subspace P is called a Baer subspace of P* (for a synthetic definition of Baer subspaces cf. 2.1 and 2.5).

The following property of Baer subspaces is fundamental: If P is a Baer subspace of P* and if x is a point of P* outside of P, then there is exactly one

38 A. Beutelspacher and J. Ueberberg

line l through x (for a proof cf. 2.6). We say that the line l is induced by the point x. We are now able to present the above mentioned theorem of BRUCK.

Theorem 1.1. (BRucK 1969) Let P* = PG(3,q 2) be a 3-dimensionalfinite projective space of order q2, and let P be a Baer subspace of P*. Note that P is of dimension 3 and of order q.

a) I f l* is a line of P* disjoint to P, then the family of lines of P induced by the points of l* is a regular spread of P.

b) I f ~ is a regular spread of P, then there are exactly two lines l~ and l~ of P* disjoint to P such that ~9 ~ is induced by the points of l~ and by the points of l~.

The aim of this paper is to give a generalization of the theorem of BRUCK to infinite pappian projective spaces of dimension 2t + 1. The generalization from the 3-dimensional case to the case of dimension 2t + 1 is rather straight forward whereas the generalization to infinite projective spaces requires some new techniques. Our main result reads as follows (cf. 5.2 and 5.3):

Theorem 1.2. a) Let P* = PG(2t + 1, F ' ) be a 2t + 1-dimensional pappian projective space such that the field F" admits a subfield F of index 2. Let P be a Baer subspace of P*. I f U* is a t-dimensional subspace of P* disjoint to P, then the family SP of lines of P induced by the points of U* is a regular spread of P.

b) Let P = PG(2t + 1, F) be a 2t + 1-dimensional pappian projective space admitting a regular spread o~. Then there exists a quadratic extension F" of the field F fulfilling the following properties:

�9 The projective space P* = PG(2t + 1, F*) admits P as a Baer subspace.

�9 I f F* is a separable field extension of F, then there are exactly two t-dimensional subspaces U~ and U~ of P* disjoint to P such that 5 v is induced by the point set of U~ and the point set U~.

I f F* is a non-separable field extension of F, then there is exactly one t- dimensional subspace U* of P* disjoint to P such that ~9 ~ is induced by the point set of U*.

In particular we obtain that the existence of a regular spread implies that the field F admits a quadratic field extension. As an application of the above theorem we give a geometric proof of the following known theorem (cf. [17]).

Theorem 1.3. Let P = PG(2t+ 1, F) be a (2t+ 1)-dimensional pappian projective space admitting a regular spread ~ . Let F be the following rank 2 geometry: The points of F are the lines of 6# and the lines of F are the 3-dimensional subspaces of P containing at least two lines of 5 '~. Then F is a pappian projective space. The field coordinatizing F is a quadratic field extension of F.

Bruck's Vision of Regular Spreads 39

Section 2 is devoted to the study of Baer subspaces of a projective space. Following DEMBOWSKI [13] we give a synthetic definition of a Baer subspace of a projective space, and we discuss some of its elementary properties. The subject of section 3 are the regular spreads of a 3-dimensional pappian projective space P. The main result of this section is Theorem 1.2 in the case of dimension 3. As a corollary it follows that if ~ is a regulus of P and if g is a line disjoint to ~ , then there is exactly one regular spread through and g. In section 4 we present the very nice proof of BRUCK of Theorem 1.1. In section 5 we shall prove Theorem 1.2. Furthermore we give a proof of Theorem 1.3.

2 Baer Subspaces of Projective Spaces

Definition 2.1. Let P* be a (not necessarily finite) projective plane. A subplane P of P* is called a Baer subplane if it fulfils the following conditions:

(i) Each line of P* has at least one point in common with P.

(ii) Every point of P* is incident with at least one line of P.

Baer subplanes of projective planes have been introduced by BAER [2] in order to determine the set of fixed points of automorphisms of order 2 of a projective plane.

If P* is a pappian projective plane, then the Baer subplanes of P* may be described as follows.

Proposition 2.2. Let P* be a projective plane over a commutative field F* and let P be a subplane of P*. Then the following assertion holds: P is a Baer subplane i f and only i f P is a projective plane over a subfield F of F* of degree 2.

Proof. First, suppose that P is a projective plane over a subfield F of degree 2 of F*. Let

ax + by + cz = 0

be the equation of a line l* in P*. Since F* is a 2-dimensional vector space over the field F the set {a, b, c} is linearly dependent over F. Hence there are three elements x, y and z of F with (x, y, z) :fi (0, 0, 0) such that ax + by + cz = O. In other words the point (x, y, z) is a point of P on the line r of P*.

In a similar way one shows that every point of P* is incident with at least one line of P.

In order to show the converse direction of the proposition, let P be a Baer subplane of P*. Then there is a subfield F of F* such that P can be coordinatized over F. Assume that F* is a vector space of dimension at least 3 over F. Let a, b and c be three linearly independent elements of F* and consider the line l* with the equation

ax + b y + c z = 0 .


Since a, b and c are linearly independent, there are no three elements x, y and z of F with (x, y, z) :p (0, 0, 0) fulfilling the above equality. In other words, the line l* is disjoint to the Baer subplane P, a contradiction.

Hence F has to be a subfield of F* of degree 2. []

Let us consider for the moment the case of a finite projective plane.

Proposition 2.3. Let P* be a finite projective plane of order q, and let P be a subplane of P'. Then the following four conditions are equivalent:

a) P is a Baer subplane of P*.

b) Each line of P* contains at least one point of P.

c) Every point of P* is incident with at least one line of P.

d) The order q of P* is a square and P is a subplane of order x/q"

This proposition is due to BRUCK [7]. For a proof cf. also BEUTELSPACHER [5], Kap. VII, Satz 1.

In the definition of a Baer subplane we requested both properties, (i) and (ii). In view of Proposition 2.3 this is not necessary for finite projective planes. Also, one easily deduces from the proof of Proposition 2.2 that in any pappian projective plane the conditions (i) and (ii) are equivalent. However, in general the conditions (i) and (ii) are independent, as has been observed by BARLOTTI [3].

Theorem 2.4. (BARLOTTI [3]) There exists an infinite projective plane P* admitting a subplane P fulfilling the following properties.

�9 Every line of P* intersects P in at least one point.

�9 There are points of P* which are not incident with any line of P.

This construction of BARLOTTI is a synthesis of the work of ANDRE [1] and of BRUEN and FISHER [12]. ANDRE described how to construct a translation plane from a given spread and BRUEN and FtSI-tER constructed a spread that is not a dual spread.

We are now going to discuss higher-dimensional Baer subspaces.

Definition 2.5. Let P* be a projective space of dimension at least 3. A projective subgeometry P of P* is called a Baer subspace if every point of P* is incident with at least one line of P.

If P* is of dimension d and if t < d, then a t-dimensional Baer subspace of P" is a Baer subspace of a t-dimensional subspace of P*.

Proposition 2.6. Let P* be a projective space over a commutative field F* and let P be a projective subspace of P*. Then the following assertions are equivalent.


a) P is a Baer subspace of P*.

b) I f ~c* is a plane of P* containing three non-collinear points of P, then P A zt* is a Baer subplane of ~*.

c) There is a subfield F of index 2 of F* such that P can be coordinatized over F.

Proof The proof is similar to the proof of Proposition 2.2 and will be omitted here. []

In finite projective spaces the following assertion holds: If P* is a finite projective space of order q and if P is a projective subspace of P ' , then P is a Baer subspace if and only if q is a square and P is of order x/-q" (This is an immediate consequence of 2.3 and 2.6.)

If P is a Baer subspace of the d-dimensional projective space P* we say that an s-dimensional subspace U* of P* is a subspace of P if the subspace P n U" is an s-dimensional subspace of P. We note the remarkable property that every point of P" is incident with a line of P (by definition) and that every hyperplane of P" intersects P in a subspace of dimension at least d - 2 (cf. 2.7). In fact, the following more general result is true.

Proposition 2.7. Let P* be a d-dimensional pappian projective space admitting a Baer subspace P.

a) Every ( d - t)-dimensional subspace of P* intersects P in a subspace whose dimension is at least d - 2 t .

b) Every s-dimensional subspace of P" is contained in an at most (2s + 1)- dimensional subspace of P.

Proof In view of the definition of a Baer subplane we may suppose that dim P" _ 3.

a) We proceed by induction on t. Step I. Let H* be a hyperplane of P*. By induction we will show that H* AP

contains a subspace of dimension i for all i E {0, 1 . . . . . d - 2}. For this pupose let p be a point of P outside H' . Consider two lines l and m of P passing through p and denote by l* and m* the corresponding lines of P*. Let x" and y" be the intersection points of l" and m* with H*. Since l and m are two lines of P, the lines l* and m* span a plane re* of P* containing a subplane of P. By Proposition 2.6 the line x'y" of n" contains a point of P. Hence the hyperplane H" contains at least one point of P.

Now suppose that H" intersects P in a subspace U of dimension i - 1 with i < d - 2 . Let p be a point of P outside of H* and let ll, 12 . . . . . la be d lines of P through p intersecting H* in d independent points xb x2 . . . . , xa such that Xl . . . . . xi are points of U. Since i < d - 2 , the lines ld-1 and ld are disjoint to U. The lines Id-1 and ld span a Baer subplane rc of P. By construction,


and U are disjoint. Since z is a Baer subplane of P ' , the line joining the points xa-1 and xa has a point z in common with P. Since the point z is contained in H" but not in U, the hyperplane H" contains the subspace (U, z) of dimension i of P. So every hyperplane of P* intersects P in a subspace whose dimension is at least d - 2.

Step 2. Let S" be a (d -O-dimensional subspace of P*. Then S* intersects P in a subspace of dimension at least d - 2t. Consider a (d - t + l)-dimensional subspace T" of P" through S'. By induction, T" contains a ( d - 2t + 2)- dimensional subspace U of P. Denote by U" the corresponding subspace in p*.

If U is contained in S' , then S" contains even a (d - 2t + 2)-dimensional subspace of P. From now on we suppose that U is not contained in S'. Then the intersection V" := S" n U" of U ~ and S" is a subspace of dimension d ' - I where d' = d - 2 t + 2 = d i m T ' . Hence V is a hyperplane in U and by induction, it follows that V* intersects the Baer subspace U of U" in a subspace of dimension at least d' - 2 = d - 2t. By construction, this (d - 2t)-dimensional subspace is contained in S' .

b) Let S* be an s-dimensional subspace of P" and let Xl, x2 . . . . . Xs+l be s + 1 independent points of S'. By definition of the Baer subspace there are s + l lines 11 . . . . , ls+l of P such that the line li is incident with the point xi for i = l . . . . . s § 1. The lines ll . . . . , ls+l span a Baer subspace of P" through S* of dimension at most 2s + 1. []

3 Regular Spreads in PG(3, F)

Definition 3.1. Let P be a 3-dimensional projective space.

a) A regulus ~ is a set of skew lines of P satisfying the following conditions.

�9 If a line I of P meets three lines of ~, then l intersects every line of ~.

�9 If a line l of P meets three lines of ~ , then every point of l is incident with (exactly) one line of ~.

b) A line of P intersecting all lines of ~ is called a transversal.

c) Note that if ~ is a regulus of P, then the set of transversals of ~ is itself a regulus, the so-called opposite regulus of ~.

Let us recall the following results concerning reguli (for a proof cf. [9] and [17]).

Theorem 3.2. Let P be a 3-dimensional projective space over a field F.

a) I f F is a commutative field (i.e. i f P is pappian), then the following assertion holds: Let la, 12 and 13 be three disjoint lines of P. Then there exists exactly one regulus of P containing the lines ll, 12 and 13.


b) I f F is a non-commutative field, there is no regulus in P.

c) I f ~1 is a regulus of P, then the point set o f P covered by the lines o f forms a hyperbolic quadric.

Proposition 3.3. Let P" be a pappian projective space of dimension 3 admitting a Baer subspace P, and let ~1 be a regulus o f P.

a) There is exactly one regulus ~* of P" containing the lines of ~1.

b) Let l* be a line of the regulus ~*. Then l* is either an element o f ~ or l* is disjoint to P.

c) Let t" be a transversal o f ~ ' . Then t* is either a transversal o f ~1 or t* is disjoint to P.

Proof a) Let ll, 12 and 13 be three lines of ~ and let l[, l 2 and l~ be the corresponding lines of P' . Let ~" be the regulus of P" through the lines l;, l~ and I~. Since each transversal of :~ is also a transversal of ~ ' , it follows that :~ is contained in :~'.

b) Let .~ be the hyperbolic quadric belonging to ~ and let l ~ be a line of the regulus ~" not belonging to ~. Assume that l" and P intersect in a point p. The point p is not contained in .~ (otherwise l" would be the line of through p, contradicting the choice of l').

Let a and b be two lines of :~ and let t be the line of P through p intersecting both lines a and b. Since t intersects the three lines a, b and l" of ~ ' , t is a transversal of ~ ' . Since t is a line of P, it follows that t is a transversal of ~ and therefore contained in & in contradiction to the fact that the point p on t is not contained in 2.

c) Consider the opposite regulus ~* of ~ and apply b). []

Definition 3.4. Let P be a 3-dimensional pappian projective space.

a) A spread of P is a set 6 e of skew lines such that every point of P is incident with (exactly) one line of ~ .

b) A spread ~ of P is regular if for any three lines of if' the regulus through these three lines is contained in 6 e.

Since the subject of this paper are regular spreads, we may, in view of Theorem 3.2, restrict ourselves to pappian projective spaces.

Lemma 3.5. Let P be a Baer subspace of a 3-dimensional pappian projective space P*. Let x and y be two points o f P~ outside of P and let lx and ly be the lines of P through x and y, respectively. Then the lines lx and ly are skew if and only i f the line joining x and y is disjoint to P.


Proof Firstly, suppose that the line xy is disjoint to P. Assume that lx and ly intersect in a point. Then the plane n spanned by lx and ly is a Baer subplane of P" containing the line xy. Hence, there is a point of P on the line xy, a contradiction.

On the other hand, suppose that the line xy contains exactly one point of P. The plane n generated by the lines lx and xy is a Baer subplane of P~ If ly is contained in n, the lines lx and ly intersect in a point what we wanted to show. Assume that ly is not contained in n. Then n and ly intersect in the point y which implies that y is a point of P, a contradiction.

If the line xy is a line of P, then we have lx = l y = xy. []

Theorem 3.6. Let P* be a 3-dimensional pappian projective space and let P be a Baer subspace of P. Let l ~ be a line of P" disjoint to P and let 5r be the set of lines of P intersecting l*. Then S,e(l *) is a regular spread of P.

Proof Step I. 6e(l ") is a spread. By Lemma 3.5 it follows that the lines of 5e(1 ") are skew. Let x be a point of P and let n* be the plane generated by x and l*. By 2.7 a), there is a line m of P in z ' . Assuming that x and m are not incident we obtain that n" is a plane of P, contradicting the fact that l" does not contain any point of P. Hence x and m are incident. Since m and l" are both contained in n ' , they intersect and m is therefore a line of 5P(l ") through x. So 5r ") is a spread of P.

Step 2. 5P(l*) is a regular spread. Let a, b and c be three lines of 5r By 3.2 a) these three lines determine a regulus ~t in P and a regulus ~" in P~ containing Y/. Since the line l" intersects each of these lines a, b and c, l" is a transversal of the regulus ~*. Let d be a fourth line of the regulus ~. Then d is also a line of the regulus ~t" and therefore intersects l" in a point. It follows that d belongs to 5e(l*). Hence 6P(l ") is a regular spread. []

Definition 3.7. Let P be a Baer subspace of the 3-dimensional pappian projective space P' . Let l* be a line of P* disjoint to P. The regular spread 5e(l ") consisting of the lines of P intersecting l ~ is called the regular spread induced by l ' .

Lemma 3.8. Let P = PG(3, F) be a 3-dimensional pappian projective space, and let ~ be a regulus of P. Then there exists a line g of P disjoint to ~ if and only if the polynomial ring F[x] contains an irreducible polynomial of degree 2.

Proof Step 1. Let .~ be the hyperbolic quadric corresponding to the regulus ~. By a suitable choice of coordinates we may suppose that

-~ {(x,, ' ' I ' ' E F and ' ' = 0 } X2, X 1, X2) X1X 2 X1, X2, X 1, X 2 - - XIX2

(for a detailed description cf. [8], w pp. 494-496). Let nl and n2 be the two planes given by the equations Xl = 0 and x2 = 0,

and let l be the intersection line of nl and n2. Then l is a line contained


in 3' (hence it is a line of ~ or a transversal of ~), and rq and ~2 are two planes, each intersecting 3" in two lines. Besides l, nl contains the line of 3' with equation

X l = O, Xtl = 0

and z~2 contains the line of 3' with equation

x2 = 0 , x~ = 0 .

In the following we shall determine all lines of P that are disjoint to 3'. Since l is contained in 3' we may restrict ourselves to the lines intersecting the planes rq and r~2 in points outside 3".

A point p = (O, p2,p~,p'2) of nl is outside 3' if and only if p~ :p 0 and P2 =P 0. A point q = (qb O, q'l, q'9) of n2 is outside 3. if and only if ql :P 0 and

! t q~ =p 0. Let p = (O, p2,Pl,P2) and q = (qbO, q'l,q'2) be two points of nl and n2, respectively, and let g be the line of P joining p and q. Finally, let x be a point on g different from p and q. Then x has the coordinates

p + ),q = (2ql,p2,P'l + ),q'l,P'2 + ),q2)

for some 0 ~ 2 ~ F. The line g and the quadric 3" have a point in common if and only if the

equation

0 = 2ql(p~ + 2q i ) - (p] + 2q~)p2

qlq~2 2 + (P~ql ' ' p2ql)2-- (*) = - - P l P 2

has a solution in F.

Step 2. If every polynomial of degree 2 is reducible over F, then the equation (*) has a solution 20 in F. Hence the point (~,ql, P2, Pl Jr-~q'l, P2 + ~,q2) is a common point of g and 3'.

Step 3. Suppose that some polynomial

2 2 + a 2 + b

is irreducible over F. Then b ~ 0. Consider the two points p :-- (0, -1 , -b , a) and q := (1,0,0, 1). Then p and q are in the planes Z~l and 7z2, respectively. Furthermore, p and q are outside of 3". By Step 1, the line g joining p and q intersects 3' if and only if the equation

2 2 + a ~ , + b = 0

has a solution in F. Since the polynomial 32 + a2 + b is irreducible over F, it follows that the line g is disjoint to .~. []

Corollary. There is no regular spread in a 3-dimensional projective space over a quadratically closed field.


Another way to prove the above corollary is the following. If P is a projective plane over an algebraically closed field, and if c~ is a conic in P, then every line of P intersects ~ in at least one point.

Theorem 3.9. Let P --- PG(3, F) be a 3-dimensional pappian projective space, and let ~ be a regulus of P. Suppose that there exists a line g of P disjoint to ~. Then there exists a field extension F ~ of degree 2 of F such that the following conditions are fulfilled.

(i) The projective space P* := PG(3,F*) over the field F* contains P as a Baer subspace. (P* is a Baer superspace of P.)

(ii) I f F* is a separable field extension (in particular i f char F 7~ 2), then there are exactly two lines t~ and t~ of P* disjoint to P such that the regular spreads 6~(t~) and 6~(t[) induced by t~ and t~ contain the regulus ~ and the line g. (As we shall see later, there is exactly one regular spread through ~ and g, in particular we have 6e(t~) = 5e(t~).) The lines t~ and t~ are skew.

(iii) I f char F = 2 and if F* is an inseparable field extension of F, then there is exactly one line t* of P* disjoint to P inducing a regular spread through the regulus ~ and the line g.

Proof Denote by .~ the hyperbolic quadric corresponding to ~. As in the proof of 3.7, we may suppose that

! / [ f ? / ? = t " "~(XI 'X2 'XI 'X2] �9 F and x lx 2 - xzx I = 0~, X1,X2~ X 1,X2

As in the proof of 3.7 we consider the planes nl and r~2 with the equations xl = 0 and x2 = 0, respectively. Let l be the intersection line of nl and re2.

Since the line g is disjoint to the quadric ~, it intersects the plane re1 in a point p = (0, P2, Pfl, p~) with P'I --P O, P2 :P 0 and it intersects the plane rr2 in a point q = (q~, 0, q~, q~) with ql ~ 0 and q~ ~ 0.

Consider the polynomial

f(2) := qlq~,~ 2 + (P~2ql -- P2ql) 2 ' --.fflP2.

Since g is disjoint to .~, it follows as in the proof of 3.7 that f is irreducible. Let F* be the field extension of degree 2 of F with respect to the polynomial

f . Since F* is a field extension of degree 2 of F, by 2.6, the projective space P* = PG(3, F ~ contains P as a Baer subspace.

Let ~* be the regulus through ~ and let ~* be the quadric corresponding to ~*. Then

,,~* = {(Xl t ' [ X l , X 2 , X l , X 2 and x l x 2 ,x2, xl,x2) ' ' E F* ' - x2x' 1 = 0}.

We first consider the case where char F :fi 2 or char F = 2 and F* is a separable field extension of F. In both cases, F* is a separable field extension


and we obtain the existence of two (different) roots 21 and 22 of F ~ of the polynomial f . Then the points

x := (21ql,P2,P'l + 21q'l,P~ + 21q~)

and

! !

y := (22ql,P2, Prl -4- 22ql,P2 -+" J.2q2)

are the common points of g and 2". Let t~ and t~ be the transversals of ~* through the points x and y,

respectively. Note that t~ and t~ are different transversals (otherwise we would obtain t~ = t~ = g, contradicting the fact that g is a line of P). By construction, the regular spreads 5:(t~) and 5e(t~) contain the regulus :~ and the line g. Since t~ and t~ are two transversals of :~', they are skew.

It remains to show that there is no further line of P* disjoint to P inducing a regular spread through :~ and g. Assume the contrary that t ~ is such a line. Since the regular spread 5:(t') induced by t" contains ~ , t" is a transversal of ~ ' . Hence the line g intersects the three transversals t~, t~ and t" which implies that g belongs to the regulus :~* in contradiction to 3.3 b).

Let us now consider the case where F" is an inseparable field extension of F (in particular char F = 2). In this case the equation

qlq2 ,;],2 + (P~zql - PZq'l) 2 + PtlP2 = 0

has exactly one solution over F ' , say 20. The point

! ! !

z := (2oqbPZ, Pl + 2oql,P2 + 20q~)

is the only common point of g and .,~*. The transversal t" of the regulus ~" through z induces a regular spread through :~ and g in P.

It remains to show that there is no further line of P" disjoint to P inducing a regular spread through ~ and g. Assume to the contrary that there is such a line, say s ~ Since each line of ~ intersects s', the line s" is a transversal of ~~ Therefore s" is contained in 2". Furthermore it follows that g and s" intersect in a point, say z'. Since g n ~." = z, we obtain z = z'. Hence s" is a transversal of the regulus ~" through z which implies s" -- t ' , a contradiction to the assumption. []

Theorem 3.10. Let P = PG(3, F) be a 3-dimensional pappian projective space such that P contains a regulus ~ and a line g disjoint to ~ . Then there exists exactly one regular spread through ~ and g.

Proof By Theorem 3.9, there exists at least one regular spread 6el through and g. Assume that 6~2 is a further regular spread containing ~ and g.

The lines and the reguli of the regular spreads ~x and ,Sf 2 define two MiSbius planes Ml and M2. Let l0 be an arbitrary line of the regulus ~. Then l0 is a


line of '-~1 and 5"2. The derivations of the M6bius planes M1 and M2 with respect to the line l0 are two affine planes A1 and A2. The points of Al and of A2 are the reguli of 5el and 5e2 through 10, respectively. The regulus corresponds to a line R' of the affine plane A1. The line g corresponds to a point g' of AI not incident with R'.

In the following we prove that 5"1 is contained in 5e2. Let h be a line of 5el. W.l.o.g. we may suppose that h :~ g and that h does not belong to ~.

Case 1. Suppose that h is a line such that the corresponding point of the affine plane A1 lies on a line of the form g'x' where x' is a point of R'. Since the point x' corresponds to a line l of the regulus ~ , the line h is contained in the regulus generated by the lines 10, g and I. Since these three lines are also contained in 5"2, it follows that h is contained in 5e2.

Case 2. Let h be a line of 5el such that the corresponding point h' in A1 lies on the parallel line of R' through the point g'. If h' = g', then h = g E 5e2. From now on, suppose that h' ~ g'. Let 1' be a point on the line R' and let k' be a third point on the line joining 1' and h t. Because of Case 1, the point k t corresponds to a line k of 5el N 5e2. Furthermore the point I t corresponds to a line of the regulus ~. Therefore h is a line of the regulus generated by 10, k and I. Since this regulus is contained in 5el and in 5"2, it follows that h is an element of 502.

From the cases 1 and 2 it follows that 5el is contained in 5e2. Since 5el and 5e2 are both spreads, we obtain equality. []

In the following theorem we summarize the results of this section.

Theorem 3.11. Let P = PG(3, F) be a 3-dimensional pappian projective space and let Jt be a regulus of P. Suppose that there exists a line g of P disjoint to ~t.

a) There is exactly one regular spread through ~l and g.

b) The field F admits a quadratic field extension F ~ and P is a Baer subspace of P" = PG(3, F'). Furthermore, one of the following possibilities occurs.

(i) F" is a separable field extension o fF , and there are exactly two lines t~ and t~ of P" disjoint to P inducing the regular spread through JI and g. The lines t~ and t~ are skew.

(ii) F" is a non-separable field extension of F, and there is exactly one line t* of P" disjoint to P inducing the regular spread through ~ and g.

Remark 3.12. In the situation of 3.11 b)(i) there are two field automorphisms of the field F ~ fixing the field F, namely the identity and an automorphism z of order 2. The field automorphism z induces a collineation ~ of P" fixing all points of the Baer subspace P. Furthermore we have

~(t ; ) = t~.


This fact explains, why there are two lines of P* inducing the regular spread through the regulus ~ and the line g. In the situation of 3.11 b)(ii) the only field automorphism of F ~ fixing the elements of F is the identity, explaining the fact that there is one line of P" inducing the regular spread through ~ and g.

4 Regular Spreads in PG(3, q)

The motivation of the present paper is the following theorem of BRUCK [8].

Theorem 4.1. Let P be a Baer subspace of P* = PG(3,q2), and let ~ be a regulus of P. Furthermore let g be a line of P disjoint to ~.

a) There is exactly one regular spread of P through ~t and g.

b) There are exactly two lines of P* disjoint to P inducing the regular spread of P through ~1 and g.

In this finite situation BRUCK [8] gave a very nice and purely synthetic proof of this result. We cannot resist to present his proof.

Lemma 4.2. Let P = PG(3, q), and let ~1 be a regulus of P. Then there are exactly

~q 2(q--1)2

lines of P disjoint to ~1.

Proof Let .~ be the quadric corresponding to ~'. There a r e tq+l := 2(q + 1) lines of P contained in .~. If p is a point of -~, there are exactly q2 lines through p intersecting .~ in two points, q - 1 lines through p intersecting .~ in the point p and two lines through p contained in _~. It follows that there are t2 := l(q + 1)2q2 lines intersecting .~ in two points and tl := (q + 1)2(q - 1) tangent lines (that is lines intersecting ~ in exactly one point). Let

b : = ( q 2 + l ) ( q 2 + q + l )

be the total number of lines of P. For the number d of lines of P disjoint to it follows

" 1

= b - tq-1 - t2 - tl = ~q2(q _ d 1) 2 . [] Z .

Lemma 4.3. Let P be a Baer subspace of P" = PG(3, q2). Let ~ be a regulus of P, and let ~" be the regulus of P* through ~t. Then every line of P disjoint to ~ intersects exactly two transversals of P*.


Proof. Let .~ and .~* be the quadrics corresponding to ~ and ~*, respectively. Furthermore denote by T* the set of all transversals of ~* that are not transversals of ~t. By 3.3 all lines of T* are disjoint to P. Note that [T'I = q2 _ q.

Step 1. Each line of P disjoint to .~ intersects at most two lines of T*. Assume to the contrary that there is a line g of P disjoint to ~ intersecting at least three lines of T*. Then g is a line of ~*. In particular g is either a line of ~ or a line disjoint to P. Both cases contradict the choice of g.

Step 2. Every line of P disjoint to ~ intersects two transversals of P* Con- sider the rank 2 geometry (D(~), S(T*), ~_), where D(~) is the set of lines of P disjoint to ~ and where S(T ~ is the set of regular spreads of P induced by the lines of T ~ For 3: E S(T*) let [~] be the number of lines of D(6:) contained in 3e, and for g ~ D(~) let [g] be the number of spreads of S(T*) containing g. Counting the incidences of (D(~), S(T')) we obtain

E [if'] = (q2 _ q)(q2 _ q) ~ES(T*)

and

[g] ~ 21D(~)l = q2(q _ 1)2 _- (q2 _ q)(q2 _ q), gEO(~t)

where the last equality follows from Lemma 4.2. It follows that [g] = 2 for all g E D(~t) ; in other words each line of D(~) intersects exactly two lines of T ~ . []

We are now able to prove Theorem 4.1.

Proof. a) The uniqueness follows from Theorem 3.10.

b) Every line of P* inducing a regular spread of P through ~ is a transversal of the regulus N" of P" through N. It follows from Lemma 4.3 that there are exactly two lines of P* disjoint to P inducing a regular spread through ~ and g. []

Note that all lemmas and theorems used for the proof of Theorem 4.1 (that is 3.6, 3.6 and 3.10) have been proved using purely synthetic arguments.

5 Regular Spreads in PG(2t + 1, F)

Definition 5.1. Let P = PG(2t + 1,F) be a pappian projective space.

a) A spread of P is a set 6: of skew lines such that each point of P is incident with (exactly) one line of 6:.

b) A spread 6# is called regular if the following property is fulfilled. If T is a 3-dimensional subspace of P containing at least two lines of 6:, then the lines of 6: contained in T are a regular spread of T.


Theorem 5.2. Let P* be a (2t + 1)-dimensional pappian projective space admitting a Baer subspace P. Let U* be a t-dimensioanl subspace of P* disjoint to P. Then the set 5~(U *) consisting o f all lines of P intersecting U* is a regular spread o f P.

Proof. Step I. 5r ~ is a partial spread, that is the lines of 5r are skew. Assume to the contrary that there are two lines g and g' of 5~(U *) intersecting in a point x. Since g and g' are both lines of P, it follows that the point x belongs to P, as well. Therefore x ~ U*. Consider the plane n of P generated by the two lines g and g' and denote by n* the corresponding plane of P*. Then n* and U* intersect in a line h*. Since n is a Baer subplane of n*, the line h ~ has at least one point in common with P, contradicting the assumption U * A P = ~ .

Step 2. 5a(U *) is a spread. For, let x be a point of P and let V* be the (t + 1)-dimensional subspace of P* generated by x and U*. By Proposition 2.7 a) there is a line g of P in V ~ Assuming that x and g are not incident, 7t ~ = (x, g) is a plane of P*, which intersects U ~ in a line. Since n* contains the Baer subplane n := r t*n P, this line would contain a point of P, a contradiction to the choice of U*.

Hence X and g are incident, and g is a line of 5r *) through x. It follows that 6e(U*) is a spread.

Step 3. 6r is a regular spread. Let ll and 12 be two lines of 5~(U*), and let T be the 3-dimensional subspace of P spanned by Ii and 12. Let xl and x2 be the intersection points of U ~ and Ii and 12, respectively. The line xlx2 is a line of U ~ inducing a regular spread 5P7- in T through 11 and 12. So the spread 5e(U*) induces the regular spread 6eT in T. Hence 6e(U ") is a regular spread of P. []

Theorem 5.3. Let P = PG(2t + 1, F), t > 2, be a (2t + 1)-dimensional pappian projective space. Let $1, $2 . . . . . St be t subspaces of P of dimension 3 satisfying the following properties:

(1) The subspaces $1,$2 . . . . . St have exactly one line, say m, in common.

(2) The subspaces Sl, $2 . . . . . St generate P.

Let ~1, J12 . . . . . ~tt be t reguli o f Sb $2 .. . . , St, respectively, all containing the line m. Suppose that there exists a line g of $1 disjoint to the regulus ~11.

a) There is exactly one regular spread 6r containing the reguli Yll . . . . ,~lt and the line g.

b) There is a field extension F* of degree 2 of F such that the following conditions are fulfilled:

(i) The projective space P* = PG(2t + 1,F*) is a Baer superspace of P.

(ii) I f F* is a separable field extension, then there are exactly two t- dimensional subspaces U~ and U~ of P" disjoint to P such that 6r =


(iii) I f F ~ is an inseparable field extension of F, then there is exactly one t-dimensional subspace U" of P" such that 5~(U *) = 6r

Proof Consider the 3-dimensional subspace $1 of P. By Theorem 3.11 there is exactly one regular spread 6el containing ~ and g. Furthermore there is a field extension F ~ of degree 2 of F fulfilling the following conditions:

�9 The projective space S~ := PG(3, F*) is a Baer superspace of Sl.

�9 If F" is a separable field extension of F, then there are exactly two lines h~ and h~ of S~ disjoint to $1 inducing the regular spread 6el.

�9 If F ~ is an inseparable field extension, then there is exactly one line h ~ of S~ disjoint to S1 inducing the regular spread 6el.

Let P* := PG(2t + 1, F*). Then P" is a Baer superspace of P. Let S[ be the 3-dimensional subspace of P" corresponding to Si for i = 1 . . . . . t, and let ~ be the regulus of S[ through ~i (i = 1,.. . , t). We first consider the case that F" is a separable field extension of F. Let h~ and h~ be the two lines of S~ inducing the regular spread 5el through ~ 1 and g. Let X 1 and X2 be the intersection points of m and h~ and h~, respectively. Furthermore let t~l and ti~ 2 be the transversals of the regulus ~ through the points xl and x2, respectively (i = 2 . . . . . t).

Define the subspaces U~ and U~ of P" as

U~ := ( h ~ , t ~ l l i = 2 . . . . . t)

and U~ := (h~,t~. 2 1 i = 2 . . . . . t).

By construction we have dim U~ = dim U~ = t. The regular spreads 5e(U~) and 5~(U~) induced by U~ and U~ contain the reguli N1,Nz . . . . . ~ t and the line g.

In order to show the uniqueness of U~ and U~ assume that there is a further t-dimensional subspace U" of P" inducing a regular spread through the reguli ~i and the line g. In view of Theorem 3.9 the line U ~ n S~ is either h~ or h~. W.l.o.g. suppose that U* N S~ = h~. Let k" := U ~ A S~ for some j ~ {2,..., t}. By construction k" is a line. Since U" induces a regular spread through the reguli ~1 . . . . . ~t , the line k" has to induce a regular spread of Sj through ~j . Therefore k" is a transversal of ~ . Since U" does not contain the common line m of ~ j and ~ l , it follows from k ~ ___ U" and h~ ___ U" that the lines k" and m and the lines h~ and m intersect in the same point Xl. Hence k" ~ U" " = t = dim U1, it follows U ~ = " = t j, 1. So ~ U~. Since dim U* * U 1.

The uniqueness of the regular spread through the reguli NI . . . . . Nt and a line g is a consequence of 3.11 a). In particular we have 5e(U~) = 5e(U~).

In the case that F ~ is an inseparable field extension of F, exactly the same reasoning as above shows that there is exactly one t-dimensional subspac e of P~ disjoint to P inducing the regular spread through the reguli N1, N2 . . . . . Nt and the line g. []


As an application of Theorem 5.3 we give a geometric proof of the following result due to BRUCK and BOSE [9] and SEGRE [17].

Theorem 5.4. Let P = PG(2t + 1, q, F) be a pappian projective space admitting a regular spread 6 a. Define the geometry P(5 e) of rank t as follows.

�9 The points of P(6 e) are the lines of 6e.

�9 The subspaces of dimension j of p(se) are those (2j + 1)-dimensioanl subspaces U of P such that 6e induces a (regular) spread in U ( j = 1 . . . . . t - l ) .

Then P(~) is a pappian projective space of dimension t that can be coordinatized over a suitable quadratic field extension F ~ of F.

Proof. It is easy to verify that the geometry P(6 e) is a t-dimensional projective space. By Theorem 5.3, there is a quadratic field extension F* of F such that P ' = PG(2t + 1, F*) admits P as a Baer subspace and such that P* contains a t-dimensional subspace U* disjoint to P and inducing the regular spread 6e.

We claim that U ~ and p(6e) are isomorphic. For, let

c~:U ~ P(6 p)

be the following map. For x E U let ct(x) be the line of P induced by x. Then is an isomorphism between U and P(S~). []

Probably, the most important case of Theorem 5.4 is the case t = 2. In this case Theorem 5.4 asserts that P (~) is a pappian projective plane.

Other applications of the relation between regular spreads and Baer subspaces are discussed in the context of the so-called indicator-sets, of subregular spreads and of parallelisms. The interested reader is referred to [11] and [6].

We conclude by mentioning an open problem: A necessary and sufficient condition for the existence of a regular spread in the projective space PG(3, F) is that F is a commutative field admitting a quadratic field extension. Consider the projective space P" = PG(3, ~/) over the Hamiltonians. Then P = PG(3, ~) can be embedded in P* as a Baer subspace. Let l* be a line of P* disjoint to P. As in the commutative case, for each point x of l" there is exactly one line of P induced by x, and the set 6e(l') of the lines of P induced by the points of l* is a spread of P. By the corollary to Theorem 3.8, 6e(l ") cannot be a regular spread. What are the geometric properties distinguishing such a spread 6P(l *) ?

Note that the translation plane associated to ~(l*) by ANDRk'S construction [1] is the affine plane over the Hamiltonians.

Acknowledgement. The authors would like to thank ARMIN HERZER for very valuable discussions.


References

[1] J. ANDRe, Uber nicht-desarguesche Ebenen mit transitiver Translationsgruppe, Math. Z. 60 (1954), 156-186.

[2] R. BAER, Projectivities with Fixed Points on Every Line of the Plane, Bull. Amer. Math. Soc. 52 (1946), 273-286.

[3] A. BARLOTrI, On the Definition of Baer Subplanes of Infinite Planes, J. Geom. 3 (1) (1973), 87-92.

[4] W. BENZ, Vorlesung fiber Geometrie der Algebren, Springer Verlag, Berlin- Heidelberg-New York 1973.

[5] A. BztrrELsPACIJER, Einf'tihrung in die endliche Geometrie II, Projektive R~iume, Bibliographisches Institut, Mannheim-Wien-Ziirich 1983.

[6] A. BEtrrELSPACHEg, On Parallelisms in Finite Projective Spaces, Geom. Ded. 3 (1974), 35-40.

[7] R.H. BgUCK, Difference Sets in a Finite Group, Trans. Amer. Math. Soc. 78 (1955), 464~81.

[8] R.H. BgUCK, Construction Problems of Finite Projective Planes, in: Proceedings of the Conference in Combinatorics held at the University of North Carolina at Chapel Hill, 10-14 April 1967, University of North Carolina Press (1969), 426-514.

[9] R.H. BRUCK and R.C. BosE, The Construction of Translation Planes from Pro- jective Spaces, J. Algebra 1 (1964), 85-102.

[10] R.H. BRUCK and R.C. BOSE, Linear Representations of Projective Spaces, J. Algebra 4 (1966), 117-172.

[11] A.A. BgUEN, Spreads and a Conjecture of Bruck and Bose, J. Algebra 23 (1972), 519-537.

[12] A.A. BRtrEN and J.C. FISHER, Spreads which are not Dual Spreads, Canad. Math. Bull. 12 (1969), 801-803.

[13] P. DEMBOWSKI, Finite Geometries, Springer Verlag, Berlin-Heidelberg-New York 1968.

[14] A. HEgZEg, Charakterisierung regul~irer Faserungen durch SchlieBungss~itze, Arch. Math. 25 (1974), 662-672.

[15] A. HERZEg, Synthetische Konstruktion affiner M6biusebenen, Arch. Math. 46 (1986), 91-96.

[16] G. Lt~ARDON, Insiemi indicatori proiettivi e fibrazioni planari di uno spazio proiettivo finito, Boll. Un. Mat. It. Ser. 6 3-B (1984), 717-735.

[17] B. SEGRE, Teoria di Galois, fibrazioni proiettivi e geometrie non desarguesiane, Ann. Mat. pura appl. 64 (1964), 1-76.

Eingegangen am: 10.01.1992

Authors' address: Albrecht Beutelspacher and Johannes Ueberberg, Mathematisches Institut, ArndtstraBe 2, W-6300 Giessen, Germany.

Bruck’s vision of regular spreads or what is the use of a baer superspace

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