Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 1, 7-21.

Reguli and Chains over Skew Fields

Andrea Blunck

Fachbereich Mathematik der TU Darmstadt Schloßgartenstraße 7, D-64289 Darmstadt, Germany

Abstract. We give a geometric definition of a regulus in a not necessarily pappian projective space of arbitrary dimension, such that the geometry of all reguli is a certain chain geometry. MSC 1991: 51A30, 51B05 Keywords: regulus, non-pappian projective space, chain geometry, skew field

The usual definitions of reguli ([5], [9]) and of chains ([1], [8]) only make sense over commutative fields: In a non-pappian projective space reguli do not exist. A chain geometry Σ(K, R) is only defined if the field K is contained in the center of the ring R.

In this paper we generalize both notions to skew fields, i.e., not necessarily commutative fields.

In Section 1 we collect some preliminary results. Then, in Section 2, we define reguli in a geometric way that works also for non-pappian spaces, thus generalizing both the usual definition and a definition due to B. Segre for three-dimensional projective spaces. We derive some properties of these reguli, using mainly linear algebra.

We want to remark here that H. Havlicek in [6] introduces Segre manifolds over skew fields, and thus also obtains a notion of reguli. It is not clear so far whether his notion coincides with ours.

In Section 3 we introduce a chain geometry Σ(K, R) associated to the set of all reguli in the not necessarily pappian projective space ℙ(K, V). We know that V ≅ U × U for some left vector space U over K. The ring R is the endomorphism ring of this vector space U. We prove that our reguli correspond to the chains of Σ(K, R). Thus we generalize a theorem proved in [2] for the pappian case.

In Section 4 we use the description of the reguli as chains in order to investigate the intersections of all reguli through three given subspaces.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}2000 Heldermann Verlag

1. Preliminaries

The usual definition of a regulus in a projective space of arbitrary dimension is as follows (see [2], [5], [9]):

1.1. Definition. Let ℙ be a projective space. A set R of at least three pairwise complementary subspaces of ℙ is called a regulus, if the following two conditions are satisfied: (R1) Any line of ℙ, transversal to three elements of R, is transversal to every element

of R. (R2) Each point on a line transversal to R belongs to an element of R.

Here we call a line L transversal to a subspace W of ℙ, if L meets W in a unique point. Then we also say that W is transversal to L. Moreover, we call L transversal to a family of subspaces, if it is transversal to each member of the family. Analogously we use the formulation that a subspace W is transversal to a family of lines.

Note that a regulus can only exist if ℙ is spanned by two isomorphic complementary subspaces. Hence ℙ = ℙ(K, V) for some left vector space V over a not necessarily commutative field K, where V ≅ U × U for another left vector space U over K. In the case that V is finite-dimensional this means that dim V must be even.

Moreover, apart from the trivial situation that dim V = 2 (where the whole set of points of the projective line ℙ(K, V) is a regulus), reguli exist only in pappian projective spaces, i.e., over commutative fields (see [5]):

1.2. Remark. Let dim V > 2. Then the following statements hold:

(i) If K is not commutative, then there are no reguli in ℙ(K, V).

(ii) If K is commutative, then any three pairwise complementary subspaces of ℙ(K, V) are contained in exactly one regulus.

Let us consider the special case that V is four-dimensional over the commutative field K: Then the elements of a regulus in ℙ(K, V) are lines. If G, H, L are pairwise complementary (i.e., pairwise skew) lines, then the unique regulus R containing them consists of all lines that meet the set R′ of all lines transversal to G, H, L. The set R′ is itself a regulus, called the regulus opposite to R. In particular, any regulus in the three-dimensional pappian projective space ℙ(K, V) can be described as the set of all lines that meet three fixed pairwise skew lines.

This observation was utilized by Segre in [10] in order to define reguli in such a way that they exist also in non-pappian spaces (see also [3]):

1.3. Definition. (B. Segre) Let ℙ be a three-dimensional projective space with line set G. Let T_{0}, T_{1}, T_{2} ∈ G be pairwise skew. Then the set

R = {G ∈ G | G is transversal to T_{0}, T_{1}, T_{2}}

is called a regulus.

Obviously the two definitions of a regulus coincide if the ground field K is commutative. Segre shows in [10] that only in the commutative case three pairwise skew lines belong to exactly one regulus. Otherwise there are many reguli through three given pairwise skew lines (compare Section 4 below).

In Section 2 we are going to generalize Segre’s definition to higher dimensions. In Section 3 we shall show that our reguli are the chains of a chain geometry. This was already shown for the commutative case in [2], [7].

Before stating our definitions, we need some terminology. We recall the following notions and results from [2]:

Since reguli again shall consist of at least three pairwise complementary subspaces of ℙ(K, V), we assume that V = U × U for some left vector space U.

We shall always identify a subspace of the vector space V with the associated projective subspace of ℙ(K, V).

Obviously, U ×{0}, {0}×U, and {(u, u) | u ∈ U} are pairwise complementary subspaces. A subspace W ≤ V is part of a triple of pairwise complementary subspaces exactly if it belongs to the set

G := {W ≤ V | W ≅ V/W (≅ U)}.

If dim V = 2n , then G consists of all n-dimensional vector subspaces of V, and thus corresponds to the set of (n−1)-dimensional projective subspaces of ℙ(K, V). In particular, if n = 2 these are the lines in ℙ(K, V).

The reguli we are going to define are always subsets of the set G.

Let R = End_{K}(U) be the ring of K-linear mappings U → U. As usual, by R^{∗} we denote the group of units of R. The projective line over R is the set

ℙ(R) := \{R(α, β) ≤ R^{2} | ∃γ, δ ∈ R : ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ GL_{2}(R)\} .

Two elements p = R(α, β) and q = R(γ, δ) of ℙ(R) are called distant, if the matrix

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) belongs to GL_{2}(R).

In [2] the following statements are shown:

1.4. Remark.

(i) Each W ∈ G has the form U^{(α,β)} := {(u^{α}, u^{β}) | u ∈ U} for a suitable pair (α, β) ∈ R^{2} with R(α, β) ∈ ℙ(R).

(ii) The mapping Φ : ℙ(R) → G : R(α, β) ↦ U^{(α,β)} is a well-defined bijection.

(iii) Two points p, q ∈ ℙ(R) are distant exactly if their images p^{Φ}, q^{Φ} ∈ G are complementary.

In particular, we have U × {0} = U^{(1,0)}, {0} × U = U^{(0,1)}, and {(u, u) | u ∈ U} = U^{(1,1)}, where 1 ∈ R is the identity on U and 0 is the zero mapping.

By definition, the group GL_{2}(R) operates transitively on ℙ(R) and leaves the relation “distant” invariant. It even acts transitively on the set of triples of pairwise distant points.

The group Aut_{K}(V) of all K-linear bijections V → V is isomorphic to GL_{2}(R); each linear bijection ψ ∈ Aut_{K}(V) has the form (u, w) ↦ (u, w)^{ψ} = (u^{α} + w^{γ}, u^{β} + w^{δ}), where

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) is a uniquely determined matrix in GL_{2}(R).

This leads directly to the following:

1.5. Remark. The actions of GL_{2}(R) on ℙ(R) and of Aut_{K}(V) on G are equivalent via Φ. In particular, the group Aut_{K}(V) acts transitively on the set of triples of pairwise complementary subspaces of ℙ(K, V).

In the sequel we shall always identify the element ψ ∈ GL_{2}(R) with the associated linear bijection in Aut_{K}(V). Moreover, by abuse of notation, we shall even denote the projective collineation of ℙ(K, V) induced by this linear bijection with ψ.

In Section 3 we need that the field K is embedded into the ring R as a subfield. This can only be done with respect to a fixed given basis. Assume that (b_{i})_{i∈I} is a basis of U. Then for k ∈ K we let λ_{k} be the linear mapping U → U defined by b_{i}^{λ_{k}} := kb_{i}. The mapping k ↦ λ_{k} is an embedding of K into R. Note that things become easier when restricting to the finite-dimensional case because then one might use (scalar) matrices. However, one has to keep in mind that also in that case λ_{k} is not the left multiplication u ↦ ku if K is not commutative.

2. A new definition of a regulus

In this section we generalize Segre’s definition of a regulus to higher dimensions. We give two definitions, one of them only makes sense in finite-dimensional projective spaces. Later we shall prove that both definitions coincide in this case.

Our first definition is a rather natural generalization of Segre’s.

2.1. Definition. Let V be of dimension 2n over K. Let T_{0}, T_{1}, …, T_{n} be n + 1 lines in ℙ(K, V) that are in general position (i.e., any n of them span ℙ(K, V)). Then the set

R := {W ∈ G | W is transversal to T_{0}, T_{1}, …, T_{n}}

is called a regulus of type (1) in ℙ(K, V).

We want to remark here that for commutative K this construction was already used by W. Burau in [4, IV, §33, 5] in the more general context of Segre manifolds. In case dim V = 2n a regulus as defined in 1.1 is a family of generators of a Segre manifold S_{1,n−1} according to Burau’s notation.

The following proposition collects some more or less immediate consequences of the definition.

Recall that a frame in an m-dimensional projective space ℙ is a family of m + 2 points such that any m + 1 of them span ℙ (and hence are a basis of ℙ).

2.2. Proposition. Let R be the regulus of type (1) transversal to T_{0}, …, T_{n}.

(i) For W ∈ R and i ∈ {0, …, n} let Q_{i} := W ∩ T_{i}. Then (Q_{0}, …, Q_{n}) is a frame of ℙ(K, W).

(ii) Through each point Q on any of the lines T_{i} there is exactly one W ∈ R.

(iii) Any two (different) elements W, W′ of R are complementary.

Proof. (i) is clear because the vector space W is n-dimensional and the T_{i} are in general position. (ii): Without loss of generality consider Q =: Q_{0} ∈ T_{0}. The subspace spanned by Q and T_{2}, …, T_{n} meets the line T_{1} in exactly one point, Q_{1} say. By induction one obtains points Q_{i} ∈ T_{i} (1 ≤ i < n) lying in the subspace spanned by Q_{0}, …, Q_{i−1} and T_{i+1}, …, T_{n}. The points Q_{0}, …, Q_{n−1} span an (n − 1)-dimensional projective subspace W that meets T_{n} and thus belongs to R. This W ∈ R is uniquely determined by Q_{0} because the induction process does not depend on the order of the lines T_{i}. (iii) is a consequence of (i) and (ii): Let Q_{i} = T_{i} ∩ W and Q′_{i} = T_{i} ∩ W′. By (1) we have W = ⨁_{i=1}^{n} Q_{i} and W′ = ⨁_{i=1}^{n} Q′_{i}. From (2) we know that T_{i} = Q_{i} ⊕ Q′_{i}, and hence V = ⨁_{i=1}^{n} T_{i} = ⨁_{i=1}^{n}(Q_{i} ⊕ Q′_{i}) = (⨁_{i=1}^{n} Q_{i}) ⊕ (⨁_{i=1}^{n} Q′_{i}) = W ⊕ W′. □

Now we come to our second definition. It is valid for projective spaces ℙ(K, V) of arbitrary dimension. Recall that we always assume that V = U × U for some vector space U.

2.3. Definition. Let V = U × U be a left vector space over K. Let (T_{i})_{i∈I} be a minimal family of lines spanning ℙ(K, V). Moreover, let H_{1}, H_{2} be two hyperplanes in ℙ(K, V) such that each line T_{i} (i ∈ I) meets H_{1} ∪ H_{2} in exactly two points (not belonging to H_{1} ∩ H_{2}). We say that a subspace W of ℙ(K, V) is regular with respect to ((T_{i})_{i∈I}, H_{1}, H_{2}), if

(i) W is transversal to (T_{i})_{i∈I} and spanned by the points (W ∩ T_{i})_{i∈I}, and (ii) W ⊆ H_{1} or W ⊆ H_{2} or W ∩ (H_{1} ∪ H_{2}) ⊆ H_{1} ∩ H_{2}.

The set

R = {W ≤ V | W is regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2})}

is called a regulus of type (2) in ℙ(K, V).

The first two cases of condition 2.3(ii) give two special elements of the regulus R.

2.4. Lemma. Let ((T_{i})_{i∈I}, H_{1}, H_{2}) and R be as in 2.3. For k ∈ {1, 2} we denote the subspace of ℙ(K, V) spanned by the points (T_{i} ∩ H_{k})_{i∈I} by X_{k}. Then the following statements hold:

(i) The subspaces X_{1} and X_{2} belong to the regulus R. They are isomorphic and complementary, hence they belong to the set G as well.

(ii) If a subspace W ∈ R contains one of the points T_{i} ∩ H_{k} (i ∈ I, k ∈ {1, 2}), then W = X_{k}.

Proof. (i): Obviously, X_{k} is regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}) and so belongs to R. The assumptions on ((T_{i})_{i∈I}, H_{1}, H_{2}) yield that V = X_{1}⊕X_{2}. Moreover, X_{1} ≅ X_{2} via T_{i}∩H_{1} ↦ T_{i} ∩ H_{2}. So X_{1}, X_{2} ∈ G. (ii): Assume that Q_{i} := T_{i} ∩ H_{k} belongs to W ∈ R. If W ⊆ H_{k}, then of course W = X_{k}. Otherwise 2.3(ii) implies Q_{i} ∈ H_{1} ∩ H_{2}, which is not allowed for a point of T_{i}. □

We also have some direct consequences of the second definition that are analogous to those of 2.2 above.

2.5. Proposition. Let R the regulus of type (2) regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}). (i) For W ∈ R and i ∈ I let Q_{i} := W ∩ T_{i}. Then (Q_{i})_{i∈I} is a basis of W. (ii) Through each point Q on any of the lines T_{i} there is exactly one W ∈ R. (iii) Any two elements W, W′ of R are complementary. Moreover, they are isomorphic

and thus belong to G.

Proof. (i) follows directly from the definition. (ii): Consider a point Q ∈ T_{i} for a fixed i ∈ I. By 2.4(ii) the only W ∈ R through T_{i} ∩ H_{k} is X_{k} (k ∈ {1, 2}). So without loss of generality we may assume that Q does not belong to H_{1} ∪ H_{2}. Hence H := (H_{1} ∩ H_{2}) ⊕ Q is a hyperplane in V, different from H_{1} and H_{2}. Each line T_{j}, j ∈ I, meets H in a unique point, because otherwise T_{j} would intersect H_{1} ∪ H_{2} in at most one point of H_{1} ∩ H_{2}, which is not allowed by 2.3. Let Q_{j} := T_{j} ∩ H (in particular, Q_{i} = Q). The subspace W := ⨁_{j∈I} Q_{j} belongs to R because W ⊆ H and H ∩ H_{k} = H_{1} ∩ H_{2} (for k = 1, 2), which implies W ∩ (H_{1} ∪ H_{2}) ⊆ H_{1} ∩ H_{2}.

It remains to prove that W ∈ R is uniquely determined by this construction. Assume that W′ ∈ R contains Q. Let Q′_{j} := W′ ∩ T_{j}. Since W′ = ⨁_{j∈I} Q′_{j}, we have to show that Q′_{j} = Q_{j} = H ∩ T_{j} holds for every j ∈ I. Suppose that this is not the case. Then there is a j ∈ I with Q′_{j} ∉ H, and thus V = H ⊕ Q′_{j} = (H_{1} ∩ H_{2}) ⊕ Q ⊕ Q′_{j}. In particular, the line L := Q⊕Q′_{j} meets each of the hyperplanes H_{1}, H_{2} in a point not contained in H_{1} ∩H_{2}. On the other hand, L is entirely contained in W′, a contradiction to W′ ∩(H_{1} ∪H_{2}) ⊆ H_{1} ∩H_{2} (which holds since Q ∈ W′ \ (H_{1} ∪ H_{2})). (iii) can be shown similarly to 2.2(iii). □

Note that so far it is not clear that the reguli defined above coincide with those of 1.1 in the commutative case. This will be shown at the end of this section.

By 2.2 and 2.5 any two elements of a regulus of type (1) and of type (2), respectively, are complementary. Moreover, we know that each regulus is a subset of G and consists of at least three elements.

Now we consider three arbitrary pairwise complementary subspaces W_{1}, W_{2}, W_{3} in ℙ(K, V). We want to show that there exist reguli of type (1) (if dim V < ∞) and of type (2) containing W_{1}, W_{2}, W_{3}.

The direct decompositions V = W_{1} ⊕ W_{2} = W_{1} ⊕ W_{3} = W_{2} ⊕ W_{3} give rise to a linear bijection π : W_{1} → W_{2}, uniquely determined by W_{3}, namely w_{1} = w_{2} + w_{3} ↦ w_{2}. We denote this linear mapping and also the induced projectivity with center W_{3} by π(W_{1}, W_{2}, W_{3}). It will be used in the subsequent propositions in order to describe the reguli through W_{1}, W_{2}, W_{3}.

2.6. Proposition. Let dim V = 2n, and let W_{1}, W_{2}, W_{3} ∈ G be pairwise complementary. Then there is at least one regulus of type (1) containing W_{1}, W_{2}, W_{3}. Every regulus of type (1) through W_{1}, W_{2}, W_{3} can be obtained as follows: Let (P_{0}, P_{1}, …, P_{n}) be a frame in W_{1}, and let π = π(W_{1}, W_{2}, W_{3}). Then (P_{0}^{π}, P_{1}^{π}, …, P_{n}^{π}) is a frame in W_{2}, and the lines T_{0} = P_{0} ⊕ P_{0}^{π}, …, T_{n} = P_{n} ⊕ P_{n}^{π} are in general position. The regulus R with transversals T_{0}, …, T_{n} contains W_{1}, W_{2}, W_{3}.

Proof. Let (P_{0}, P_{1}, …, P_{n}) be a frame in W_{1} such that P_{i} = Kb_{i}, b_{i} = c_{i} + d_{i} ∈ W_{1}, c_{i} ∈ W_{2}, d_{i} ∈ W_{3} (for i ∈ {0, …, n}). Then T_{i} meets W_{2} in the point Kc_{i} = P_{i}^{π} and W_{3} in the point Kd_{i}. So the construction yields a regulus through W_{1}, W_{2}, W_{3}.

Every regulus of type (1) through W_{1}, W_{2}, W_{3} can be obtained in this way, because for given transversals T_{0}, …, T_{n} the points P_{i} = W_{1} ∩ T_{i} (i ∈ {0, …, n}) build a frame of W_{1}, and P′_{i} = W_{2} ∩ T_{i} equals P_{i}^{π} since T_{i} is contained in the subspace spanned by P_{i} and the center W_{3} of π. □

A similar statement holds for the reguli of type (2). Before stating it we need the definition of a fundamental figure of a projective space ℙ. This is a pair ((P_{i})_{i∈I}, H) consisting of a basis (P_{i})_{i∈I} of ℙ and hyperplane H in ℙ such that no point P_{i} (i ∈ I) belongs to H.

The notion of a fundamental figure in a projective space is due to H. Brauner [3]. It serves as a substitute for the notion of a frame (called fundamental set by Brauner) in the infinite-dimensional case. In particular, the group of projective collineations of a projective space ℙ acts transitively on the set of fundamental figures in ℙ, and it acts sharply transitively exactly if ℙ is pappian (see [3, Satz 4.6]).

2.7. Proposition. Let W_{1}, W_{2}, W_{3} ∈ G be pairwise complementary. Then there is at least one regulus of type (2) containing W_{1}, W_{2}, W_{3}.

Every regulus of type (2) through W_{1}, W_{2}, W_{3} satisfying W_{1} ⊆ H_{1}, W_{2} ⊆ H_{2} (with H_{1}, H_{2} as in 2.3) can be obtained as follows:

Let ((P_{i})_{i∈I}, H) be a fundamental figure in W_{1}, and let π = π(W_{1}, W_{2}, W_{3}). Then ((P_{i}^{π})_{i∈I}, H^{π}) is a fundamental figure in W_{2}. The family (T_{i} = P_{i} ⊕ P_{i}^{π})_{i∈I} is a minimal family of lines spanning ℙ(K, V), and the union of the two hyperplanes H_{1} = W_{1} ⊕ H^{π}, H_{2} = H ⊕ W_{2} is met by each line T_{i} in exactly two points. The regulus regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}) contains W_{1}, W_{2}, W_{3} and satisfies W_{1} ⊆ H_{1}, W_{2} ⊆ H_{2}.

Proof. Let ((P_{i})_{i∈I}, H) be a fundamental figure in W_{1}. As in the proof of 2.6 one can see that every line T_{i} = P_{i} ⊕ P_{i}^{π} meets W_{k} (k ∈ {1, 2, 3}). The points of intersection span W_{k} since (P_{i})_{i∈I} is a basis of W_{1}. Moreover, we have W_{1} ⊆ H_{1} and W_{2} ⊆ H_{2}. Next we show that also W_{3} satisfies 2.3(ii): Consider a point Kw_{3} in W_{3} ∩ H_{1} with w_{3} = w_{1} + w_{2} for unique w_{1} ∈ W_{1}, w_{2} ∈ W_{2}, w_{1} ≠ 0 ≠ w_{2}. By definition of π we have −w_{2} = w_{1}^{π}. From w_{1} − w_{1}^{π} = w_{3} ∈ H_{1} = W_{1} ⊕ H^{π} we conclude w_{1}^{π} ∈ H^{π}, hence we have w_{1} ∈ H and w_{3} ∈ H ⊕ H^{π} = H_{1} ∩ H_{2}. Similarly, a point in W_{3} ∩ H_{2} must belong to H_{1} ∩ H_{2}. So W_{3} ∩ (H_{1} ∪ H_{2}) ⊆ H_{1} ∩ H_{2}. Altogether, the regulus regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}) contains W_{1}, W_{2}, W_{3} and satisfies W_{1} ⊆ H_{1}, W_{2} ⊆ H_{2}.

Every regulus R of type (2) through W_{1}, W_{2}, W_{3} with W_{1} ⊆ H_{1}, W_{2} ⊆ H_{2} can be obtained in this way: Let R be regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}). Then ((P_{i})_{i∈I}, H) with P_{i} = W_{1} ∩ T_{i}, H = W_{1} ∩ H_{2} is a fundamental figure in W_{1}: First of all, by definition of the regulus, (P_{i})_{i∈I} is a basis of W_{1}. Moreover, no P_{i} belongs to H because P_{i} ∈ H_{1}\(H_{1}∩H_{2}). This also implies that H is a hyperplane in W_{1}. As in the proof of 2.6 one sees that T_{i} = P_{i} ⊕ P_{i}^{π}. In addition, the hyperplane H ⊕ W_{2} = (W_{1} ∩ H_{2}) ⊕ (W_{2} ∩ H_{2}) of V is contained in the hyperplane H_{2}, so H ⊕ W_{2} = H_{2}. Analogously, H_{1} = W_{1} ⊕ H^{π}. So R equals the regulus obtained from the fundamental figure ((P_{i})_{i∈I}, H). □

Our next aim is to show that the reguli of type (1) and type (2) coincide in the finitedimensional case. In order to make calculations easier, we want to consider reguli containing the three pairwise complementary subspaces U^{(1,0)} = U × {0}, U^{(0,1)} = {0} × U, and U^{(1,1)} = {(u, u) | u ∈ U}.

This is possible without loss of generality because of the next lemma.

2.8. Lemma. Let ℜ_{1} and ℜ_{2} be the sets of all reguli of type (1) and (2), respectively, in ℙ(K, V). Then the following statements hold:

(i) Every collineation of ℙ(K, V) acts on the sets ℜ_{1} and ℜ_{2}, i.e., maps reguli to reguli (of the same type).

(ii) The group Aut_{K}(V) acts transitively on both sets ℜ_{1} and ℜ_{2}, i.e., any regulus can be mapped to any other regulus (of the same type) by a projective collineation.

Proof. (i) is clear by the definitions of the reguli. (ii): By 2.6 and 2.7 each regulus is determined by three pairwise complementary subspaces W_{1}, W_{2}, W_{3} and a frame (or a fundamental figure) in W_{1}. By 1.5, the group Aut_{K}(V) acts transitively on the set of triples of pairwise complementary subspaces. So without loss of generality we may assume W_{1} = U^{(1,0)}, W_{2} = U^{(0,1)}, and W_{3} = U^{(1,1)}. Now let F, F′ be two frames (or two fundamental figures) in W_{1}. Then there is a projective collineation of the projective space ℙ(K, W_{1}) mapping F to F′, and hence there is a linear bijection

α ∈ R = End_{K}(U) such that ( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ) ∈ GL_{2}(R) ≅ Aut_{K}(V) maps F to F′ (and fixes

W_{1}, W_{2}, W_{3}). □

We need two more lemmas in order to arrive at our desired result that ℜ_{1} = ℜ_{2} in the finite-dimensional case.

2.9. Lemma. Let (b_{1}, …, b_{n}) be a basis of U, and let T_{i} = K(b_{i}, 0) ⊕ K(0, b_{i}) for i ∈ {1, …, n}. Then W ∈ G is transversal to (T_{1}, …, T_{n}) exactly if

W = {(∑_{i=1}^{n} x_{i}k_{i}b_{i}, ∑_{i=1}^{n} x_{i}l_{i}b_{i}) | x_{1}, …, x_{n} ∈ K} for fixed pairs (k_{i}, l_{i}) ∈ K^{2} \ {(0, 0)}.

Proof. This is clear because the intersection point of W with T_{i} is some K(k_{i}b_{i}, l_{i}b_{i}) with (k_{i}, l_{i}) ≠ (0, 0). The linearly independent vectors (k_{1}b_{1}, l_{1}b_{1}), …, (k_{n}b_{n}, l_{n}b_{n}) span the n-dimensional vector space W. □

In the formulation of the next lemma we make use of the embedding k ↦ λ_{k} of K into R = End_{K}(U) introduced at the end of Section 1.

2.10. Lemma. Let (b_{1}, …, b_{n}) be a basis of U, let T_{i} = K(b_{i}, 0) ⊕ K(0, b_{i}) for i ∈ {1, …, n}, and let W ∈ G be transversal to (T_{1}, …, T_{n}). Moreover, let b_{0} = ∑_{i=1}^{n} b_{i},

T_{0} = K(b_{0}, 0) ⊕ K(0, b_{0}), and H = {∑_{i=1}^{n} x_{i}b_{i} | x_{1}, …, x_{n} ∈ K, ∑_{i=1}^{n} x_{i} = 0}, H_{1} = U × H,

H_{2} = H × U. Then the following statements are equivalent:

(i) W = U^{(1,0)} or W = U^{(λ_{k},1)} = {(u^{λ_{k}}, u) | u ∈ U} for some k ∈ K.

(ii) W is transversal to T_{0}.

(iii) W ⊆ H_{1} or W ⊆ H_{2} or W ∩ (H_{1} ∪ H_{2}) ⊆ H_{1} ∩ H_{2}.

Proof. One can easily check that (i) implies (ii) and (iii).

Now we consider the implications (ii) ⇒ (i) and (iii) ⇒ (i). Since W is transversal to

(T_{1}, …, T_{n}), we know from 2.9 that W = {(∑_{i=1}^{n} x_{i}k_{i}b_{i}, ∑_{i=1}^{n} x_{i}l_{i}b_{i}) | x_{1}, …, x_{n} ∈ K} for

fixed pairs (k_{i}, l_{i}) ∈ K^{2}\{(0, 0)}. What we have to show is that the equations k_{1} = … = k_{n} and l_{1} = … = l_{n} hold.

First let (ii) be satisfied. Then W contains a point K(kb_{0}, lb_{0}) with suitable (k, l) ≠ (0, 0). Since ((k_{1}b_{1}, l_{1}b_{1}), …, (k_{n}b_{n}, l_{n}b_{n})) is a basis of W, we conclude k = k_{1} = … = k_{n} and l = l_{1} = … = l_{n}, as desired.

Now assume that (iii) holds. If W ⊆ H_{1}, then W = U × {0}. Next we show that otherwise l_{i} ≠ 0 holds for each i ∈ {1, …, n}: If l_{i} = 0, then the point K(k_{i}b_{i}, l_{i}b_{i}) = K(k_{i}b_{i}, 0) belongs to H_{1} = U × H, and from (iii) and b_{i} ∉ H we conclude that k_{i} = 0, a contradiction.

So without loss of generality we may assume l_{1} = … = l_{n} = 1. For i, j ∈ {1, …, n} with i ≠ j we consider the point K(k_{i}b_{i} − k_{j}b_{j}, b_{i} − b_{j}) in W, which again belongs to H_{1}, and thus we have k_{i}b_{i} − k_{j}b_{j} ∈ H by (3), which means k_{i} = k_{j}. □

Using the notation from above, we now consider the points P_{i} =K(b_{i}, 0) (for i ∈ {0, …, n}) and the hyperplane H ×{0} in U ×{0}. Obviously (P_{1}, …, P_{n}) is a basis of the projective space corresponding to U × {0}. Moreover, (P_{0}, …, P_{n}) is a frame and ((P_{1}, …, P_{n}), H × {0}) is a fundamental figure. Hence Propositions 2.6 and 2.7 together with Lemmas 2.8 and 2.10 imply the following:

2.11. Corollary. Let V be finite-dimensional. Then each regulus of type (1) in ℙ(K, V) is also a regulus of type (2) and vice versa.

So we do not have to distinguish between both types of reguli any more (note that if dim V = ∞ we only had one type of reguli from the beginning). In the sequel we shall simply speak of reguli.

We know already (from 2.6, 2.7) that through any three pairwise complementary subspaces W_{1}, W_{2}, W_{3} there is at least one regulus. Our next aim is to show that this regulus is uniquely determined exactly if the ground field K is commutative. In order to prove this, we need again several lemmas.

According to Proposition 2.7 one can construct a (unique) regulus from the given data (W_{1}, W_{2}, W_{3}) and F, where F is a fundamental figure in W_{1} (note that W_{3} is involved because it gives rise to the projectivity π(W_{1}, W_{2}, W_{3})). We call this regulus R(W_{1}, W_{2}, W_{3}, F). It is clear that R(W_{1}, W_{2}, W_{3}, F) contains W_{1}, W_{2}, W_{3}. It is not yet clear, however, that all reguli containing W_{1}, W_{2}, W_{3} look like this (with a suitable fundamental figure F). We shall show that in fact this is true (no matter whether K is commutative or not).

2.12. Lemma. Let R = R(W_{1}, W_{2}, W_{3}, F) and let π = π(W_{1}, W_{2}, W_{3}). Then

R = R(W_{2}, W_{1}, W_{3}, F^{π}) = R(W_{1}, W_{3}, W_{2}, F).

Proof. We use the notation of 2.7. So R = {W ∈ G | W is regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2})}. The first equation of our assertion is clear, because obviously R(W_{2}, W_{1}, W_{3}, F^{π}) consists of all W ∈ G that are regular w.r.t. ((T_{i})_{i∈I}, H_{2}, H_{1}).

It remains to show that R = R(W_{1}, W_{3}, W_{2}, F). Let π̃ = π(W_{1}, W_{3}, W_{2}). Let H̃_{1} = W_{1} ⊕ Hπ̃ and H̃_{2} = H ⊕ W_{3}. An easy calculation (using that each w_{1} ∈ W_{1} has the form

w_{1} = w_{1}^{π} + wπ̃ ; 1) shows that H̃_{1} = H_{1} and H̃_{1} ∩ H̃_{2} = H_{1} ∩ H_{2}.

We know that the regulus R̃ = R(W_{1}, W_{3}, W_{2}, F) consists of all W ∈ G regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H̃_{2}). Of course, W_{1}, W_{2}, W_{3} belong to R̃. In particular, the subspace X̃_{1} spanned by all points T_{i} ∩ H̃_{1} (i ∈ I) equals W_{1}, and, analogously, X̃_{2} equals W_{3}.

Consider W ∈ R, W ≠ W_{1}, W_{2}, W_{3}. We have to show that W is regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H̃_{2}). The only thing that is not yet clear is that W ∩ H̃_{2} ⊆ H_{1} ∩ H̃_{2}. The subspace W′ = W ∩ H̃_{2} is a hyperplane in W (since otherwise W = X̃_{2} = W_{3} by 2.4(ii) because all T_{i} meet W). Moreover, W′ contains W′′ = W ∩ H_{1} ∩ H̃_{2} = W ∩ H_{1} ∩ H_{2}, which equals W ∩ H_{2} because W is regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}). But W′′ = W ∩ H_{2} also is a hyperplane in W and hence must coincide with W′. □

2.13. Lemma. Let R = R(W_{1}, W_{2}, W_{3}, F), and let R̃ = R(W_{1}, W_{2}, W̃_{3}, F). Moreover, let π = π(W_{1}, W_{2}, W_{3}) and π̃ = π(W_{1}, W_{2}, W̃_{3}). Then the following statements are equivalent:

(i) F^{π} = F π̃, (ii) R = R̃, (iii) W̃_{3} ∈ R.

Proof. (i) ⇒ (ii) is clear because W_{3} and W̃_{3} are only needed in the construction of the reguli in order to determine F^{π} and F π̃, respectively. (ii) ⇒ (iii) is trivial.

(iii) ⇒ (i): Assume that W̃_{3} ∈ R, and let F = ((P_{i})_{i∈I}, H). Since every T_{i} = P_{i} ⊕ P_{i}^{π} meets W̃_{3} we have P_{i}^{π} = P π̃ ; i. Consider h ∈ H. We show that hπ̃ ∈ H^{π}. Then Hπ̃ is a hyperplane in W_{2} contained in the hyperplane H^{π} and hence equal to H^{π}. By definition of π̃ there is a w ∈ W̃_{3} with w = h − hπ̃. So w ∈ W̃_{3} ∩ (H ⊕ W_{2}) = W̃_{3} ∩ H_{2}, which, by assumption, is contained in H_{1} ∩ H_{2} = H ⊕ H^{π}. This implies hπ̃ ∈ H^{π}, because hπ̃ is unique in W_{2}. □

Now we come to the announced statement that the reguli through W_{1}, W_{2}, W_{2} only depend on the choice of a fundamental figure F in W_{1}.

2.14. Theorem. Let R be a regulus containing the pairwise complementary subspaces W_{1}, W_{2}, W_{3} ∈ G. Then R = R(W_{1}, W_{2}, W_{3}, F) for a suitable fundamental figure F in W_{1}.

Proof. We know that R = R(W′_{1}, W′_{2}, W′_{3}, F′) for W′_{1}, W′_{2}, W′_{3} ∈ R and a fundamental figure F′ in W′_{1}. First we want to substitute W′_{1} by W_{1} and F′ by a fundamental figure F in W_{1}. We assume that W_{1} ≠ W′_{1}, W′_{2}, W′_{3} (otherwise an application of 2.12 leads to the desired result). Because of W_{1} ∈ R we know from 2.13 that R = R(W′_{1}, W′_{2}, W_{1}, F′). Applying 2.12 yields R = R(W_{1}, W′_{1}, W′_{2}, F) for a suitable F. A similar procedure then substitutes W′_{2} by W_{2} and W′_{3} by W_{3}. □

Now we turn to the commutative case. Recall that a projective collineation of a projective space ℙ(K, W) is uniquely determined by its action on a fundamental figure exactly if K is commutative.

2.15. Theorem. Let W_{1}, W_{2}, W_{3} ∈ G be pairwise complementary. Then the equation R(W_{1}, W_{2}, W_{3}, F) = R(W_{1}, W_{2}, W_{3}, F′) holds for any two fundamental figures F, F′ of W_{1} if, and only if, K is commutative.

Proof. Let W ∈ G be a subspace complementary to W_{1} and W_{2}. Moreover, let π = π(W_{1}, W_{2}, W_{3}) and π̃ = π(W_{1}, W_{2}, W). By 2.13 we know that W ∈ R = R(W_{1}, W_{2}, W_{3}, F) is equivalent to F^{π} = F π̃. Analogously, W ∈ R′ = R(W_{1}, W_{2}, W_{3}, F′) is equivalent to F′^{π} = F′π̃. Now the condition F^{π} = F π̃ implies π = π̃, and hence also F′^{π} = F′π̃ for every F′, exactly if K is commutative. This yields the assertion. □

So in the commutative case a regulus only depends on three of its elements. It also is a regulus in the usual sense (of Definition 1.1):

2.16. Corollary. Let K be commutative, and let W_{1}, W_{2}, W_{3} ∈ G be pairwise complementary. Then there is exactly one regulus containing W_{1}, W_{2}, W_{3}. This regulus coincides with the unique regulus in the sense of Definition 1.1 that contains W_{1}, W_{2}, W_{3}.

Proof. The first statement is clear because of the preceding theorems.

By 2.5(iii) any two elements of a regulus are complementary. It remains to show that the regulus R through W_{1}, W_{2}, W_{3} satisfies conditions (R1) and (R2) of 1.1.

Since R is independent of the choice of the fundamental figure in W_{1}, according to 2.7 any line T meeting W_{1}, W_{2}, W_{3} can be chosen as one of the transversals T_{i} that determine R. So (R1) is satisfied, and, moreover, 2.5(ii) yields (R2). □

3. Reguli and chains

In this section we are going to interpret the reguli from above as so-called chains. First we recall what a chain geometry is. Note that our definition here is broader than the usual one that can be found, e.g., in [1] or [8].

As mentioned in Section 1, the set G of all subspaces of ℙ(K, V) that possess an isomorphic complement can be identified with the projective line ℙ(R) over the ring R = End_{K}(U) via Φ : R(α, β) ↦ U^{(α,β)}.

Moreover, after fixing a basis (b_{i})_{i∈I} of U, one obtains an embedding k ↦ λ_{k} of the field K into R. Hence the projective line over K may be embedded into ℙ(R) as well, via K(k, l) ↦ R(λ_{k}, λ_{l}).

The image of ℙ(K) under this embedding will be denoted by C_{0}. It is our standard chain. All the other chains in ℙ(R) are images of C_{0} under the group GL_{2}(R). So the chain set ℭ(K, R) is the orbit C_{o}^{GL_{2}(R)} = {C_{0}^{γ} | γ ∈ GL_{2}(R)}.

The incidence structure Σ(K, R) := (ℙ(R), ℭ(K, R)) is called the chain geometry over (K, R). One can easily see that two points of ℙ(R) are joined by a chain exactly if they are distant (in the sense of Section 1).

Up to now, chain geometries Σ(K, R) have mainly been investigated in the case that R is a K-algebra, i.e., the field K is contained in the center of R. Of course then K has to be commutative. In that case the chain geometry Σ(K, R) satisfies the extra condition that any three pairwise distant points are joined by a unique chain. See [1] and [8] for more results in this case.

In [2, Thm. 2.8] it is shown that if the ground field K of the projective space ℙ(K, V) is commutative, then the incidence structure (G, ℜ), where ℜ is the set of all reguli (in the sense of Definition 1.1), is isomorphic to the chain geometry Σ(K, R) over the ring R = End_{K}(U), which in this case is a K-algebra. The isomorphism Σ(K, R) → (G, ℜ) is our bijection Φ.

We want to show now that this result can be carried over to the non-commutative case, using our new definition of a regulus.

Let ℜ be the set of all reguli (according to Definitions 2.1 and 2.3) in ℙ(K, V), where K need not be commutative. As before, we assume V = U × U. For the rest of this section, we fix a basis (b_{i})_{i∈I} of U and embed K into R = End_{K}(U) with respect to this basis.

In order to show that the incidences structure (G, ℜ) is isomorphic to Σ(K, R), we shall proceed as follows: Our crucial result will say that the image of the standard chain C_{0} under Φ is a standard regulus R_{0} = R(U^{(1,0)}, U^{(0,1)}, U^{(1,1)}, F_{0}), with respect to some standard fundamental figure F_{0}. The announced isomorphism theorem then follows easily by using 1.5 and 2.8.

We begin with the definition of the standard fundamental figure in U^{(1,0)} = U × {0}. We assume that the index set I of our basis (b_{i})_{i∈I} of U contains the element 0. For i ∈ I_{0} := I \ {0} we define a_{i} := b_{i} − b_{0}. Then (a_{i})_{i∈I_{0}} is a basis of a hyperplane H in U, such that no b_{i} (i ∈ I) belongs to H. For i ∈ I let P_{i} := K(b_{i}, 0). Moreover, let H_{0} := H × {0}. Then F_{0} := ((P_{i})_{i∈I}, H_{0}) is a fundamental figure in U × {0}.

The next statement is our decisive theorem. In the finite-dimensional case it is an immediate consequence of 2.10.

3.1. Theorem. The image C_{0}^{Φ} of the standard chain C_{0} is a regulus, namely C_{0}^{Φ} = R_{0} := R(U^{(1,0)}, U^{(0,1)}, U^{(1,1)}, F_{0}).

Proof. We recall that the regulus R_{0} = R(U^{(1,0)}, U^{(0,1)}, U^{(1,1)}, F_{0}) consists of all subspaces W ∈ G regular w.r.t. ((T_{i})_{i∈I}, H_{1}, H_{2}), where T_{i} = K(b_{i}, 0) ⊕ K(0, b_{i}), H_{1} = U × H and H_{2} = H × U.

First we show that Φ maps each element of C_{0} into R_{0}. This is clear for R(1, 0) and R(0, 1), because their images are U^{(1,0)} and U^{(0,1)}, respectively. So it remains to consider R(λ_{k}, 1) for k ∈ K^{∗} = K \ {0}. Its image is U^{(λ_{k},1)} = {(u^{λ_{k}}, u) | u ∈ U}. This subspace satisfies condition 2.3(i) because it contains exactly one point of each line T_{i}, namely, K(kb_{i}, b_{i}) = K(b_{i}^{λ_{k}}, b_{i}), and it is spanned by these points. Moreover, U^{(λ_{k},1)} satisfies 2.3(ii): We have U^{(λ_{k},1)} ∩ H_{1} = {(h^{λ_{k}}, h) | h ∈ H}. Since H is invariant under λ_{k} by construction, this means U^{(λ_{k},1)} ∩ H_{1} ⊆ H × H = H_{1} ∩ H_{2}. Analogously one sees that U^{(λ_{k},1)} ∩ H_{2} ⊆ H_{1} ∩ H_{2}, because U^{(λ_{k},1)} = U^{(1,λ_{k}^{−1})}.

Now we consider an arbitrary W ∈ R_{0} \ {U^{(1,0)}}. Then W = U^{(α,β)} = R(α, β)^{Φ} for some R(α, β) ∈ ℙ(R). By 2.5(iii) we know that W is complementary to U^{(1,0)} and so we may assume that β = 1. Hence we have to prove that α = λ_{k} for some k ∈ K.

Since W = U^{(α,1)} is transversal to (T_{i})_{i∈I}, for each i ∈ I there are elements u_{i} ∈ U \{0} and (k_{i}, l_{i}) ∈ K^{2} \ {(0, 0)} such that u_{i}^{α} = k_{i}b_{i} and u_{i} = l_{i}b_{i}. Because of u_{i} ≠ 0 we may even assume that all l_{i} are equal to 1. So we have b_{i}^{α} = k_{i}b_{i} (i ∈ I), which determines α uniquely.

It remains to show that k_{i} = k_{j} holds for all pairs (i, j) ∈ I × I. This will be done by using the fact that W fulfils 2.3(ii). The subspaces X_{1} and X_{2} of 2.4 are U × {0} and {0} × U, respectively. So obviously W ≠ X_{1}, and if W = X_{2} our assertion is shown. Otherwise 2.3(ii) means (by 2.4) that the intersection W ∩ (H_{1} ∪ H_{2}) = {(u^{α}, u) | u ∈ H or u^{α} ∈ H} is contained in H_{1} ∩ H_{2} = H × H. This implies H^{α} ⊆ H. Now let i ∈ I_{0}.

Then a_{i} = b_{i} − b_{0} ∈ H, and hence a_{i}^{α} = k_{i}b_{i} − k_{0}b_{0} ∈ H. Since (a_{i})_{i∈I_{0}} is a basis of H and (b_{i})_{i∈I} is a basis of U, this means k_{i} = k_{0}, as desired. □

Now we know that the chain C_{0} is mapped to a regulus. This can easily be carried over to the other chains.

3.2. Corollary. Let C ∈ ℭ(K, R). Then C^{Φ} is a regulus.

Proof. By definition C = C_{0}^{γ} for some γ ∈ GL_{2}(R) ≅ Aut_{K}(V). Thus by 3.1 we have C^{Φ} = C_{0}^{γΦ} = C_{0}^{Φγ} = R_{o}^{γ}, where γ ∈ Aut_{K}(V) by 2.8(i) maps reguli to reguli. This yields the assertion. □

By 3.1 we also know that the standard regulus R_{0} = R(U^{(1,0)}, U^{(0,1)}, U^{(1,1)}, F_{0}) is the image of a chain. This holds for every other regulus as well, because of the action of Aut_{K}(V) ≅ GL_{2}(R) on the set ℜ.

3.3. Corollary. Let R ∈ ℜ be any regulus. Then R = C^{Φ} for some C ∈ ℭ(K, R). Hence ℭ(K, R)^{Φ} = ℜ.

Proof. By 2.8(ii) there is a γ ∈ Aut_{K}(V) ≅ GL_{2}(R) mapping the standard regulus R_{0} to R. So by 3.1 we have R = R_{0}^{γ} = C_{0}^{Φγ} = C_{0}^{γΦ}, where C_{0}^{γ} is a chain. □

Altogether, we now have completed the proof of the isomorphism theorem generalizing [2, Thm. 2.8]:

3.4. Main Theorem. The incidence structure (G, ℜ) of all reguli in ℙ(K, V) is isomorphic to the chain geometry Σ(K, R).

After having established Theorem 3.4 we can make use of the algebraic description of (G, ℜ) as the chain geometry Σ(K, R) for our investigation of the geometric properties of the reguli in ℙ(K, V).

The first result of 2.16, e.g., now follows directly from the fact that in a chain geometry Σ(K, R) over a K-algebra R any three pairwise distant points are joined by a unique chain (note that of course R = End_{K}(U) is a K-algebra if K is commutative). Moreover, also the second part of 2.16 now is immediately clear, namely, that in the commutative case the old and the new reguli coincide, because in [2] we showed the same theorem as 3.4 for the old reguli (and we used the same isomorphism Φ).

4. Intersections of reguli

In this section we want to determine the intersection I = I(W_{1}, W_{2}, W_{3}) of all reguli containing three given pairwise complementary subspaces W_{1}, W_{2}, W_{3} of ℙ(K, V). If K is commutative this is the unique regulus through W_{1}, W_{2}, W_{3}. Otherwise the center Z = Z(K) of the field K comes into play.

By 2.14 we know that I = ⋂{R(W_{1}, W_{2}, W_{3}, F) | F fundamental figure in W_{1}}. We could use this description and some linear algebra for the explicit determination of I. However, we prefer taking the chain-geometric point of view.

But before, we consider Segre’s case of the reguli in the three-dimensional projective space ℙ(K, K^{4}). Here both the transversals and the elements of a regulus are lines. One can easily see that in this case our intersection I = I(W_{1}, W_{2}, W_{3}) is the line set I = {L ∈ G | L transversal to R̃}, where R̃ is the regulus determined by the transversal lines W_{1}, W_{2}, W_{3}. Segre proves in [10] that this set I is a regulus over the center Z = Z(K).

In order to make this more explicit, we introduce some terminology that will also be used later for stating our generalization of Segre’s result.

Let V be a vector space (of arbitrary dimension) over K, and let Z = Z(K). We call v ∈ V a Z-vector with respect to the basis (c_{j})_{j∈J} of V if all coordinates of v (w.r.t. this basis) belong to Z. A subspace W ≤ V is a Z-subspace w.r.t. (c_{j})_{j∈J} if it possesses a basis of Z-vectors w.r.t. (c_{j})_{j∈J}. The lattice of all Z-subspaces of V w.r.t. some fixed basis gives rise to a projective space over Z, contained in ℙ(K, V) and with the same dimension as ℙ(K, V). Such a projective space will be called a projective Z-subspace of ℙ(K, V).

So Segre’s result reads as follows: The intersection I is a regulus in some projective Z-subspace of ℙ(K, K^{4}) (with respect to a suitable basis of the vector space K^{4}).

We want to show that a similar result holds in higher dimensions. We use the notation of the preceding sections. Instead of I we first consider the corresponding intersection of chains.

4.1. Lemma. Let Î_{0} be the intersection of all chains containing the points R(1, 0), R(0, 1), R(1, 1). Then Î_{0} = {R(λ_{z}, 1) | z ∈ Z} ∪ {R(1, 0)}, i.e., Î_{0} is the projective line ℙ(Z) embedded into ℙ(R) with respect to k ↦ λ_{k}.

Proof. Since the standard chain C_{0} is one of the chains under consideration, each element of Î_{0} \ {R(1, 0), R(0, 1)} has the form R(λ_{k}, 1) for some k ∈ K^{∗}. Let C = C_{0}^{γ} (γ ∈ GL_{2}(R)) be another chain through R(1, 0), R(0, 1), R(1, 1). We may assume that γ fixes these three points (because the subgroup GL_{2}(K) of GL_{2}(R) acts triply transitively on C_{0}). So γ =

( \begin{matrix}α & 0 \\ 0 & α \end{matrix} ) for some α ∈ R^{∗}. A point p = R(λ_{k}, 1) ∈ C_{0} (with k ∈ K^{∗}) also belongs to

C = C_{0}^{γ} exactly if p = R(λ_{l}, 1)^{γ} = R(λ_{l}α, α) for an element l ∈ K^{∗}. So p belongs to the intersection Î_{0} if, and only if, for all α ∈ R^{∗} there is an l ∈ K^{∗} such that αλ_{k} = λ_{l}α.

In case k ∈ Z this condition is fulfilled because then λ_{k} is the mapping u ↦ ku and hence commutes with all α ∈ R.

In order to show that on the other hand the condition above implies k ∈ Z we consider special mappings α: For s ∈ K let α_{s} ∈ R^{∗} be defined by b_{i}^{α_{s}} = b_{i} (i ∈ I_{0}), b_{0}^{α_{s}} = b_{0} + sb_{j}, where j ∈ I_{0} is some fixed index. Let α_{s}λ_{k} = λ_{l}α_{s} be satisfied for some l ∈ K^{∗}. Then for i ∈ I_{0} we have kb_{i} = b_{i}^{α_{s}λ_{k}} = b_{i}^{λ_{l}α_{s}} = lb_{i}, and so l = k. Moreover, kb_{0} + skb_{j} = b_{0}^{α_{s}λ_{k}} = b_{0}^{λ_{l}α_{s}} = b_{0}^{λ_{k}α_{s}} = kb_{0} + ksb_{j}, and hence sk = ks holds. Since s ∈ K was chosen arbitrarily, this means k ∈ Z. □

The result of this lemma now can be carried over (using the action of GL_{2}(R)) to the intersection of the chains through any other three pairwise distant points.

Our formulation of the subsequent proposition uses the chain geometry Σ(Z, R), i.e., the incidence structure (ℙ(R), ℭ(Z, R)). By 4.1 the elements of ℭ(Z, R) are the images of Î_{0} under GL_{2}(R).

4.2. Proposition. Let R = End_{K}(U) and Z = Z(K). Let p, q, r ∈ ℙ(R) be pairwise distant. Then the intersection Î = Î(p, q, r) of all chains through p, q, r is a chain of the chain geometry Σ(Z, R).

Of course this means that Î(p, q, r) is the unique chain in Σ(Z, R) containing p, q, r. Recall that this chain is uniquely determined because Z is contained in the center of R (after identifying z ∈ Z with λ_{z} ∈ R) and hence R is a Z-algebra.

Intersections of chains as above are considered by Benz in [1] for chain geometries Σ(K, L) where both K and L are skew fields. Benz calls them traces (“Fährten”); he proves a theorem similar to 4.2 for these traces ([1, IV 2.3.3]).

Using 3.4 one can now derive from 4.1 the following generalization of Segre’s result:

4.3. Proposition. Let W_{1}, W_{2}, W_{3} ∈ G be pairwise complementary. Then the intersection I = I(W_{1}, W_{2}, W_{3}) of all reguli containing W_{1}, W_{2}, W_{3} is a regulus in a projective Zsubspace of ℙ(K, V).

Proof. We may assume that W_{1} = U^{(1,0)}, W_{2} = U^{(0,1)}, W_{3} = U^{(1,1)}, because the group GL_{2}(R) ≅ Aut_{K}(V) maps Z-subspaces to Z-subspaces (as it permutes the bases of V). By 3.4 the intersection I_{0} of all reguli through these three subspaces is Î_{0}^{Φ}, with Î_{0} defined as in 4.1. So by 4.1 we have I_{0} = {U^{(λ_{z},1)} | z ∈ Z} ∪ {U^{(1,0)}}. Obviously (compare 3.1) this is the standard regulus in the projective Z-subspace of ℙ(K, V) with respect to the basis ((b_{i}, 0))_{i∈I}, ((0, b_{i}))_{i∈I} of V = U × U. □

References

[1] Benz, W.: Vorlesungen über Geometrie der Algebren. Springer Verlag, Berlin 1973.

[2] Blunck, A.: Regular spreads and chain geometries. Bull. Belg. Math. Soc. (to appear).

[3] Brauner, H.: Geometrie projektiver Räume I. Bibl. Institut, Zürich 1976.

[4] Burau, W.: Mehrdimensionale projektive und höhere Geometrie. Dt. Verlag d. Wissenschaften, Berlin 1961.

[5] Grundhöfer, T.: Reguli in Faserungen projektiver Räume. Geom. Dedicata 11 (1981), 227–237.

[6] Havlicek, H.: Baer subspaces within Segre manifolds. Results Math. 23 (1993), 322– 329.

[7] Herzer, A.: Über niedrigdimensionale projektive Darstellungen von Kettengeometrien. Geom. Dedicata 19 (1985), 287–293.

[8] Herzer, A.: Chain Geometries. In: Handbook of Incidence Geometry. F. Buekenhout ed., Elsevier, Amsterdam 1995.

[9] Knarr, N.: Translation Planes. Lecture Notes in Math. 1611, Springer Verlag, Berlin 1995.

[10] Segre, B.: Lectures on Modern Geometry. Cremonese, Roma 1961.

Received November 19, 1998