Point-affine quadrangles
Bernhild Stroppel
Mathematisches Institut B, Universit¨ at Stuttgart, D-70550 Stuttgart, Germany, bernhild@mathematik.uni-stuttgart.de
Received: 15 October 1999; accepted: 15 October 1999.
Abstract. We give an axiomatic description of point-affine quadrangles as obtained by deleting a point star from a generalized quadrangle. We define a completion and observe that isomorphisms can be extended to isomorphisms between completions.
Keywords: generalized quadrangles, affine generalized quadrangles, elation quadrangles MSC 2000 classification: primary 51A10, secondary 51A50, 51E12, 51E14
1. Introduction
We consider incidence structures Q = (P, L, F ) where P and L are disjoint sets and F ⊆ P ×L is the incidence relation. The elements of P and L are called points and lines respectively, the elements of F are called flags. The (incidence geometrical) distance will be denoted by d and we use the following notation:
D i (x) := {y ∈ P ∪ L | d(x, y) = i} , D i :=
(x, y) ∈ (P ∪ L) 2 d(x, y) = i .
Definition 1. An incidence structure Q = (P, L, F ) is called generalized quadrangle if, and only if, the following axioms are satisfied:
(V 1 ) |D 1 (x) | ≥ 3 for all x ∈ P ∪ L.
(V 2 ) There are neither binangles nor triangles in Q.
(V 3 ) The diameter of Q is 4, that is sup {d(x, y) | x, y ∈ P ∪ L} = 4.
The following properties are checked easily:
() Take any two elements x and y of P ∪ L. Then there exists an ordinary quadrangle containing x and y.
( ⊥) For each anti-flag (x, g) ∈ (P ×L)\F there exists exactly one flag (y, h) ∈ F
such that (x, h) ∈ F and (y, g) ∈ F hold. We denote y and h by π(x, g)
and λ(x, g) respectively.
2. Affine derivation of quadrangles
In this section we consider a generalized quadrangle Q = (P, L, F ) with distance d and a chosen point p ∈ P . Our aim is to motivate the axioms in Section 3 and to prepare for Theorem 2.
Definition 2. The affine derivation of Q in p is the incidence structure A p ( Q) = (A, G, I), where A := D 4 (p), G := D 3 (p) and I is the restriction of F to A p ( Q). We will denote the distance in A p ( Q) by δ, and put
∆ i (x) := {y ∈ A ∪ G | δ(x, y) = i} ,
∆ i :=
(x, y) ∈ (A ∪ G) 2 δ(x, y) = i .
We call the elements of A and G affine points and affine lines respectively.
Example. The affine derivation of the smallest generalized quadrangle is the graph of the cube.
+-+
++-
+-- ---
-++ +++
--+
-+-
+++
-+- -++
+-+
+-- --+
---
++-
Note that there are points at distance 6 in this example. If Q is anti-regular (for instance if Q is an orthogonal quadrangle over a field of characteristic different from 2) then A p Q does not contain points at distance 6.
Remarks 1.
1. For any a ∈ A we have ∆ 1 (a) = D 1 (a). That means, for every pair (a, g) ∈ (A × G) \ I there exists λ(a, g) ∈ G.
2. For g ∈ G there exists q ∈ D 2 (p) such that ∆ 1 (g) ∪ {q} = D 1 (g), namely q = π(p, g).
3. G = {a ∨ b | a, b ∈ A; d(a, b) = 2}, where a ∨ b is the joining line of a and b.
(In fact, assertion 2 above gives us ∆ 1 (g) ≥ 2 for all g ∈ G. So we have a 1 , a 2 ∈ A with g = a 1 ∨ a 2 . The other direction follows from assertion 1.)
Now we consider the distances in A := A p ( Q). Let A also denote A ∪ G.
Lemma 1. We have δ ≥ d. More exactly:
1. For (x, y) ∈ A 2 ∩ (D 2 \ ∆ 2 ) we have: (x, y) ∈ ∆ 6 and x, y ∈ G.
2. For (a, g) ∈ A 2 ∩ (D 3 \ ∆ 3 ) we have: (a, g) ∈ ∆ 5 .
3. For (x, y) ∈ A 2 ∩ (D 4 \ ∆ 4 ) we have: (x, y) ∈ ∆ 6 and x, y ∈ A.
Proof. As incidence in A is induced from Q we have δ(x, y) ≥ d(x, y) for all x, y ∈ A.
(1.) Let us take (x, y) ∈ A 2 ∩ (D 2 \ ∆ 2 ). Then we have x, y ∈ G by Re- marks 1.3 and there exists z ∈ D 1 (x) ∩ D 1 (y) ∩ D 2 (p). With () we get a chain (x, a, k, b, l, c, y) with a = z = c. By Remarks 1.2 we have {a, c} ⊆ A. Now if b ∈ A then k, l ∈ A by Remarks 1.3. If b ∈ A we choose a ∈ D 1 (x) \ {a, z} such that π(a , l) ∈ A. This is possible by Remarks 1.2. So we have a chain of length 6 from x to y in A. Any shorter chain would give binangles or triangles in Q.
(2.) Take (a, g) ∈ A 2 ∩ (D 3 \ ∆ 3 ), say a ∈ A and g ∈ G. So we have π(a, g) ∈ A. From Remarks 1.2 we get the existence of an affine point b ∈ ∆ 1 (h) for a line h ∈ ∆ 1 (a) such that λ(b, g) ∈ A and π(b, g) ∈ A. Then the chain (a, h, b, λ(b, g), π(b, g), g) has length 5 in A.
(3.) Let (x, y) ∈ A 2 ∩ (D 4 \ ∆ 4 ). Then we have x, y ∈ A by Remarks 1.2 and D 2 (x) ∩ D 2 (y) ⊆ D 2 (p). Choose g ∈ ∆ 1 (x) and h ∈ ∆ 1 (y) \ {λ(y, g)}. By Remarks 1.2 there exists a ∈ (∆ 1 (g) ∩A)\{x} with π(a, h) ∈ A. Thus the chain
(x, g, a, λ(a, h), π(a, h), h, y)
is a chain of length 6 in A.
QEDWe want to reconstruct the generalized quadrangle from its affine derivation.
To this end, we describe points of D 2 (p) in terms of A.
Definition 3. Let a pseudo-pencil be a maximal set H ⊆ G of lines with δ(g, h) ∈ {0, 6} for all g, h ∈ H.
Lemma 2. For q ∈ D 2 (p) we have that Π q := D 1 (q) ∩ A is a pseudo-pencil.
Moreover, for every pseudo-pencil H there exists a point x ∈ D 2 (p) such that H = D 1 (x) ∩ A. Thus the system of pseudo-pencils is a partition of G.
Proof. Take two distinct lines g and h in D 1 (q) ∩ A. By Lemma 1 we have δ(g, h) = 6. Let l ∈ G \ Π q , that means d(q, l) = 3. If π(q, l) ∈ A holds, we have δ(λ(q, l), l) = 2 = 6. If π(q, l) ∈ A, there exists by () and Remarks 1.2 a line h ∈ Π q \ {λ(q, l)} with δ(h, l) = 4 = 6.
Now let H be any pseudo-pencil and g = h ∈ H. Lemma 1 yields d(g, h) = 2.
So we have that all lines of H meet in x := π(p, g) by Remarks 1.2 and thus we
have H = D 1 (x) ∩ A.
QEDOur next aim is to describe the lines through p with δ. Let us regard the
following assertions about x, y ∈ D 2 (p).
A: x and y are not collinear in P.
B: x and y are collinear in P.
C: For all g ∈ Π x there exists an h ∈ Π y with δ(g, h) = 2.
D: For all g ∈ Π x and all h ∈ Π y we have δ(g, h) = 4.
Lemma 3. Let x, y ∈ D 2 (p) be distinct points. Then we have A ⇐⇒ C and B ⇐⇒ D
Proof. Obviously we have implications
E: A ⇐⇒ ¬B and C = ⇒ ¬D.
If x and y are collinear points, assertion D holds because there are no triangles in Q. So we have B =⇒ D and C =⇒ A by E. In order to show A =⇒ C we choose two not collinear points x, y ∈ D 2 (p), that is d(x, y) = 4 and we have d(g, y) = 3 for all lines g ∈ Π x . Absence of triangles in Q yields π(y, g) ∈ A. Thus we have δ(g, λ(y, g)) = 2 and assertion C holds. Now E implies D = ⇒ B.
QEDNow we define an equivalence relation ∼ on the set of all pseudo-pencils in A, as follows:
Π x ∼ Π y ⇐⇒ δ(g, h) = 2 for all g ∈ Π x and all h ∈ Π y
We see that Π x ∼ Π y means d(x, y) ≤ 2. Thus the equivalence classes of ∼ are exactly the sets D 1 (l) \ {p} for lines l ∈ D 1 (p).
In addition we consider the equivalence relation whose equivalence classes are the pseudo-pencils of A. Using both equivalence relations we can reconstruct the quadrangle Q from its affine derivation A.
Proposition 1. We have Q ∼ = ( ˆ P , ˆ L, ˆ F ),where
P := A ˆ ∪ G ∪ {p} and L := G ˆ ∪ G
∼
The incidence ˆ F is described as follows:
F := I ˆ ∪ {([g] , g) | g ∈ G} ∪
(x, [x] ∼ ) x ∈ G
∪
(p, [x] ∼ ) x ∈ G , where I ⊆ A × G is the incidence relation of A.
For the distance γ in ( ˆ P , ˆ L, ˆ F ) we obtain from 3:
Corollary 1. If x, y ∈ D 2 (p) are two different points then we have γ(Π x , Π y )
= 6 − min {δ(g, h) | (g, h) ∈ Π x × Π y }.
3. Axioms for point-affine quadrangles
Our aim is to give an axiomatic description of a point-affine quadrangle A such that we can construct a generalized quadrangle Q with A being the affine derivation of Q. A main motivation was the observation that several con- structions of generalized quadrangles in fact start with building a point-affine quadrangle. Prominent examples are elation generalized quadrangles, see Sec- tion 5.
There is a more general notion of “affine quadrangle”, denoting incidence structures obtained from a generalized quadrangle by deleting a geometric hy- perplane. Pralle [6] gives an axiomatic characterization of this general class, based on the notion of totally connected pairs of lines. Our approach, in con- trast, is based on distances.
Definition 4. An incidence structure A = (A, G, I) with distance δ is called a point-affine quadrangle if the following axioms hold.
(A 1 ) For a ∈ A we have |∆ 1 (a) | ≥ 3, for g ∈ G we have |∆ 1 (g) | ≥ 2.
(A 2 ) There are neither binangles nor triangles in A.
(A 3 ) The diameter of A is 6.
(A 4 ) We require that g h ⇐⇒ δ(g, h) ∈ {0, 6} is an equivalence relation on G and that each equivalence class of contains at least 2 elements. The equivalence class of g is denoted by [g] , and the canonical map will be denoted by α : g → [g] .
(A 5 ) On G ,
[x] ∼ [y] ⇐⇒ δ(g, h) = 2 for all g ∈ [x] and all h ∈ [y]
defines an equivalence relation with at least 3 equivalence classes and at least 2 elements in each class.
(A 6 ) For g ∈ G we have that α| ∆
2(g) is a surjection onto {[x] | [x] ∼ [g] }.
(A 7 ) For a ∈ A we have that α| ∆
1(a) is a surjection onto G \∆
3(a) . (A 8 ) Let β : G →
[x] ∼ x ∈ G
: g → [[g] ] ∼ . Then the restriction β | ∆
1(a) is surjective for each a ∈ A.
Example. If Q is a generalized quadrangle then the affine derivation at any
point of Q is a point-affine quadrangle, see Section 2. In fact, we shall see that
there are no other examples, cf. Theorem 1 below.
Remark 1. Note the analogy with affine planes (“line-affine triangles”), where one would require (A 1 ), absence of binangles, diameter 4, transitivity of the relation δ(g, h) ∈ {0, 4} (parallelism of lines) and the corresponding variant of (A 7 ) (existence of parallels). As, for instance, the Pappos figure shows, we have to require also that there are no pairs of points at distance 4 (existence of joining lines).
Now we want to reverse affine derivation. To this end, we construct a com- pletion C(A). Let
P := A ∪ G ∪ {∞} and L := G ∪ G
∼
and define incidence by
F = I ∪ {([g] , g) | g ∈ G} ∪
(x, [x] ∼ ) x ∈ G
∪
( ∞, [x] ∼ ) x ∈ G . This means that affine points are only incident with affine lines and to an affine line we add exactly one new point. Moreover the point ∞ is incident only with the new lines.
Let us denote the distance on C(A) = (P, L, F ) by γ.
Definition 5. The incidence structure C(A) = (P, L, F ) described above is called the completion of the point-affine quadrangle A = (A, G, I).
Remark 2. Let Q be a generalized quadrangle. Then Proposition 1 says that Q ∼ = C(A p ( Q)) holds for each point p ∈ Q. Thus completion reverses derivation.
Theorem 1. The completion C(A) = (P, L, F ) of a point-affine quadrangle is a generalized quadrangle.
Proof. (V 1 ) For a ∈ A we have |Γ 1 (a) | ≥ 3 by (A 1 ). For x := [g] ∈ P \ A we have |Γ 1 (x) | ≥ 3 by (A 4 ) and [x] ∼ ∈ Γ 1 (x). Axiom (A 5 ) ensures |Γ 1 ( ∞)| ≥ 3.
Let g ∈ G. By (A 1 ) and [g] ∈ Γ 1 (g) we have |Γ 1 (g) | ≥ 3. Let l ∈ L \ G. By (A 5 ) we get at least two elements in ∆ 1 (l), and ∞ ∈ Γ 1 (l) yields |Γ 1 (l) | ≥ 3.
(V 2 ) We have neither binangles nor triangles in A by (A 2 ). In C(A) \ A the point ∞ is the only one which meets more than one line of L \ G. If there would be a triangle with two distinct collinear points x, y ∈ P \ A, we would have δ(g, h) = 4 for all (g, h) ∈ Π x ×Π y by definition of collinearity. So we cannot get any triangle. If there would be a binangle with points x ∈ P \ A and a ∈ A, this would mean that x = [a] . But (a, g) ∈ I and (a, h) ∈ I for two lines g, h ∈ Π x
means that δ(g, h) = 0 and thus there exists no binangle.
(V 3 ) First we show γ(x, y) ≤ 4 for all x, y ∈ C(A).
(i) Let us consider the point ∞. For x ∈ A the chain (x, g, [g] , [[g] ] ∼ , ∞) has
length 4 and γ(x, ∞) ≤ 4 holds. For x ∈ G we have γ(x, ∞) = 2 by definition
of F . For x ∈ G we have the chain (x, [x] , [[x] ] ∼ , ∞) and so γ(x, ∞) ≤ 3. For x ∈ L \ G we have γ(x, ∞) = 1 again by definition of F .
(ii) Let p ∈ G . For x ∈ G we have the chain (x, [x] ∼ , ∞, [p] ∼ , p), that is γ(x, p) ≤ 4. For x ∈ L \ G we have the chain (x, ∞, [p] ∼ , p), that is γ(x, p) ≤ 3. For x ∈ G either we have [x] ∼ p and γ(x, p) ≤ 3 or there exists a line in ∆ 2 (x) incident with p by (A 6 ). So we have γ(x, p) ≤ 3 again. For x ∈ A choose a line g ∈ ∆ 1 (x) and get γ(x, p) ≤ 4 with the same argument as above.
(iii) Let a ∈ A. For x ∈ ∆ 6 (a) choose g ∈ ∆ 1 (x) and get δ(a, g) ≤ 5. If δ(a, g) = 5 holds, we get a line l ∈ ∆ 1 (a) with [l] = [g] by (A 7 ). Thus we have γ(x, a) ≤ 4.
For x ∈ G and δ(a, x) > 4 we have again by (A 7 ) a line l ∈ ∆ 1 (a) with [l] = [x] and so γ(x, a) ≤ 3 holds. For x ∈ L \ G the axiom (A 8 ) yields a line in ∆ 1 (a) which meets x, and we have γ(x, a) ≤ 3.
(iv) It remains to look at distances between lines. Let g ∈ G. For x ∈ ∆ 6 (g) we have [x] = [g] , that is γ(x, g) ≤ 2. For x ∈ L \ G we have the chain (x, ∞, [[g] ] ∼ , [g] , g) and so γ(x, g) ≤ 4 holds. For two new lines l, h ∈ L \ G we have γ(l, h) ≤ 2 because of (∞, x) ∈ F for all x ∈ L \ G.
Finally, we observe that (A × Γ 1 ( ∞)) ∩ F = ∅ and so γ(x, ∞) > 2 for all x ∈ A. Thus we have Γ 4 ( ∞) = ∅ and (V 3 ) is proved.
QED4. Isomorphisms
Definition 6. Let Q = (P, L, F ) and Q = (P , L , F ) be incidence struc- tures with distance functions d and d , respectively. A bijection ϕ : P ∪ L → P ∪ L is called an isomorphism (from Q onto Q ) if we have P ϕ = P and F ϕ = F .
Remark 3. A surjection ϕ : Q → Q is an isomorphism if, and only if, we have P ϕ ⊆ P and d(x, y) = d (x ϕ , y ϕ ) for all x, y ∈ Q.
From Definition 3 and Corollary 1 we infer:
Lemma 4. Each isomorphism ϕ : A → A between point-affine quadran- gles maps pseudo-pencils to pseudo-pencils, and for Π x , Π y ∈ C(A) we have:
γ ((Π x ) ϕ , (Π y ) ϕ ) = γ(Π x , Π y ), where γ and γ are the distances in C(A) and C(A ) respectively.
Now let A and A be point-affine quadrangles and let ϕ be an isomorphism from A onto A . Then we can define an extension Φ from C(A) onto C(A ) of ϕ as follows. Let Φ | A = ϕ and ([g] ) Φ := [g ϕ ] , ([x] ∼ ) Φ :=
x Φ
∼ , ∞ Φ := ∞ . Then Φ is an isomorphism from C(A) onto C(A ).
Proposition 1 gives isomorphisms from Q 1 onto C(A p
1( Q 1 )) and from
C(A p
2( Q 2 )) onto Q 2 that induce the identity on A 1 and A 2 , respectively. Com-
bining these with Φ, we obtain:
Theorem 2. Let A 1 = A p
1( Q 1 ) and A 2 = A p
2( Q 2 ) be affine derivations of generalized quadrangles and let ϕ : A 1 → A 2 be an isomorphism. Then there exists an isomorphism ˆ ϕ : Q 1 → Q 2 with ˆ ϕ | A
1= ϕ and p ϕ 1 ˆ = p 2 .
Corollary 2. For each generalized quadrangle Q and any two points p, q ∈ Q we have A p ( Q) ∼ = A q ( Q) if, and only if, there exists α ∈ Aut Q with p α = q.
Proof. Apply Theorem 2 to find α and use Remark 3 to conclude D 4 (p) α =
D 4 (q) and D 3 (p) α = D 3 (q) from p α = q.
QED5. Elation generalized quadrangles
In [4] Kantor constructs finite elation quadrangles as group coset geometries.
This construction was extended to the infinite case in [2], [5]. See Remarks 2 below for more remarks on the literature. We give an alternative proof of these results, which serves as an example how the axioms for point-affine quadrangles can be verified. Note also the analogies with the treatment of affine translation planes by Andr´ e [1].
Proposition 2. Let E be a group and F a family of subgroups. To each T ∈ F, let a subgroup T ∗ be associated such that the following conditions are satisfied: For all T, S, V ∈ F we have
(i) {1} < T < T ∗ < E,
(ii) T = S implies E = T ∗ S(= ST ∗ ) and T ∗ ∩ S = {1}, (iii) T = S = V = T implies T S ∩ V = {1},
(iv) E = T ∗ ∪
S ∈F ST . Then A := (E,
T ∈F E/T, ∈) is a point-affine quadrangle and C(A) is an elation quadrangle with elation group E and elation center ∞.
Before we start with the proof of Proposition 2, we note the following Lemma 5. Under the assumptions of Proposition 2, one also has for all T, V ∈ F:
(v) T = V implies E =
S ∈F T SV .
Proof. Consider T, V ∈ F. For any x ∈ E = V ∗ ∪
S ∈F SV we then have x ∈ V ∗ or x ∈
S ∈F SV . For x ∈ V ∗ we pick t ∈ T \{1}, then (ii) yields tx ∈ V ∗ , and tx ∈
S ∈F SV follows. Thus x ∈
S ∈F T SV holds for all x ∈ E.
QEDRemark 4. Assertions (i)–(v) hold in each elation quadrangle. However,
assertion (iv) is independent of (i), (ii), (iii), (v), see [3].
Proof of Proposition. We remark that the line T x determines T ∈ F, in view of
T x = Sy ⇐⇒ T = S and yx −1 ∈ T.
Multiplication (from the right) with a ∈ E induces an automorphism of the incidence structure. Therefore, it suffices to describe distances from 1. For x ∈ E we have:
δ(1, x)
= 0
≤ 2 ≤ 4
≤ 6
⇐⇒
x = 1 x ∈
V ∈F V x ∈
S,V ∈F SV x ∈
T,S,V ∈F T SV
δ(1, T x)
= 1
≤ 3 ≤ 5
⇐⇒
x ∈ T x ∈
S ∈F T S x ∈
S,V ∈F T SV
(A 1 ) We have ∆ 1 (1) = F. From (iv) and (i) we deduce |F| ≥ 2. If we have |F| = 2, then (v) and (ii) yield E = T V = T ∗ V and so T = T ∗ , which contradicts (i). For T ∈ F the inequality |∆ 1 (T ) | = |T | ≥ 2 is a consequence of T = {1}.
(A 2 ) If there is a 2-gon with lines T and S, then there exists x ∈ E \{1} with x ∈ T ∩ S ⊆ T ∗ ∩ S, contradicting (ii). If there is a 3-gon with lines S, T, V x, we have V t = V x = V s for some t ∈ T and s ∈ V t∩S ⊆ V T ∩S, which contradicts (iii).
(A 3 ) For any x ∈ E we have x ∈
S ∈F T SV by (v) and therefore δ(1, x) ≤ 6.
The same argument shows that δ(1, T x) ≤ 5 for all T ∈ F. The proof of (A 4 ) will also show that there exist lines at distance 6.
(A 4 ) Here we will show
δ(T x, Sy) ∈ {0, 6} ⇐⇒ T = S and xy −1 ∈ T ∗ . First we choose x = 1, so we must prove
δ(T, Sy) ∈ {0, 6} ⇐⇒ T = S and y ∈ T ∗ .
Starting from T = S and y ∈ T ∗ , we obtain δ(T, Sy) = δ(T, T y) = 0 if y ∈ T . For y ∈ T ∗ \T we have T ∩T y = ∅ and therefore δ(T, T y) ≥ 4. But δ(T, T y) = 4 is equivalent to the existence of some z ∈ T y ⊆ T ∗ and some V ∈ F \ {T } such that V z ∩ T = ∅. So we take u ∈ V z ∩ T and get uz −1 ∈ V ∩ T z −1 ⊆ V ∩ T ∗ . This yields z = u ∈ T and T y = T z = T , a contradiction. We have thus proved that y ∈ T ∗ implies δ(T, T y) ∈ {0, 6}.
Now we assume δ(T, Sy) = 6. For T = S we would have E =
V ∈F SV T
by (v) and for each y ∈ E there exist s ∈ S, t ∈ T, v ∈ V such that y = s −1 vt.
Thus we have vt ∈ Sy and (Sy, vt, V t, t, T ) is a chain of length 4 from Sy to T . So we have S = T . Moreover δ(T, T y) ≥ 6 implies δ(1, T y) ≥ 5, that is y ∈ E \
S ∈F T S ⊆ T ∗ by (iv).
To show transitivity we take T x Sy and Sy V z and obtain first T = S = V . Without loss of generality we take x = 1 and get y ∈ T ∗ and there exists w ∈ E such that y, z ∈ T ∗ w, that is T ∗ w = T ∗ y = T ∗ and therefore we have z ∈ T ∗ . We have shown [T ] = T ∗ /T and [T x] = (T ∗ /T )x = {T tx | t ∈ T ∗ }. The inequality [T ] = {T } follows from T ∗ > T .
(A 5 ) We claim
(T ∗ /T )x ∼ (S ∗ /S)y ⇐⇒ T = S By definition (T ∗ /T )x ∼ (S ∗ /S)y means
∀t ∗ ∈ T ∗ ∀s ∗ ∈ S ∗ : δ(T t ∗ x, Ss ∗ y) = 2 (1) that is, we have T t ∗ x = Ss ∗ y or T t ∗ x ∩ Ss ∗ y = ∅.
For T = S we have T S ∗ = E xy −1 , that is xy −1 ∈ T s ∗ for some s ∗ ∈ S ∗ , which implies s ∗ y ∈ T x ∩ Ss ∗ y. This means that (1) is false, and (T ∗ /T )x ∼
(S ∗ /S)y. For T = S we have T u ∩ T v = ∅ whenever T u = T v and so the corresponding pseudo-pencils are collinear. This establishes the claim. It follows that ∼ is an equivalence relation with |F| ≥ 3 equivalence classes and |E/T ∗ | ≥ 2 elements in each equivalence class.
(A 6 ) Let Sx be a given line and (T ∗ /T )y ∼ (S ∗ /S)x, that is T = S. Then we have yx −1 ∈ E = T ∗ S. Take t ∗ ∈ T ∗ , s ∈ S such that sx = t ∗ y ∈ Sx ∩ T t ∗ y.
Thus we have T t ∗ y ∈ ∆ 2 (Sx) and T t ∗ y ∈ [T y] .
(A 7 ) Let a ∈ E be arbitrary. Consider lines T y with δ(a, T y) = 3.
If δ(a, T y) = 1 holds, we see that T a = T y ∈ ∆ 1 (a) and T a T y. If δ(a, T y) = 5, we have ∀S ∈ F : Sa∩T y = ∅. So we see that ay −1 ∈
S ∈F ST and therefore ay −1 ∈ T ∗ by (iv). This is equivalent to a ∈ T ∗ y and so T a is the line with T a T y we searched for.
(A 8 ) If [T ∗ /T ] ∼ and x ∈ E are given, then T x is the pre-image we need.
QED
