An algebraic structure for Moufang quadrangles

Tom De Medts

Author address:

Department of Pure Mathematics and Computer Algebra, Ghent University, Galglaan 2, B-9000 Gent, Belgium

E-mail address: tdemedts@cage.UGent.be

Contents

Chapter 1. Introduction 1 Acknowledgment 2

Chapter 2. Definition 3

Chapter 3. Some Identities 7

Chapter 4. From Quadrangular Systems To Moufang Quadrangles 15

Chapter 5. From Moufang Quadrangles To Quadrangular Systems 23

Chapter 6. Some Remarks 35

Chapter 7. Examples 37 7.1. Quadrangular Systems of Quadratic Form Type 37 7.2. Quadrangular Systems of Involutory Type 37 7.3. Quadrangular Systems of Indifferent Type 38 7.4. Quadrangular Systems of Pseudo-quadratic Form Type 39 7.5. Quadrangular Systems of Type E_{6}, E_{7} and E_{8} 40 7.6. Quadrangular Systems of Type F_{4} 43

Chapter 8. The Classification 47 8.1. Quadrangular Systems of Involutory Type 49 8.2. Quadrangular Systems of Quadratic Form Type 60 8.3. Quadrangular Systems of Indifferent Type 63 8.4. Quadrangular Systems of Pseudo-quadratic Form Type, I 65 8.5. Quadrangular Systems of Type F_{4} 70 8.6. Quadrangular Systems of Pseudo-quadratic Form Type, II 81 8.7. Quadrangular Systems of Type E_{6}, E_{7} and E_{8} 93

Appendix A. Abelian Quadrangular Systems 95 A.1. Reduced Quadrangular Systems 96 A.2. Indifferent Quadrangular Systems 96 A.3. Radical Quadrangular Systems 97

Appendix. Bibliography 99

Abstract

Very recently, the classification of Moufang polygons has been completed by Tits and Weiss. Moufang n-gons exist for n ∈ {3, 4, 6, 8} only. For n ∈ {3, 6, 8}, the proof is nicely divided into two parts: first, it is shown that a Moufang n-gon can be parametrized by a certain interesting algebraic structure, and secondly, these algebraic structures are classified. The classification of Moufang quadrangles (n=4) is not organized in this way due to the absence of a suitable algebraic structure. The goal of this article is to present such a uniform algebraic structure for Moufang quadrangles, and to classify these structures without referring back to the original Moufang quadrangles from which they arise, thereby also providing a new proof for the classification of Moufang quadrangles, which does consist of the division into these two parts. We hope that these algebraic structures will prove to be interesting in their own right.

Received by the editor September 19, 2002.

2000 Mathematics Subject Classification. 51E12, 16W10, 20E42.

Key words and phrases. Moufang quadrangles, quadrangular systems.

The author is a Research Assistant of the Fund for Scientific Research - Flanders (Belgium)

(F.W.O.-Vlaanderen).

CHAPTER 1 Introduction

The irreducible spherical buildings of rank at least three have been classified by J. Tits in 1974 [15]. The irreducible spherical buildings of rank two – which are called generalized polygons – are too numerous to classify, but in the addenda of [15], the Moufang condition for spherical buildings was introduced, and it was observed that every thick irreducible spherical building of rank at least three as well as every irreducible residue of such a building satisfies the Moufang condition. In this sense, the Moufang polygons are the “building bricks” of any spherical building of rank at least three.

Very recently, the classification of Moufang polygons has been completed by J. Tits and R. Weiss in [20]. It was first shown by J. Tits (see [17] and [18]) that Moufang n-gons exist for n ∈ {3, 4, 6, 8} only; see also [21]. For n ∈ {3, 6, 8}, the proof is divided into two parts, namely (A) it is shown that a Moufang n-gon can be parametrized by a certain algebraic structure, and (B) these algebraic structures are classified.

More precisely, it was already shown in 1933 (but in a slightly different form; see [2] or [5]) by R. Moufang (see [11]) that all Moufang triangles can be described by an alternative division ring, a notion which had been introduced by M. Zorn (see [22]). These alternative division rings were classified by R. Bruck and E. Kleinfeld in 1951; see [3]. The Moufang hexagons are described by unital quadratic Jordan division algebras of degree three, also known as anisotropic cubic norm structures (see [16]). These structures have been classified in its full generality in 1986 by H. Petersson and M. Racine (see [13] and [14]), whose proof is built on earlier work by A. Albert [1], F.D. Jacobson and N. Jacobson [6], N. Jacobson [7], [8] and K. McCrimmon [9], [10]. The Moufang octagons, finally, can be described by a so-called octagonal system, as was shown by J. Tits in 1983 (see [19]); since these systems have a very simple description, there is no need for part (B) in this case.

The classification of Moufang quadrangles (n=4) in [20] is not organized in this way due to the absence of a suitable algebraic structure. Instead, there are six different parameter systems, and even then, the division of the proof into parts (A) and (B) is missing in the two cases which lead to the exceptional quadrangles. Surprisingly, one of these classes, namely the exceptional quadrangles of type F_{4}, had only recently been discovered by R. Weiss during the classification process; see also [12].

The goal of this article is to present a uniform algebraic structure for Moufang quadrangles. These “quadrangular systems” reveal some of the structure of Moufang quadrangles which is hard to see without them. For example, we have successfully used them to answer a basic question about the automorphism group of the Moufang quadrangles of type F_{4} left open in (37.38) of [20]; see [4]. Moreover, it is possible to classify these structures without referring back to the original Moufang quadrangles from which they arise, thereby providing a new proof for the classification of Moufang quadrangles, which does consist of the division into parts (A) and (B).

The Moufang hexagons all arise from forms of algebraic groups of type G_{2}, 3D_{4}, E_{6} or E_{8}, or they are of mixed type associated with groups of type G_{2}. The Moufang quadrangles arise either from certain classical groups or from forms of algebraic groups of type E_{6}, E_{7} or E_{8} or are of mixed type associated with groups of type B_{2} or F_{4}. The quadrangular systems parametrize the Moufang quadrangles in the same way that the Jordan algebras mentioned above parametrize the Moufang hexagons, and it is our hope that the quadrangular systems will turn out to be equally interesting objects of study.

We start by giving the (ad hoc) definition of the quadrangular systems. Considering the background of the Moufang quadrangles, it should not be too surprising that we need a large number of axioms to describe these systems. In the next chapter, we give some elementary properties of these systems. In chapter 4, we explain how to construct a Moufang quadrangle starting from an arbitrary quadrangular system. In chapter 5, we show that every Moufang quadrangle arises in this way. After a couple of remarks in chapter 6, we present a list of 6 examples of quadrangular systems, which corresponds to the 6 different classes of Moufang quadrangles as described in [20]. Finally, chapter 8 is devoted to the classification of the quadrangular systems. We conclude with an appendix in which we restate the axiom system for abelian quadrangular systems and for some specific subclasses of those.

Acknowledgment

I am very grateful to Richard Weiss, for providing me a copy of his book “Moufang Polygons” [20] prior to publication, and for the many illuminating discussions we have had on this topic.

CHAPTER 2 Definition

Throughout this article, we will use the following notation. If S is a group, then we define S^{∗} := S \ {neutral element}. If S is a set which contains an element called “0”, then we define S^{∗} := S \ {0}. It will always be clear from the context which definition we mean.

Consider an abelian group (V, +) and a (possibly non-abelian) group (W, ⊞). The inverse of an element w ∈ W will be denoted by ⊟w, and by w_{1} ⊟ w_{2}, we mean w_{1} ⊞ (⊟w_{2}). Suppose that there is a map τ_{V} from V × W to V and a map τ_{W} from W × V to W, both of which will be denoted by · or simply by juxtaposition, i.e. τ_{V} (v, w) = vw = v · w and τ_{W} (w, v) = wv = w · v for all v ∈ V and all w ∈ W. Consider a map F from V × V to W and a map H from W × W to V, both of which are “bi-additive” in the sense that

F (v_{1} + v_{2}, v) = F (v_{1}, v) ⊞ F (v_{2}, v) ;
F (v, v_{1} + v_{2}) = F (v, v_{1}) ⊞ F (v, v_{2}) ;
H(w_{1} ⊞ w_{2}, w) = H(w_{1}, w) + H(w_{2}, w) ;
H(w, w_{1} ⊞ w_{2}) = H(w, w_{1}) + H(w, w_{2}) ;

for all v, v_{1}, v_{2} ∈ V and all w, w_{1}, w_{2} ∈ W. Suppose furthermore that there exists a fixed element ϵ ∈ V^{∗} and a fixed element δ ∈ W^{∗}, and suppose that, for each v ∈ V^{∗}, there exists an element v^{−1} ∈ V^{∗}, and for each w ∈ W^{∗}, there exists an element κ(w) ∈ W^{∗}, such that, for all w, w_{1}, w_{2} ∈ W and all v, v_{1}, v_{2} ∈ V, the following axioms are satisfied. We define

\overline{v} := ϵF (ϵ, v) − v
Rad(F) := {v ∈ V | F (v, V) = 0}
Rad(H) := {w ∈ W | H(w, W) = 0}
Im(F) := F (V, V)
Im(H) := H(W, W)

(Q_{1}) wϵ = w.

(Q_{2}) vδ = v.

(Q_{3}) (w_{1} ⊞ w_{2})v = w_{1}v ⊞ w_{2}v.

(Q_{4}) (v_{1} + v_{2})w = v_{1}w + v_{2}w.

(Q_{5}) w(−ϵ) · v = w(−v).

(Q_{6}) v · w(−ϵ) = vw.

(Q_{7}) Im(F) ⊆ Rad(H).

(Q_{8}) [w_{1}, w_{2}v]_{⊞} = F (H(w_{2}, w_{1}), v).

(Q_{9}) δ ∈ Rad(H).

(Q_{10}) If Rad(F) ≠ 0, then ϵ ∈ Rad(F).

(Q_{11}) w(v_{1} + v_{2}) = wv_{1} ⊞ wv_{2} ⊞ F (v_{2}w, v_{1}).

(Q_{12}) v(w_{1} ⊞ w_{2}) = vw_{1} + vw_{2} + H(w_{2}, w_{1}v).

(Q_{13}) (v^{−1})^{−1} = v (if v ≠ 0).

(Q_{14}) κ(⊟κ(⊟w)) = w(−ϵ) (if w ≠ 0).

(Q_{15}) wv · v^{−1} = w (if v ≠ 0).

(Q_{16}) v^{−1} · wv = −\overline{v(⊟w)} (if v ≠ 0).

(Q_{17}) F (v_{1}^{−1}, \overline{v_{2}})v_{1} = F (v_{1}, v_{2}) (if v_{1} ≠ 0).

(Q_{18}) vκ(w) · (⊟w) = −v (if w ≠ 0).

(Q_{19}) w · vκ(w) = κ(w)v (if w ≠ 0).

(Q_{20}) H(κ(w_{1}), w_{2})w_{1} = H(w_{1}, w_{2}) (if w_{1} ≠ 0).

Then we call the system (V, W, τ_{V} , τ_{W} , ϵ, δ) a quadrangular system. Note that we omit the maps F and H in our notation, as well as the maps v ↦ v^{−1} and w ↦ κ(w). The reason is that they are uniquely determined by V, W, τ_{V}, τ_{W}, ϵ and δ; see Theorem 3.7.

Remark 2.1. We will sometimes think about the maps τ_{V} from V × W to V and τ_{W} from W × V to W as “actions”, since it will turn out that, for every w ∈ W^{∗}, the map from V to itself which maps v to vw for every v ∈ V is an automorphism of V; similarly, for every v ∈ V^{∗}, the map from W to itself which maps w to wv for every w ∈ W is an automorphism of W; see Theorem 3.6. Note, however, that these maps are no group actions in the proper sense of the word, since v(w_{1} ⊞ w_{2}) ≠ vw_{1} · w_{2} and w(v_{1} + v_{2}) ≠ wv_{1} · v_{2} in general.

Remark 2.2. In writing down these axioms, we used the convention that the maps which are denoted by juxtaposition preceed those which are denoted by “·”. Note, however, that there is no danger of confusion, since we have not defined a multiplication on V or on W. Hence we will often write wvv^{−1} instead of wv · v^{−1}, for example.

We will show in Theorem 3.8 below that the following two identities are satisfied for every quadrangular system, for all v_{1}, v_{2} ∈ V and all w_{1}, w_{2} ∈ W.

(Q_{21}) F (v_{1}, v_{2}) = F (v_{2}, v_{1}).

(Q_{22}) H(w_{1}, w_{2}) = −\overline{H(w_{2}, w_{1})}.

Remark 2.3. These two identities show that, in some sense, F is a symmetric form and H is a skew-hermitian form. Note, however, that V and W are not vector spaces in general.

Moreover, we will show in Theorem 6.1 that the following four identities are satisfied for every quadrangular system, for all v, c ∈ V and all w, z ∈ W. We first introduce the notion of a reflection, which is a direct generalization of the classical notion of a reflection in a quadratic space:

π_{v}(c) := c − vF (v^{−1}, \overline{c})  (if v ≠ 0)
Π_{w}(z) := z ⊞ w(−H(κ(w), z))  (if w ≠ 0) .

Then

(Q_{23}) v · Π_{w}(z) = −\overline{\overline{\overline{\overline{v}(⊟w)}z}κ(w)} (if w ≠ 0).

(Q_{24}) w · \overline{π_{v}(ϵ)^{−1}} · \overline{π_{v}(c)} = wvcv^{−1} (if v ≠ 0).

(Q_{25}) π_{v}(\overline{c · δv})w = π_{v}(\overline{c · wv}) (if v ≠ 0).

(Q_{26}) Π_{⊟z}(w · ϵz)v = Π_{⊟z}(w · vz) (if w ≠ 0).

Let Ω := (V, W, τ_{V} , τ_{W} , ϵ, δ) and Ω′ := (V′, W′, τ_{V′}, τ_{W′}, ϵ′, δ′) be two quadrangular systems. We say that (ϕ, ψ) is a morphism from Ω to Ω′ if and only if ϕ is a morphism from V to V′ and ψ is a morphism from W to W′ such that ϕ(ϵ) = ϵ′, ψ(δ) = δ′, ϕ(vw) = ϕ(v)ψ(w) and ψ(wv) = ψ(w)ϕ(v), for all v ∈ V and all w ∈ W.

A morphism (ϕ, ψ) is called an monomorphism (respectively epimorphism, isomorphism) if and only if both ϕ and ψ are monomorphisms (respectively epimorphisms, isomorphisms). We call Ω and Ω′ isomorphic if and only if there exists an isomorphism (ϕ, ψ) from Ω to Ω′.

CHAPTER 3 Some Identities

We will now prove some identities which we will use in the construction of the Moufang quadrangles in chapter 4, and which will also be used in the classification of quadrangular systems.

Definition 3.1. For each w ∈ W^{∗}, we define λ(w) := ⊟κ(⊟w). Using this definition, (Q_{14}) can be rephrased as κ(λ(w)) = w(−ϵ).

Lemma 3.2. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w ∈ W^{∗} and all v ∈ V, we have that

(i) vwκ(⊟w) = −v;

(ii) κ(w)(v(⊟w)) = w(−v) .

Proof. If we plug in λ(w) for w in (Q_{18}), then it follows from (Q_{14}) that v(w(−ϵ))(⊟λ(w)) = −v, and by (Q_{6}) and the definition of λ, this is equivalent to vwκ(⊟w) = −v, which proves (i).

If we plug in λ(w) for w in (Q_{19}), then we get, again by (Q_{14}), that λ(w)(v · w(−ϵ)) = w(−ϵ)v. By (Q_{6}), (Q_{5}) and the definition of λ, this is equivalent to ⊟κ(⊟w)(vw) = w(−v). Replacing w by ⊟w now yields (ii). □

Lemma 3.3. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the following holds, for all w ∈ W and all v ∈ V :

(i) wv = 0 ⟺ w = 0 or v = 0 ; (ii) vw = 0 ⟺ v = 0 or w = 0 .

Proof. We will only prove statement (i); because of Lemma 3.2(i), the proof of (ii) is completely similar. By choosing v_{1} = v_{2} = 0 in (Q_{11}), we get w0 = w0 ⊞ w0, from which it follows that w0 = 0. Similarly, it follows from (Q_{3}) that 0v = 0.

On the other hand, suppose that wv = 0. If v ≠ 0, then it follows from (Q_{15}) that w = wvv^{−1} = 0v^{−1} = 0. □

Lemma 3.4. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w ∈ W and all v ∈ V, we have :

(i) (⊟w)v = ⊟(wv) ;

(ii) (−v)w = −(vw) .

It follows that the notations ⊟wv and −vw are unambiguous.

Proof. By putting w_{1} = w and w_{2} = ⊟w in (Q_{3}), we get 0v = wv ⊞ (⊟w)v, from which it follows that (⊟w)v = ⊟(wv). Similarly, (ii) follows from (Q_{4}). □

Lemma 3.5. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the following holds, for all w ∈ W and all v ∈ V :

(i) w_{1}v = w_{2}v ⟺ w_{1} = w_{2} or v = 0 ;

(ii) v_{1}w = v_{2}w ⟺ v_{1} = v_{2} or w = 0 .

Proof. By (Q_{3}) and Lemma 3.4(i), we have (w_{1} ⊟ w_{2})v = w_{1}v ⊟ w_{2}v, and so (i) is an immediate consequence of Lemma 3.3(i). Similarly, (ii) follows from Lemma 3.3(ii). □

Theorem 3.6. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then

(i) for every w ∈ W^{∗}, the map from V to itself which maps v to vw for every v ∈ V is an automorphism of V ;

(ii) for every v ∈ V^{∗}, the map from W to itself which maps w to wv for every w ∈ W is an automorphism of W .

Proof. We will only show (i), the proof of (ii) being completely similar. So let w ∈ W^{∗} be arbitrary, and let α be the map from V to itself which maps v to vw for every v ∈ V. By Lemma 3.3(ii), we have that α(0) = 0, and it follows from (Q_{4}) that α(v_{1} + v_{2}) = α(v_{1}) + α(v_{2}) for all v_{1}, v_{2} ∈ V, so α is a group morphism. Since w ≠ 0, it follows from Lemma 3.5(ii) that α is injective. Finally, it follows from (Q_{18}) that α(−vκ(⊟w)) = v for all v ∈ V, hence α is surjective as well, and we are done. □

Theorem 3.7. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the maps F and H and the maps v ↦ v^{−1} and w ↦ κ(w) are uniquely determined.

Proof. By (Q_{11}), F (v_{2}, v_{1}) = ⊟δv_{2} ⊟ δv_{1} ⊞ δ(v_{1} + v_{2}), so F is uniquely determined. Note that this implies that the map v ↦ \overline{v} is uniquely determined as well. By (Q_{12}), H(w_{2}, w_{1}) = −ϵw_{2} −ϵw_{1} +ϵ(w_{1} +w_{2}), so H is uniquely determined. Suppose that v^{∗} were another “inverse” of v. Then it would follow from (Q_{16}) that v^{∗}(wv) = v^{−1}(wv), but then Lemma 3.5 would imply that v^{∗} = v^{−1} after all. Similarly, it follows from Lemma 3.2(ii) that the map κ is uniquely determined. □

Theorem 3.8. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the identities (Q_{21}) and (Q_{22}) are satisfied for all v_{1}, v_{2} ∈ V and all w_{1}, w_{2} ∈ W.

Proof. We will first show that (Q_{21}) follows from (Q_{8}), (Q_{9}) and (Q_{11}). Since V is abelian, δ(v_{1} + v_{2}) = δ(v_{2} + v_{1}), and hence, by (Q_{11}), we have that

δv_{1} ⊞ δv_{2} ⊞ F (v_{2}, v_{1}) = δv_{2} ⊞ δv_{1} ⊞ F (v_{1}, v_{2}) ,

for all v_{1}, v_{2} ∈ V. In order to show (Q_{21}), it thus suffices to show that δv_{1} and δv_{2} commute for all v_{1}, v_{2} ∈ V. By (Q_{8}) and (Q_{9}),

[δv_{1}, δv_{2}]_{⊞} = F (H(δ, δv_{1}), v_{2}) = 0

for all v_{1}, v_{2} ∈ V, and hence (Q_{21}) holds.

Similarly, we will show that (Q_{22}) follows from (Q_{12}), (Q_{15}) and (Q_{16}). By substituting ⊟w for w and ϵ for v in (Q_{16}), we have that \overline{ϵw} = −ϵ^{−1}(⊟w) for all w ∈ W. Moreover, by (Q_{15}), wϵ^{−1} = w for all w ∈ W. By (Q_{12}) and the fact that H is additive in both variables, we thus have that

\overline{H(w_{2}, w_{1})} = \overline{ϵ(w_{1} ⊞ w_{2})} − \overline{ϵw_{1}} − \overline{ϵw_{2}}
= −ϵ^{−1}(⊟w_{2} ⊟ w_{1}) + ϵ^{−1}(⊟w_{1}) + ϵ^{−1}(⊟w_{2})
= −H(⊟w_{1}, ⊟w_{2}ϵ^{−1})
= −H(w_{1}, w_{2})

for all w_{1}, w_{2} ∈ W, hence (Q_{22}) holds. □

Lemma 3.9. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the map v ↦ \overline{v} is additive. In particular, we have that \overline{−v} = −\overline{v} for all v ∈ V. Moreover, for all c ∈ V^{∗}, the map π_{c} is additive. In particular, we have that π_{c}(−v) = −π_{c}(v) for all v ∈ V.

Proof. It follows from (Q_{7}) that H(F (ϵ, v_{2}), F (ϵ, v_{1})) = 0, for all v_{1}, v_{2} ∈ V. Hence

ϵF (ϵ, v_{1} + v_{2}) = ϵ(F (ϵ, v_{1}) ⊞ F (ϵ, v_{2}))
= ϵF (ϵ, v_{1}) + ϵF (ϵ, v_{2}) ,

by (Q_{12}). Since \overline{v} = ϵF (ϵ, v)−v, it follows from this that the map v ↦ \overline{v} is additive. Similarly, it follows from (Q_{7}) that H(F (c^{−1}, \overline{v_{2}}), F (c^{−1}, \overline{v_{1}})c) = 0, for all c ∈ V^{∗} and all v_{1}, v_{2} ∈ V. Since the map v ↦ \overline{v} is additive, it now follows, again by (Q_{12}), that

π_{c}(v_{1} + v_{2}) = (v_{1} + v_{2}) − cF (c^{−1}, \overline{v_{1} + v_{2}})
= v_{1} + v_{2} − c(F (c^{−1}, \overline{v_{1}}) ⊞ F (c^{−1}, \overline{v_{2}}))
= v_{1} − cF (c^{−1}, \overline{v_{1}}) + v_{2} − cF (c^{−1}, \overline{v_{2}})
= π_{c}(v_{1}) + π_{c}(v_{2}) ,

which is what we wanted to show. □

Lemma 3.10. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w ∈ W^{∗} and all v ∈ V^{∗}, we have

(i) (−v)^{−1} = −(v^{−1}) ;

(ii) κ(⊟w) = ⊟λ(w) .

Proof. If we replace w by δ in (Q_{16}), we have that −(v^{−1})(δv) = \overline{v(⊟δ)}. If we replace w by δ(−ϵ) in the same identity (Q_{16}), then we get, by (Q_{5}) and (Q_{6}) that v^{−1}(δ(−v)) = −\overline{v(⊟δ)}. If we replace v by −v in this identity, then we get, using the fact that \overline{−v} = −\overline{v}, that (−v)^{−1}(δv) = \overline{v(⊟δ)}. It follows that (−v)^{−1}(δv) = −(v^{−1})(δv). Since δv is non-zero, this implies, by Lemma 3.5(i), that (−v)^{−1} = −(v^{−1}), which proves (i). Identity (ii) follows immediately from the definition of λ. □

Lemma 3.11. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w ∈ W and all v ∈ V, we have

wv(−ϵ) = w(−v) .

Proof. Note that this identity is trivial if v = 0, so assume v ≠ 0. By (Q_{15}) and Lemma 3.10(i), we have that wvv^{−1} = w(−v)(−v^{−1}). It follows, by (Q_{5}), that wv(−ϵ)(−v^{−1}) = w(−v)(−v^{−1}). By Lemma 3.5(i), this implies that wv(−ϵ) = w(−v), for all w ∈ W and all v ∈ V. □

Lemma 3.12. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all v ∈ V, we have

\overline{\overline{v}} = v .

Proof. Assume v ≠ 0. By replacing w by ⊟δ in (Q_{16}), we see that \overline{v} = −v^{−1}(⊟δv). If, on the other hand, we replace v by v^{−1} and w by δv in this same identity (Q_{16}), then we get v(δvv^{−1}) = −\overline{v^{−1}(⊟δv)}. Combining those two equalities gives us the required identity \overline{\overline{v}} = v, since v(δvv^{−1}) is equal to v because of (Q_{15}) and (Q_{2}). □

Lemma 3.13. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w ∈ W and all v ∈ V, we have

(i) w(−v) = F (vw, v) ⊟ wv;

(ii) v(⊟w) = H(w, wv) − vw.

Proof. If we put v_{1} = −v and v_{2} = v in (Q_{11}), then we get that w0 = w(−v) ⊞ wv ⊞ F (vw, −v). Since F is additive in both variables, this is equivalent to w(−v) = F (vw, v) ⊟ wv, which proves (i). Similarly, (ii) follows from (Q_{12}). □

Lemma 3.14. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all v ∈ V^{∗}, we have

(i) κ(δv) = δ(\overline{v})^{−1} ; (ii) \overline{v^{−1}} = (\overline{v})^{−1} .

Proof. If we substitute δv for w and −v^{−1} for v in Lemma 3.2(ii), then we get that

κ(δv) · (−v^{−1}(⊟δv)) = δvv^{−1} ,

and hence, by (Q_{16}) and (Q_{15}),

κ(δv) · \overline{v} = δ .

By (Q_{15}), it thus follows that κ(δv) = δ(\overline{v})^{−1}, which shows (i). Note that it follows from Lemma 3.13(ii) that v(⊟δ) = −v for all v ∈ V, since δ ∈ Rad(H) by (Q_{9}). By Lemma 3.2(i) with ⊟δ\overline{v} in place of w, (Q_{16}) with \overline{v} in place of v and ⊟δ in place of w, Lemma 3.12, and (i) with \overline{v} in place of v, we now have that

(\overline{v})^{−1} = −(\overline{v})^{−1} · (⊟δ\overline{v}) · κ(δ\overline{v})
= v · δv^{−1} = −\overline{v^{−1}(⊟δ)} = \overline{v^{−1}} ,

which shows (ii). □

Lemma 3.15. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all v_{1}, v_{2} ∈ V and all w_{1}, w_{2} ∈ W, we have that

(i) F (v_{1}, v_{2})(−ϵ) = F (v_{1}, v_{2}) ;

(ii) H(κ(w_{1}(−ϵ)), w_{2}) = −H(κ(w_{1}), w_{2}) .

Proof. If we substitute \overline{v_{2}} for v_{2} in (Q_{17}), then we get, using (Q_{21}), that F (v_{2}, v_{1}^{−1})v_{1} = F (v_{1}, \overline{v_{2}}). Replacing v_{1} by −v_{1} in this last identity and applying Lemma 3.10(i) yields F (v_{2}, v_{1}^{−1})(−v_{1}) = F (v_{1}, \overline{v_{2}}). Thus, by (Q_{5}), F (v_{2}, v_{1}^{−1})(−ϵ)v_{1} = F (v_{2}, v_{1}^{−1})v_{1}, and it follows from Lemma 3.5(i) that F (v_{2}, v_{1}^{−1})(−ϵ) = F (v_{2}, v_{1}^{−1}). Replacing v_{1} by v_{1}^{−1} and using (Q_{13}) completes the proof of (i).

The proof of (ii) is similar. If we substitute w_{1}(−ϵ) for w_{1} in (Q_{20}), then we get that H(κ(w_{1}(−ϵ)), w_{2}) · w_{1}(−ϵ) = H(w_{1}(−ϵ), w_{2}). On the other hand, since Im(F) ⊆ Rad(H) by (Q_{7}), it follows from Lemma 3.13(i) that H(w_{1}(−ϵ), w_{2}) = H(⊟w_{1}, w_{2}) = −H(w_{1}, w_{2}). Hence

H(κ(w_{1}(−ϵ)), w_{2}) · w_{1}(−ϵ) = −H(w_{1}, w_{2}) ,

and it follows from (Q_{6}) and (Q_{20}) that

H(κ(w_{1}(−ϵ)), w_{2}) · w_{1} = −H(κ(w_{1}), w_{2}) · w_{1} .

It now follows from Lemma 3.5(ii) that (ii) holds. □

Lemma 3.16. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w ∈ W and all v ∈ V, we have

(i) F (\overline{vw}, v^{−1}) ⊟ w(−v)v^{−1} = w (if v ≠ 0) ;
(ii) H(w, κ(w)(−v)) + vκ(w)w = v (if w ≠ 0) .

Proof. Putting v_{1} = v and v_{2} = vw in (Q_{17}), and using (Q_{21}), yields F (\overline{vw}, v^{−1})v = F (vw, v), from which it follows, by (Q_{15}), that F (\overline{vw}, v^{−1}) = F (vw, v)v^{−1}. It follows from Lemma 3.13(i) and from (Q_{3}) that

w(−v)v^{−1} = (F (vw, v) ⊟ wv)v^{−1}
= F (vw, v)v^{−1} ⊟ wvv^{−1}
= F (\overline{vw}, v^{−1}) ⊟ w ,

from which (i) follows, since Im(F) ⊆ Z(W) by (Q_{7}) and (Q_{8}).

If we plug in vκ(w) for v in Lemma 3.13(ii), we get

vκ(w)(⊟w) = H(w, w · vκ(w)) − vκ(w)w ,

and applying (Q_{18}) and (Q_{19}) yields −v = H(w, κ(w)v) − vκ(w)w. Replacing v by −v gives us the required identity (ii). □

Theorem 3.17. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all w_{1} ∈ W^{∗}, w_{2} ∈ W, v_{1} ∈ V^{∗} and v_{2} ∈ V, we have

(i) F (2v_{2} − \overline{v_{1}F (v_{2}, v_{1}^{−1})}, v_{1}^{−1}) = 0 ;

(ii) H(κ(w_{1}) ⊞ λ(w_{1}), w_{2}) + H(λ(w_{1}), w_{1}(−H(κ(w_{1}), w_{2}))) = 0 .

Proof. By (Q_{5}), (Q_{15}) and Lemma 3.15(i), we have

F (v_{2}, v_{1}^{−1})(−v_{1})v_{1}^{−1} = F (v_{2}, v_{1}^{−1})(−ϵ)v_{1}v_{1}^{−1}
= F (v_{2}, v_{1}^{−1}) .

If we put v = v_{1} and w = F (v_{2}, v_{1}^{−1}) in Lemma 3.16(i), we get

F (\overline{v_{1}F (v_{2}, v_{1}^{−1})}, v_{1}^{−1}) = F (v_{2}, v_{1}^{−1}) ⊞ F (v_{2}, v_{1}^{−1})(−v_{1})v_{1}^{−1}
= F (v_{2}, v_{1}^{−1}) ⊞ F (v_{2}, v_{1}^{−1})
= F (2v_{2}, v_{1}^{−1}) ,

from which (i) follows.

To prove (ii), we first observe that, by (Q_{7}), it follows from Lemma 3.13(i) that

H(w_{1}(−v), w_{2}) = H(F (vw_{1}, v) ⊟ w_{1}v, w_{2})
= −H(w_{1}v, w_{2}) ,

for all w_{1}, w_{2} ∈ W and all v ∈ V. We also observe that κ(λ(w))(−v) = w(−ϵ)(−v) = wv because of (Q_{14}) and (Q_{5}), and that vκ(λ(w)) = v(w(−ϵ)) = vw because of (Q_{14}) and (Q_{6}), for all w ∈ W^{∗} and all v ∈ V.

If we substitute λ(w_{1}) for w and −H(κ(w_{1}), w_{2}) for v in Lemma 3.16(ii), then we get, using these remarks, that

H(λ(w_{1}), w_{1}(−H(κ(w_{1}), w_{2}))) = −H(κ(w_{1}), w_{2}) + H(κ(w_{1}), w_{2})w_{1}λ(w_{1})
= −H(κ(w_{1}), w_{2}) + H(w_{1}, w_{2})λ(w_{1})
= −H(κ(w_{1}), w_{2}) − H(w_{1}(−ϵ), w_{2})λ(w_{1})
= −H(κ(w_{1}), w_{2}) − H(κ(λ(w_{1})), w_{2})λ(w_{1})
= −H(κ(w_{1}), w_{2}) − H(λ(w_{1}), w_{2})
= −H(κ(w_{1}) ⊞ λ(w_{1}), w_{2}) ,

where we have used identity (Q_{20}) twice. This completes the proof of (ii). □

Lemma 3.18. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then, for all v ∈ V^{∗}, c ∈ V, w ∈ W^{∗} and z ∈ W, we have that

(i) π_{v}(c) = c − vF (v^{−1}, \overline{c}) = c − \overline{v^{−1}F (v, c)} ;

(ii) Π_{w}(z) = z ⊞ w(−H(κ(w), z)) = z ⊞ λ(w)H(w, z) .
Proof. By (Q_{17}), (Q_{15}) and (Q_{16}), we have that
vF (v^{−1}, \overline{c}) = v · F (v, c)v^{−1}
= −\overline{v^{−1}(⊟F (v, c))} .

Since Im(F) ⊆ Rad(H) by (Q_{7}), it follows from Lemma 3.13(ii) that v^{−1}(⊟F (v, c)) = −v^{−1}F (v, c), and hence

vF (v^{−1}, \overline{c}) = \overline{v^{−1}F (v, c)} ,

which shows (i).

By (Q_{20}), Lemma 3.2(i) and (Q_{19}), we have that

w(−H(κ(w), z)) = w · (H(w, z)κ(⊟w))
= ⊟κ(⊟w)H(w, z)
= λ(w)H(w, z) ,

which shows (ii). □

In the sequel, we will use both expressions as definitions of π_{v} and Π_{w}, without explicitly referring to this lemma.

Lemma 3.19. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system, and let w ∈ Rad(H) and v ∈ V. Then wv ∈ Rad(H) as well.

Proof. By (Q_{8}), [w, w_{2}]_{⊞} = 0 for all w_{2} ∈ W, hence v(w ⊞ w_{2}) = v(w_{2} ⊞ w). It follows from (Q_{12}) that H(w_{2}, wv) = H(w, w_{2}v) = 0 for all w_{2} ∈ W, since w ∈ Rad(H). By (Q_{22}), this implies that H(wv, w_{2}) = 0 for all w_{2} ∈ W, hence wv ∈ Rad(H). □

Lemma 3.20. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then we have that vw = −\overline{\overline{v}(δ\overline{v^{−1}})(⊟wv)} for all v ∈ V^{∗} and all w ∈ W.

Proof. It follows from (Q_{16}) that \overline{vw} = −v^{−1}(⊟wv) for all v ∈ V^{∗} and all w ∈ W. In particular, we have that \overline{v} = −v^{−1}(⊟δv) for all v ∈ V^{∗}, and hence that \overline{v}κ(δv) = v^{−1} by Lemma 3.2(i). If we substitute this expression for v^{−1} in the first identity, then we get that \overline{vw} = −\overline{v}κ(δv)(⊟wv) for all v ∈ V^{∗} and all w ∈ W. The result follows, since κ(δv) = δ\overline{v^{−1}} for all v ∈ V^{∗} by Lemma 3.14(i) and Lemma 3.14(ii). □

Lemma 3.21. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then π_{v}(vw) = v(⊟w) for all v ∈ V^{∗} and all w ∈ W.

Proof. Let v ∈ V^{∗} and w ∈ W be arbitrary. It follows from (Q_{5}) and (Q_{15}) that w(−v)v^{−1} = w(−ϵ). It thus follows from Lemma 3.16(i) that F (\overline{vw}, v^{−1}) = w ⊞ w(−ϵ). Hence, by (Q_{12}), (Q_{5}), (Q_{6}) and Lemma 3.13(ii),

vF (\overline{vw}, v^{−1}) = v(w ⊞ w(−ϵ))
= vw + v · w(−ϵ) + H(w(−ϵ), wv)
= vw + vw + H(w(−ϵ), w(−ϵ)(−v))
= vw + vw − v(⊟w(−ϵ)) − v(w(−ϵ))
= vw + vw − v(⊟w) − vw
= vw − v(⊟w) ,

and hence

π_{v}(vw) = vw − vF (\overline{vw}, v^{−1}) = v(⊟w) ,

which is what we had to show. □

The following two lemmas generalize some properties of reflections in an ordinary quadratic space.

Lemma 3.22. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then

(i) F (v, π_{v}(c)) = F (v, −c) for all v ∈ V^{∗} and all c ∈ V ; (ii) H(λ(w), Π_{w}(z)) = −H(κ(w), z) for all w ∈ W^{∗} and all z ∈ W .

Proof. By Lemma 3.18(i) and Theorem 3.17(i) with v^{−1} in place of v_{1} and c in place of v_{2}, we have that

F (v, π_{v}(c)) = F (v, c − \overline{v^{−1}F (v, c)})
= F (2c − \overline{v^{−1}F (c, v)}, v) ⊟ F (c, v)
= F (v, −c) ,

which shows (i). By Theorem 3.17(ii) with w_{1} = w and w_{2} = z,

H(λ(w), Π_{w}(z)) = H(λ(w), z ⊞ w(−H(κ(w), z)))

= H(κ(w) ⊞ λ(w), z) + H(λ(w), w(−H(κ(w), z))) − H(κ(w), z)

= −H(κ(w), z) ,

which shows (ii). □

Lemma 3.23. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then
(i) π_{v}(π_{v}(c)) = c for all v ∈ V^{∗} and all c ∈ V ;
(ii) Π_{⊟w}(Π_{w}(z)) = z for all w ∈ W^{∗} and all z ∈ W .

14 3. SOME IDENTITIES

Proof. By Lemma 3.22(i),

π_{v}(π_{v}(c)) = π_{v}(c) − \overline{v^{−1}F (v, π_{v}(c))}

= −π_{v}(−c) − \overline{v^{−1}F (v, −c)} = c + \overline{v^{−1}F (v, −c)} − \overline{v^{−1}F (v, −c)} = c ,

which shows (i). By Lemma 3.10(ii) and Lemma 3.22(ii),

Π_{⊟w}(Π_{w}(z)) = Π_{w}(z) ⊞ (⊟w)(−H(κ(⊟w), Π_{w}(z)))

= Π_{w}(z) ⊟ wH(λ(w), Π_{w}(z)) = Π_{w}(z) ⊟ w(−H(κ(w), z) = z ,

which shows (ii). □

CHAPTER 4 From Quadrangular Systems To Moufang Quadrangles

We will now describe how we can construct a Moufang quadrangle from a quadrangular system. We will use the method described in Tits and Weiss [20]. Therefore, we will describe 4 groups U_{1}, U_{2}, U_{3} and U_{4}, and we will implicitly define the group U_{+} := ⟨U_{1}, U_{2}, U_{3}, U_{4}⟩ by giving the commutator relations between any two of those groups. It is possible to construct a graph Q out of the data (U_{+}, U_{1}, U_{2}, U_{3}, U_{4}).

To prove that this construction will actually result in a Moufang quadrangle, it suffices to check that certain conditions A_{k}, B_{k} and C_{k} are satisfied (see [20] for details), and that we can construct groups U_{0} and U_{5}, such that, for every a_{0} ∈ U_{0}^{∗}, there exists some element μ(a_{0}) ∈ U_{4}^{∗}a_{0}U_{4}^{∗}, and for every a_{5} ∈ U_{5}^{∗}, there exists some element μ(a_{5}) ∈ U_{1}^{∗}a_{5}U_{1}^{∗}, for which certain conditions have to be satisfied (see again [20] for details).

Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Let U_{1} and U_{3} be two groups isomorphic to W, and let U_{2} and U_{4} be two groups isomorphic to V. Denote the corresponding isomorphisms by

\begin{matrix}x_{1} : & W → U_{1} : & w ↦ x_{1}(w) ; \\ x_{2} : & V → U_{2} : & v ↦ x_{2}(v) ; \\ x_{3} : & W → U_{3} : & w ↦ x_{3}(w) ; \\ x_{4} : & V → U_{4} : & v ↦ x_{4}(v) ; \end{matrix}

we say that U_{1} and U_{3} are parametrized by W and that U_{2} and U_{4} are parametrized by V. For all 1 ≤ i < j ≤ 4, we will denote the group ⟨U_{i}, …, U_{j}⟩ by U_{[i,j]}. Now, we implicitly define the group U_{+} = U_{[1,4]} by the following commutator relations:

[x_{1}(w_{1}), x_{3}(w_{2})^{−1}] = x_{2}(H(w_{1}, w_{2})) ,
[x_{2}(v_{1}), x_{4}(v_{2})^{−1}] = x_{3}(F (v_{1}, v_{2})) ,
[x_{1}(w), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv) ,
[U_{i}, U_{i+1}] = 1 ∀i ∈ {1, 2, 3} ,

for all w, w_{1}, w_{2} ∈ W and all v, v_{1}, v_{2} ∈ V. We will denote the corresponding graph by Q(V, W, τ_{V} , τ_{W} , ϵ, δ). If we define

ξ_{13}(x_{1}(w_{1}), x_{3}(w_{2})^{−1}) = x_{2}(H(w_{1}, w_{2})) ,
ξ_{24}(x_{2}(v_{1}), x_{4}(v_{2})^{−1}) = x_{3}(F (v_{1}, v_{2})) ,
ξ_{14}(x_{1}(w), x_{4}(v)^{−1}) = x_{2}(vw)x_{3}(wv) ,

then we can rephrase the conditions A_{k}, B_{k} and C_{k} as follows.

For all (i, j) ∈ {(1, 3), (2, 4), (1, 4)}, the following conditions should hold, for all a_{i}, b_{i} ∈ U_{i}, for all a_{j}, b_{j} ∈ U_{j}, and for all c ∈ U_{[i+1,j−1]}.

A_{ij}. ξ_{ij}(a_{i}b_{i}, a_{j}^{−1}) = ξ_{ij}(a_{i}, a_{j}^{−1})^{b_{i}}ξ_{ij}(b_{i}, a_{j}^{−1}).

B_{ij}. ξ_{ij}(a_{i}, (a_{j}b_{j})^{−1}) = ξ_{ij}(a_{i}, a_{j}^{−1})ξ_{ij}(a_{i}, b_{j}^{−1})^{a_{j}^{−1}}.

C_{ij}. c^{ξ_{ij}(a_{i},a_{j}^{−1})} = c^{a_{i}^{−1}a_{j}a_{i}a_{j}^{−1}}.

Theorem 4.1. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the corresponding graph Q(V, W, τ_{V} , τ_{W} , ϵ, δ) satisfies all of the conditions A_{ij}, B_{ij} and C_{ij}.

Proof. By plugging in the formulas for the functions ξ_{ij}, we get the following explicit conditions, which must hold for all v, v′, v_{1}, v_{2} ∈ V and all w, w′, w_{1}, w_{2} ∈ W.

A_{13}. x_{2}(H(w_{1} ⊞ w_{2}, w′)) = x_{2}(H(w_{1}, w′))^{x_{1}(w_{2})}x_{2}(H(w_{2}, w′)).

A_{24}. x_{3}(F (v_{1} + v_{2}, v′)) = x_{3}(F (v_{1}, v′))^{x_{2}(v_{2})}x_{3}(F (v_{2}, v′)).

A_{14}. x_{2}(v(w_{1}⊞w_{2}))x_{3}((w_{1}⊞w_{2})v) = (x_{2}(vw_{1})x_{3}(w_{1}v))^{x_{1}(w_{2})}·(x_{2}(vw_{2})x_{3}(w_{2}v)).

B_{13}. x_{2}(H(w′, w_{1} ⊞ w_{2})) = x_{2}(H(w′, w_{1}))x_{2}(H(w′, w_{2}))^{x_{3}(⊟w_{2})}.

B_{24}. x_{3}(F (v′, v_{1} + v_{2})) = x_{3}(F (v′, v_{1}))x_{3}(F (v′, v_{2}))^{x_{4}(−v_{2})}.

B_{14}. x_{2}((v_{1}+v_{2})w)x_{3}(w(v_{1}+v_{2})) = (x_{2}(v_{1}w)x_{3}(wv_{1}))·(x_{2}(v_{2}w)x_{3}(wv_{2}))^{x_{4}(−v_{1})}.

C_{13}. x_{2}(v)^{x_{2}(H(w,w′))} = x_{2}(v)^{x_{1}(⊟w)x_{3}(w′)x_{1}(w)x_{3}(⊟w′)}.

C_{24}. x_{3}(w)^{x_{3}(F (v,v′))} = x_{3}(w)^{x_{2}(−v)x_{4}(v′)x_{2}(v)x_{4}(−v′)}.

C_{14,2}. x_{2}(v′)^{x_{2}(vw)x_{3}(wv)} = x_{2}(v′)^{x_{1}(⊟w)x_{4}(v)x_{1}(w)x_{4}(−v)}.

C_{14,3}. x_{3}(w′)^{x_{2}(vw)x_{3}(wv)} = x_{3}(w′)^{x_{1}(⊟w)x_{4}(v)x_{1}(w)x_{4}(−v)}. Note that [U_{1}, U_{2}] = [U_{2}, U_{3}] = [U_{3}, U_{4}] = 1; some of the conditions can be simplified by this observation.

Condition (A_{13}) is equivalent to

x_{2}(H(w_{1} ⊞ w_{2}, w′)) = x_{2}(H(w_{1}, w′))x_{2}(H(w_{2}, w′)) ,

which is, in turn, equivalent to the fact that H is additive in the first variable. Completely similarly, (A_{24}), (B_{13}) and (B_{24}) also follow from the fact that F and H are additive in both variables.

By (Q_{12}), the left hand side of (A_{14}) can be rewritten as

x_{2}(vw_{1} + vw_{2} + H(w_{2}, w_{1}v))x_{3}((w_{1} ⊞ w_{2})v) .

Using the fact that b^{a} = [a, b^{−1}]b, we can rewrite the right hand side as

x_{2}(vw_{1})[x_{1}(w_{2}), x_{3}(w_{1}v)^{−1}]x_{3}(w_{1}v)x_{2}(vw_{2})x_{3}(w_{2}v)

which is also equal to

x_{2}(vw_{1})x_{2}(H(w_{2}, w_{1}v))x_{3}(w_{1}v)x_{2}(vw_{2})x_{3}(w_{2}v) .

Since [U_{2}, U_{2}] = [U_{2}, U_{3}] = 1, we can rewrite this once more as

x_{2}(vw_{1} + vw_{2} + H(w_{2}, w_{1}v))x_{3}(w_{1}v + w_{2}v) .

It now follows from (Q_{3}) that (A_{14}) holds.

Similarly, (B_{14}) follows from (Q_{11}) and (Q_{4}); we additionally need the fact that Im(F) ≤ Z(W), which follows from (Q_{7}) and (Q_{8}).

Since [U_{2}, U_{1}] = [U_{2}, U_{2}] = [U_{2}, U_{3}] = 1, (C_{13}) becomes trivial. Because [U_{3}, U_{2}] = [U_{3}, U_{4}] = 1, we have that (C_{24}) is equivalent to the condition [w, F (v, v′)]_{⊞} = 1. Since Im(F) ≤ Z(W), this is always satisfied.

To prove (C_{14,2}), we need to show that

x_{2}(v′) = x_{2}(v′)^{x_{4}(v)x_{1}(w)x_{4}(−v)} ,

which is the same as

(x_{2}(v′)^{x_{4}(v)})^{x_{1}(w)} = x_{2}(v′)^{x_{4}(v)} .

Since x_{2}(v′)^{x_{4}(v)} = x_{2}(v′)[x_{2}(v′), x_{4}(−v)^{−1}] = x_{2}(v′)x_{3}(F (v′, −v)), we have that

(x_{2}(v′)^{x_{4}(v)})^{x_{1}(w)} = (x_{2}(v′)x_{3}(F (v′, −v)))^{x_{1}(w)}
= x_{2}(v′)[x_{1}(w), x_{3}(F (v′, −v))^{−1}]x_{3}(F (v′, −v))
= x_{2}(v′)x_{2}(H(w, F (v′, −v)))x_{3}(F (v′, −v))
= x_{2}(v′)x_{3}(F (v′, −v))
= x_{2}(v′)^{x_{4}(v)}

since Im(F) ≤ Rad(H) by (Q_{7}). Thus (C_{14,2}) holds.

The left hand side of (C_{14,3}) is equal to

x_{3}(w′)[x_{3}(w′), x_{3}(wv)] ,

which is, by (Q_{8}), also equal to

x_{3}(w′ ⊞ F (H(w, w′), v)) .

The right hand side is equal to

x_{3}(w′)^{x_{1}(⊟w)x_{4}(v)x_{1}(w)x_{4}(−v)}

= (x_{2}(−H(w, w′))x_{3}(w′))^{x_{4}(v)x_{1}(w)x_{4}(−v)}

= (x_{2}(−H(w, w′))x_{3}(F (H(w, w′), v))x_{3}(w′))^{x_{1}(w)x_{4}(−v)}

= (x_{2}(−H(w, w′))x_{3}(w′ ⊞ F (H(w, w′), v)))^{x_{1}(w)x_{4}(−v)}

= (x_{2}(−H(w, w′))x_{2}(H(w, w′ ⊞ F (H(w, w′), v)))x_{3}(w′ ⊞ F (H(w, w′), v)))^{x_{4}(−v)}

= x_{3}(w′ ⊞ F (H(w, w′), v))^{x_{4}(−v)}

= x_{3}(w′ ⊞ F (H(w, w′), v)) ,

thus (C_{14,3}) holds. This concludes the proof of this theorem. □

Let U_{0} be a group parametrized by V (via a map x_{0}), and let U_{5} be a group parametrized by W (via a map x_{5}). We define an action of U_{0} on U_{[1,3]} by the following commutator relations.

[U_{0}, U_{1}] = 1
[x_{0}(v_{1}), x_{2}(v_{2})^{−1}] = x_{1}(F (v_{1}, \overline{v_{2}}))
[x_{0}(v), x_{3}(w)^{−1}] = x_{1}(wv)x_{2}(−\overline{v(⊟w)})

for all w ∈ W and all v, v_{1}, v_{2} ∈ V. For each x_{4}(v) ∈ U_{4}^{∗}, we define an element μ(x_{4}(v)) ∈ U_{0}^{∗}x_{4}(v)U_{0}^{∗} as

μ(x_{4}(v)) = x_{0}(v^{−1})x_{4}(v)x_{0}(v^{−1}) .

We define an action of U_{5} on U_{[2,4]} by the following commutator relations.

[x_{2}(v), x_{5}(w)^{−1}] = x_{3}(w(−v))x_{4}(−v(⊟w))
[x_{3}(w_{1}), x_{5}(w_{2})^{−1}] = x_{4}(H(w_{2}, w_{1}))
[U_{4}, U_{5}] = 1

for all w, w_{1}, w_{2} ∈ W and all v ∈ V. For each x_{1}(w) ∈ U_{1}^{∗}, we define an element μ(x_{1}(w)) ∈ U_{5}^{∗}x_{1}(w)U_{5}^{∗} as

μ(x_{1}(w)) = x_{5}(κ(w))x_{1}(w)x_{5}(λ(w)) .

Note that, by Lemma 3.10, μ(x_{4}(v)^{−1}) = μ(x_{4}(v))^{−1}, and μ(x_{1}(w)^{−1}) = μ(x_{1}(w))^{−1}.

It follows Theorem 4.1 that the graph Q(V, W, τ_{V} , τ_{W} , ϵ, δ) satisfies the conditions which are denoted by (M_{1}) and (M_{2}) in [20]. In order to obtain a Moufang quadrangle, this graph additionally has to satisfy the conditions (M_{3}) and (M_{4}). In Theorem 4.2, we will show that (M_{3}) holds; the validity of (M_{4}) will be shown in Theorem 4.3.

Theorem 4.2. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the corresponding graph Q(V, W, τ_{V} , τ_{W} , ϵ, δ), together with the group U_{0} and the map μ, satisfies the following conditions, for all v ∈ V.

(i) U_{0}^{μ(x_{4}(ϵ))} = U_{4} , considered as subgroups of Aut(U_{[1,3]}) ; (ii) U_{1}^{μ(x_{4}(v))} = U_{3} . More precisely, we have that x_{1}(w)^{μ(x_{4}(v))} = x_{3}(w(−v))

for all w ∈ W and all v ∈ V^{∗};

(iii) U_{2}^{μ(x_{4}(v))} = U_{2} . More precisely, we have that x_{2}(v′)^{μ(x_{4}(v))} = x_{2}(π_{v}(v′))

for all v′ ∈ V and all v ∈ V^{∗};

(iv) U_{3}^{μ(x_{4}(v))} = U_{1} . More precisely, we have that x_{3}(w)^{μ(x_{4}(v))} = x_{1}(wv^{−1})

for all w ∈ W and all v ∈ V^{∗};

(v) U_{4}^{μ(x_{4}(ϵ))} = U_{0} , considered as subgroups of Aut(U_{[1,3]}) .

Proof. For all w ∈ W and all v ∈ V^{∗}, we have

x_{1}(w)^{μ(x_{4}(v))} = x_{1}(w)^{x_{0}(v^{−1})x_{4}(v)x_{0}(v^{−1})}
= x_{1}(w)^{x_{4}(v)x_{0}(v^{−1})}
= (x_{1}(w)x_{2}(−vw)x_{3}(w(−v)))^{x_{0}(v^{−1})}
= x_{1}(w)x_{1}(F (−\overline{vw}, v^{−1}))x_{2}(−vw)x_{1}(w(−v)v^{−1})
· x_{2}(−\overline{v^{−1}(⊟w(−v))})x_{3}(w(−v))
= x_{3}(w(−v)) ,

where we have used Lemma 3.16, (Q_{16}) and Lemma 3.12 for the last equality. By substituting wv^{−1} for w and −v for v, we also get

x_{1}(wv^{−1})^{μ(x_{4}(−v))} = x_{3}(wv^{−1}v) ,

and since μ(x_{4}(−v)) = μ(x_{4}(v))^{−1} and by (Q_{15}), it follows that

x_{3}(w)^{μ(x_{4}(v))} = x_{1}(wv^{−1}) .

So we have proved that U_{1}^{μ(x_{4}(v))} ⊆ U_{3} and U_{3}^{μ(x_{4}(v))} ⊆ U_{1}. If we replace v by −v in those two relations, and conjugate by μ(x_{4}(v)), it also follows that U_{1} ⊆ U_{3}^{μ(x_{4}(v))} and U_{3} ⊆ U_{1}^{μ(x_{4}(v))}. So (ii) and (iv) are proved.

We will now prove (iii). For all v ∈ V^{∗} and all v′ ∈ V, we have

x_{2}(v′)^{μ(x_{4}(v))} = x_{2}(v′)^{x_{0}(v^{−1})x_{4}(v)x_{0}(v^{−1})}
= (x_{1}(F (v^{−1}, \overline{v′}))x_{2}(v′))^{x_{4}(v)x_{0}(v^{−1})}
= (x_{1}(F (v^{−1}, \overline{v′}))x_{2}(−vF (v^{−1}, \overline{v′}))x_{3}(F (v^{−1}, \overline{v′})(−v))
· x_{2}(v′)x_{3}(⊟F (v′, v)))^{x_{0}(v^{−1})}
= (x_{1}(F (v^{−1}, \overline{v′}))x_{2}(v′ − vF (v^{−1}, \overline{v′})))^{x_{0}(v^{−1})} ,

where we have used (Q_{17}) for the last equality. It follows that

x_{2}(v′)^{μ(x_{4}(v))} = x_{1}(F (v^{−1}, \overline{v′}))x_{1}(F (v^{−1}, \overline{v′ − vF (v^{−1}, \overline{v′})}))x_{2}(v′ − vF (v^{−1}, \overline{v′}))
= x_{2}(v′ − vF (v^{−1}, \overline{v′}))
= x_{2}(π_{v}(v′)) ,

where we have used Lemma 3.9 and Theorem 3.17(i). Hence U_{2}^{μ(x_{4}(v))} ⊆ U_{2}, and again by replacing v by −v and conjugating by μ(x_{4}(v)), we get that U_{2} ⊆ U_{2}^{μ(x_{4}(v))} as well, from which (iii) follows.

To prove (v), we will check that the action of μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ)) on U_{[1,3]} is the same as the action of x_{0}(v) on U_{[1,3]}, for all v ∈ V. Note that we will use the fact that wϵ^{−1} = w, which follows by choosing v = ϵ in (Q_{15}), and the fact that F (ϵ^{−1}, \overline{v}) = F (ϵ, v), which holds by substituting ϵ for v_{1} in (Q_{17}).

Using the definition of the map v ↦ \overline{v}, we see that

x_{2}(v)^{μ(x_{4}(ϵ))} = x_{2}(v − ϵF (ϵ^{−1}, \overline{v}))
= x_{2}(v − ϵF (ϵ, v))
= x_{2}(−\overline{v}) ,

for all v ∈ V. Since −\overline{(−\overline{v})} = v, replacing v by −\overline{v} and conjugating by μ(x_{4}(−ϵ)) yields

x_{2}(v)^{μ(x_{4}(−ϵ))} = x_{2}(−\overline{v})

for all v ∈ V, as well. For the action on U_{1}, we have

x_{1}(w)^{μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))} = x_{3}(w)^{x_{4}(v)μ(x_{4}(ϵ))}
= x_{3}(w)^{μ(x_{4}(ϵ))}
= x_{1}(w)
= x_{1}(w)^{x_{0}(v)} ;

for the action on U_{2}, we have

x_{2}(v′)^{μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))} = x_{2}(−\overline{v′})^{x_{4}(v)μ(x_{4}(ϵ))}
= (x_{2}(−\overline{v′})x_{3}(F (\overline{v′}, v)))^{μ(x_{4}(ϵ))}
= x_{2}(v′)x_{1}(F (v, \overline{v′}))
= x_{2}(v′)^{x_{0}(v)} .

To check the action on U_{3}, we need (Q_{6}), (Q_{5}), (Q_{22}), Lemma 3.9 and Lemma 3.13(ii) :

x_{3}(w)^{μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))} = x_{1}(w(−ϵ))^{x_{4}(v)μ(x_{4}(ϵ))}
= (x_{1}(w(−ϵ))x_{2}(−vw)x_{3}(wv))^{μ(x_{4}(ϵ))}
= x_{3}(w)x_{2}(\overline{vw})x_{1}(wv)
= x_{1}(wv)x_{2}(\overline{vw} + H(wv, w))x_{3}(w)
= x_{1}(wv)x_{2}(\overline{vw − H(w, wv)})x_{3}(w)
= x_{1}(wv)x_{2}(−\overline{v(⊟w)})x_{3}(w)
= x_{3}(w)^{x_{0}(v)} .

Thus (v) is proved.

To prove (i), we will check that the action of μ(x_{4}(ϵ))x_{4}(v)μ(x_{4}(−ϵ)) on U_{[1,3]} is the same as the action of x_{0}(v) on U_{[1,3]}. We can take a shortcut by observing that

μ(x_{4}(ϵ))x_{4}(v)μ(x_{4}(−ϵ)) = μ(x_{4}(ϵ))^{2}μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))μ(x_{4}(−ϵ))^{2} .

We just have to do a short calculation to see that

x_{1}(w)^{μ(x_{4}(ϵ))^{2}} = x_{1}(w)^{μ(x_{4}(−ϵ))^{2}} = x_{1}(w(−ϵ)) ;
x_{2}(v)^{μ(x_{4}(ϵ))^{2}} = x_{2}(v)^{μ(x_{4}(−ϵ))^{2}} = x_{2}(v) ;
x_{3}(w)^{μ(x_{4}(ϵ))^{2}} = x_{3}(w)^{μ(x_{4}(−ϵ))^{2}} = x_{3}(w(−ϵ)) .

For the action on U_{1}, we have

x_{1}(w)^{μ(x_{4}(ϵ))x_{4}(v)μ(x_{4}(−ϵ))} = x_{1}(w(−ϵ))^{μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))μ(x_{4}(−ϵ))^{2}}
= x_{1}(w(−ϵ))^{μ(x_{4}(−ϵ))^{2}}
= x_{1}(w)
= x_{1}(w)^{x_{0}(v)} ;

for the action on U_{2}, we have, by Lemma 3.15, that

x_{2}(v′)^{μ(x_{4}(ϵ))x_{4}(v)μ(x_{4}(−ϵ))} = x_{2}(v′)^{μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))μ(x_{4}(−ϵ))^{2}}
= (x_{2}(v′)x_{1}(F (v, \overline{v′})))^{μ(x_{4}(−ϵ))^{2}}
= x_{2}(v′)x_{1}(F (v, \overline{v′})(−ϵ))
= x_{2}(v′)x_{1}(F (v, \overline{v′}))
= x_{2}(v′)^{x_{0}(v)} ;

Finally, for the action on U_{3}, we have, using Lemma 3.11, that

x_{3}(w)^{μ(x_{4}(ϵ))x_{4}(v)μ(x_{4}(−ϵ))} = x_{3}(w(−ϵ))^{μ(x_{4}(−ϵ))x_{4}(v)μ(x_{4}(ϵ))μ(x_{4}(−ϵ))^{2}}
= (x_{1}(w(−v))x_{2}(−\overline{v(⊟w)})x_{3}(w(−ϵ)))^{μ(x_{4}(−ϵ))^{2}}
= x_{1}(wv)x_{2}(−\overline{v(⊟w)})x_{3}(w)
= x_{3}(w)^{x_{0}(v)} .

So we have proved (i), and this completes the proof of this theorem. □

Theorem 4.3. Let (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the corresponding graph Q(V, W, τ_{V} , τ_{W} , ϵ, δ), together with the group U_{5} and the map μ, satisfies the following conditions, for all w ∈ W.

(i) U_{5}^{μ(x_{1}(δ))} = U_{1} , considered as subgroups of Aut(U_{[2,4]}) ; (ii) U_{4}^{μ(x_{1}(w))} = U_{2} . More precisely, we have that x_{4}(v)^{μ(x_{1}(w))} = x_{2}(vw) for

all v ∈ V and all w ∈ W^{∗};

(iii) U_{3}^{μ(x_{1}(w))} = U_{3} . More precisely, we have that x_{3}(w′)^{μ(x_{1}(w))} = x_{3}(Π_{w}(w′))

for all w′ ∈ W and all w ∈ W^{∗};

(iv) U_{2}^{μ(x_{1}(w))} = U_{4} . More precisely, we have that x_{2}(v)^{μ(x_{1}(w))} = x_{4}(−vκ(w))

for all v ∈ V and all w ∈ W^{∗};

(v) U_{1}^{μ(x_{1}(δ))} = U_{5} , considered as subgroups of Aut(U_{[2,4]}) .

Proof. The proof of this theorem is very similar to the previous one, so we will skip most of the calculations.

For all w ∈ W^{∗} and all v ∈ V, we have

x_{2}(v)^{μ(x_{1}(w))} = x_{2}(v)^{x_{5}(κ(w))x_{1}(w)x_{5}(λ(w))}
= x_{4}(−vκ(w)) ,

where we have to use (Q_{19}) and Lemma 3.16(ii). By substituting −w for w and vw for v, we also get

x_{2}(vw)^{μ(x_{1}(⊟w))} = x_{4}(−vwκ(⊟w)) ,

and since μ(x_{1}(⊟w)) = μ(x_{1}(w))^{−1} and by Lemma 3.2(i), it follows that

x_{4}(v)^{μ(x_{1}(w))} = x_{2}(vw) .

So we have proved that U_{4}^{μ(x_{1}(w))} ⊆ U_{2} and U_{2}^{μ(x_{1}(w))} ⊆ U_{4}. If we replace w by ⊟w in those two relations, and conjugate by μ(x_{1}(w)), it also follows that U_{4} ⊆ U_{2}^{μ(x_{1}(w))} and U_{2} ⊆ U_{4}^{μ(x_{1}(w))}. So (ii) and (iv) are proved.

We will now prove (iii). For all w ∈ W^{∗} and all w′ ∈ W, we have

x_{3}(w′)^{μ(x_{1}(w))} = x_{3}(w′)^{x_{5}(κ(w))x_{1}(w)x_{5}(λ(w))}
= x_{3}(w′ ⊞ w(−H(κ(w), w′)))
= x_{3}(Π_{w}(w′)) ,

where we have to use (Q_{20}) and Theorem 3.17(ii). Hence U_{3}^{μ(x_{1}(w))} ⊆ U_{3}, and again by replacing w by −w and conjugating by μ(x_{1}(w)), we get that U_{3} ⊆ U_{3}^{μ(x_{1}(w))} as well, from which (iii) follows.

To prove (v), we will check that the action of μ(x_{1}(⊟δ))x_{1}(w)μ(x_{1}(δ)) on U_{[2,4]} is the same as the action of x_{5}(w) on U_{[2,4]}, for all w ∈ W. First of all, observe that it follows from Lemma 3.13(ii) and from the fact that δ ∈ Rad(H) (by (Q_{7})) that v(⊟δ) = −v. If we put w = δ in (Q_{18}), it thus follows that vκ(δ) = v; if we put w = ⊟δ in this same identity (Q_{18}), it follows that vκ(⊟δ) = −v. Furthermore, if we put w_{1} = δ in (Q_{20}), it follows from (Q_{7}) that H(κ(δ), w) = 0, for all w ∈ W.

Using these facts, we can prove that

x_{4}(v)^{μ(x_{1}(⊟δ))x_{1}(w)μ(x_{1}(δ))} = x_{4}(v)^{x_{5}(w)} ;
x_{3}(w′)^{μ(x_{1}(⊟δ))x_{1}(w)μ(x_{1}(δ))} = x_{3}(w′)^{x_{5}(w)} ;
x_{2}(v)^{μ(x_{1}(⊟δ))x_{1}(w)μ(x_{1}(δ))} = x_{2}(v)^{x_{5}(w)} ,

where we have to use (Q_{21}) and Lemma 3.13(i) as well. Thus (v) is proved.

To prove (i), we have to check that the action of μ(x_{1}(δ))x_{5}(w)μ(x_{1}(⊟δ)) on U_{[2,4]} is the same as the action of x_{5}(w(−ϵ)) on U_{[2,4]}. Again, we can take a shortcut by observing that

μ(x_{1}(δ))x_{1}(w)μ(x_{1}(⊟δ)) = μ(x_{1}(δ))^{2}μ(x_{1}(⊟δ))x_{1}(w)μ(x_{1}(δ))μ(x_{1}(⊟δ))^{2} .

First, we observe that

x_{2}(v)^{μ(x_{1}(δ))^{2}} = x_{2}(v)^{μ(x_{1}(⊟δ))^{2}} = x_{2}(−v) ;
x_{3}(w)^{μ(x_{1}(δ))^{2}} = x_{3}(w)^{μ(x_{1}(⊟δ))^{2}} = x_{3}(w) ;
x_{4}(v)^{μ(x_{1}(δ))^{2}} = x_{4}(v)^{μ(x_{1}(⊟δ))^{2}} = x_{4}(−v) .

It now follows from a short calculation that

x_{4}(v)^{μ(x_{1}(δ))x_{5}(w)μ(x_{1}(⊟δ))} = x_{4}(v)^{x_{5}(w(−ϵ))} ;
x_{3}(w′)^{μ(x_{1}(δ))x_{5}(w)μ(x_{1}(⊟δ))} = x_{3}(w′)^{x_{5}(w(−ϵ))} ;
x_{2}(v)^{μ(x_{1}(δ))x_{5}(w)μ(x_{1}(⊟δ))} = x_{2}(v)^{x_{5}(w(−ϵ))} .

So we have proved (i), and this completes the proof of this theorem. □

This completes the proof of the fact that the graph Q(V, W, τ_{V} , τ_{W} , ϵ, δ) is a Moufang quadrangle.

CHAPTER 5 From Moufang Quadrangles To Quadrangular Systems

In this chapter, we will prove that every Moufang quadrangle can be obtained from the construction described in the previous chapters. We will make intensive use of Chapter 21 “Quadrangles” in [20]. Since we are dealing with the same objects as in [20], it should not be very surprising that we need these same properties. However, after recalling these facts, our approach will very quickly diverge from the one given in [20].

Let Γ be an arbitrary Moufang quadrangle, and consider a fixed apartment Σ = (0, 1, …, 7), where the vertices are labeled modulo 8. We will denote its root groups U(i, i + 1, …, i + 4) by U_{i}, for all i ∈ ℤ. The following properties of the root groups are fundamental.

Theorem 5.1. (i) [U_{i}, U_{j}] ≤ U_{[i+1,j−1]}, if i < j < i + 4.

(ii) For each i, the product map from U_{i} × U_{i+1} × U_{i+2} × U_{i+3} to U_{[i,i+3]} is bijective.

Proof. See, for example, [20, (5.5) and (5.6)]. □

Thanks to this theorem, we can use the following notation. Let a_{i} ∈ U_{i} and a_{j} ∈ U_{j}, with j ∈ {i + 2, i + 3}. For each k such that i < k < j, we set

[a_{i}, a_{j}]_{k} = a_{k} ,

where a_{k} is the unique element of U_{k} appearing in the factorization of [a_{i}, a_{j}] ∈ U_{[i+1,j−1]}.

Let V_{i} := [U_{i−1}, U_{i+1}] ≤ U_{i} and Y_{i} := C_{U_{i}}(U_{i−2}) ≤ U_{i} for each i. It can be shown (see [20, (21.20.i)]) that Y_{i} = C_{U_{i}}(U_{i+2}) as well.

The following theorem defines the functions κ, λ and μ.

Theorem 5.2. For each i, there exist unique functions κ_{i}, λ_{i} : U_{i}^{∗} → U_{i+4}^{∗}, such that (i − 1)^{a_{i}λ_{i}(a_{i})} = i + 1 and (i + 1)^{κ_{i}(a_{i})a_{i}} = i − 1, for all a_{i} ∈ U_{i}^{∗}. The product μ_{i}(a_{i}) := κ_{i}(a_{i})a_{i}λ_{i}(a_{i}) fixes i and i + 4 and reflects Σ, and U_{j}^{μ_{i}(a_{i})} = U_{2i+4−j} for each a_{i} ∈ U_{i}^{∗} and each j.

Proof. See [20, (6.1)]. □

Since we will apply these functions only when it is clear in which U_{i}^{∗} the argument lies, we will write κ, λ and μ in place of κ_{i}, λ_{i} and μ_{i}. Note that it follows from the last statement of this theorem, that U_{i} and U_{j} are conjugate (and hence isomorphic) whenever i and j have the same parity.

Lemma 5.3. For all a_{i} ∈ U_{i}^{∗}, we have :
(i) μ(a_{i}^{−1}) = μ(a_{i})^{−1} ;

(ii) λ(a_{i}^{−1}) = κ(a_{i})^{−1} ; (iii) μ(a_{i}^{g}) = μ(a_{i})^{g} for every element g ∈ Aut(Γ) mapping Σ to itself .

Proof. See [20, (6.2)]. □

The following “Shift Lemma” is essential.

Theorem 5.4. Suppose, for some i, that [a_{i}, a_{i+3}^{−1}] = a_{i+1}a_{i+2}, with a_{k} ∈ U_{k} for each k, and with a_{i} and a_{i+3} non-trivial. Then we have:

(i) a_{i} = a_{i+2}^{μ(a_{i+3})} and a_{i+1} = a_{i+3}^{μ(a_{i})} ; (ii) [κ(a_{i+3}), a_{i+2}^{−1}] = a_{i}a_{i+1} ; (iii) [a_{i+1}, λ(a_{i})^{−1}] = a_{i+2}a_{i+3} .

Proof. See [20, (21.19)]. □

The following theorem already puts strong restrictions on the root groups.

Theorem 5.5. By relabeling the vertices of Σ by the transformation i ↦ 5 − i if necessary, we can assume the following :

(i) Y_{i} ≠ 1, [U_{i}, U_{i}] ≤ V_{i} ≤ Y_{i} ≤ Z(U_{i}), for all odd i; (ii) U_{i} is abelian, for all even i.

Proof. See [20, (21.28)]. □

From now on, we will assume that we have chosen the labeling of our apartment Σ in such a way that the statements of Theorem 5.5 hold. We will also use the following results from [20].

Theorem 5.6. (see [20, (21.29)]) If a_{1} ∈ Y_{1}^{∗}, then κ(a_{1}) and λ(a_{1}) both lie in Y_{5}^{∗}.

Theorem 5.7. (see [20, (21.33)]) Let h = μ(a_{1})^{2}, for some a_{1} ∈ Y_{1}^{∗}. Then :

(i) a_{3}^{h} = a_{3}, for all a_{3} ∈ U_{3} ; (ii) a_{4}^{h} = a_{4}^{−1}, for all a_{4} ∈ U_{4} .

Theorem 5.8. (see [20, (21.34)]) κ(a_{4}) = λ(a_{4}), for all a_{4} ∈ U_{4}^{∗}.

Theorem 5.9. (see [20, (21.36)]) Let a_{1} ∈ U_{1}^{∗}, a_{2} ∈ U_{2}, a_{3} ∈ U_{3} and a_{4} ∈ U_{4}^{∗}. Then :

(i) a_{2}^{μ(a_{4})}a_{2}^{−1} = [[λ(a_{4}), a_{2}^{−1}], a_{4}]_{2} ; (ii) [[λ(a_{4}), a_{2}^{−1}], a_{4}]_{3} = [a_{2}, a_{4}]^{−1} ; (iii) [a_{1}, [a_{3}, κ(a_{1})]^{−1}]_{2} = [a_{1}, a_{3}^{−1}]^{−1} .

Theorem 5.10. (see [20, (21.37)]) [μ(a_{4})^{2}, Y_{1}U_{2}Y_{3}U_{4}] = 1, for all a_{4} ∈ U_{4}^{∗}.

Proof. For all the proofs of these theorems, see [20], except for Theorem 5.9(iii), for which the proof is completely similar to the proof of Theorem 5.9(ii). □

We can now start to build up our quadrangular systems. We start the construction by choosing an arbitrary parametrization of the group U_{1} by some group (W, ⊞) ≅ U_{1}, and an arbitrary parametrization of the group U_{4} by some group (V, +) ≅ U_{4}. We will denote the isomorphisms from W to U_{1} and from V to U_{4} by x_{1} and x_{4}, respectively. Choose some fixed elements e_{1} = x_{1}(δ) ∈ Y_{1}^{∗} (note that Y_{1}^{∗} is non-empty because of Theorem 5.5(i)) and e_{4} = x_{4}(ϵ) ∈ U_{4}^{∗}, where we choose e_{4} in Y_{4}^{∗} if Y_{4} ≠ 1. Since U_{3} is isomorphic to U_{1}, we can also have it parametrized by the same group (W, ⊞) by some isomorphism x_{3}, which we define by setting

x_{3}(w) := [x_{1}(w), e_{4}^{−1}]_{3} ,

for all w ∈ W. Similarly, we let U_{2} be parametrized by (V, +), via the isomorphism x_{2} defined by

x_{2}(v) := [e_{1}, x_{4}(v)^{−1}]_{2} ,

for all v ∈ V. To parametrize U_{0} and U_{5}, we choose the following isomorphisms x_{0} and x_{5} from V to U_{0} and from W to U_{5}, respectively :

x_{0}(v) := x_{4}(v)^{μ(e_{4})} ,
x_{5}(w) := x_{1}(w)^{μ(e_{1})} ,

for all w ∈ W and all v ∈ V. We will now define a map F from V × V to W and a map H from W × W on V, by setting

[x_{1}(w_{1}), x_{3}(w_{2})^{−1}] = x_{2}(H(w_{1}, w_{2})) ,
[x_{2}(v_{1}), x_{4}(v_{2})^{−1}] = x_{3}(F (v_{1}, v_{2})) ,

for all w_{1}, w_{2} ∈ W and all v_{1}, v_{2} ∈ V. Furthermore, we define a map τ_{V} from V ×W to V and a map τ_{W} from W × V to W, both of which will be denoted by · or by juxtaposition, by setting

[x_{1}(w), x_{4}(v)^{−1}]_{2} = x_{2}(τ_{V} (v, w)) = x_{2}(vw) ,
[x_{1}(w), x_{4}(v)^{−1}]_{3} = x_{3}(τ_{W} (w, v)) = x_{3}(wv) ,

for all w ∈ W and all v ∈ V. Finally, for each w ∈ W^{∗}, we define two elements κ(w), λ(w) ∈ W^{∗} by setting

κ(x_{1}(w)) = x_{5}(κ(w)) ,
λ(x_{1}(w)) = x_{5}(λ(w)) ,

and for each v ∈ V^{∗}, we define an element v^{−1} ∈ V^{∗}, by setting

κ(x_{4}(v)) = x_{0}(v^{−1}) .

Note that, by Theorem 5.8, λ(x_{4}(v)) = x_{0}(v^{−1}) as well.

If we can now prove that these data satisfy all of the axioms (Q_{1}) – (Q_{20}), then we have proved that every Moufang quadrangle can actually be obtained from the construction in the previous chapters, since we have started from an arbitrary Moufang quadrangle. At the same time, however, we will show that the identities (Q_{21}) – (Q_{26}) hold; see Theorem 6.1.

Remark 5.11. It is interesting to observe that the choice of δ and ϵ is arbitrary (up to some restrictions about the radical). This gives us some freedom for the choice of the base points for the parametrizing structure of an arbitrary Moufang quadrangle. See also Remark 6.4.

By Theorem 5.5(ii), the group U_{4} is abelian. Since U_{4} is parametrized by (V, +), we have that V is abelian as well.

By the definition of the isomorphism x_{3} and the definition of the map from V × W to V, we have, for all w ∈ W, that x_{3}(w) = [x_{1}(w), e_{4}^{−1}]_{3} = [x_{1}(w), x_{4}(ϵ)^{−1}]_{3} = x_{3}(wϵ), from which it follows that w = wϵ, which proves (Q_{1}). Similarly, we can prove that (Q_{2}) holds.

We now take a look at the subgroups V_{3} and Y_{3} of U_{3}. By definition, we have V_{3} = [U_{2}, U_{4}] = [x_{2}(V), x_{4}(V)^{−1}] = x_{3}(F (V, V)) = x_{3}(Im(F)). The elements of U_{3} which commute with every element of U_{1}, are exactly those elements x_{3}(w) ∈ U_{3} such that [x_{1}(w′), x_{3}(w)] = 1, for all w′ ∈ W, this is, such that x_{2}(H(w′, w)) = 1 or equivalently H(w′, w) = 0, for all w′ ∈ W. This means that Y_{3} = C_{U_{3}}(U_{1}) = x_{3}(Rad(H)). It now follows from Theorem 5.5(i) that [W, W] ≤ Im(F) ≤ Rad(H) ≤ Z(W). In particular, we have proved (Q_{7}). We have also proved that [Im(F), W] = 1.

Completely similarly as in the previous paragraph, it follows from the definitions that Y_{1} = C_{U_{1}}(U_{3}) = x_{1}(Rad(H)) and that Y_{4} = C_{U_{4}}(U_{2}) = x_{4}(Rad(F)). It thus follows from e_{1} = x_{1}(δ) ∈ Y_{1}^{∗} that δ ∈ Rad(H)^{∗}, and it follows from the fact that e_{4} = x_{4}(ϵ) was chosen to lie in Y_{4}^{∗} if Y_{4} ≠ 1 that ϵ ∈ Rad(F)^{∗} if Rad(F) ≠ 0. Hence we have shown (Q_{9}) and (Q_{10}).

Using the identity [ab, c^{−1}] = [a, c^{−1}]^{b}[b, c^{−1}] and the fact that U_{1} and U_{2} commute (because of Theorem 5.1(i)), we can deduce that

x_{2}(H(w_{1} ⊞ w_{2}, w′)) = [x_{1}(w_{1} ⊞ w_{2}), x_{3}(w′)^{−1}]
= [x_{1}(w_{1})x_{1}(w_{2}), x_{3}(w′)^{−1}]
= [x_{1}(w_{1}), x_{3}(w′)^{−1}]^{x_{1}(w_{2})}[x_{1}(w_{2}), x_{3}(w′)^{−1}]
= x_{2}(H(w_{1}, w′))^{x_{1}(w_{2})}x_{2}(H(w_{2}, w′))
= x_{2}(H(w_{1}, w′))x_{2}(H(w_{2}, w′))
= x_{2}(H(w_{1}, w′) + H(w_{2}, w′)) ,

for all w_{1}, w_{2}, w′ ∈ W, so H is additive in the first variable. Similarly, it follows from the identity [a, (bc)^{−1}] = [a, b^{−1}][a, c^{−1}]^{b^{−1}} that H is additive in the second variable. In the same way, we can deduce from those two identities that F is additive in both variables. Since we will use this fact very often from now on, we will not mention it explicitly anymore.

Using the same identity [ab, c^{−1}] = [a, c^{−1}]^{b}[b, c^{−1}] and the fact that [U_{2}, U_{2}] = 1 (since V is abelian) and [U_{2}, U_{3}] = 1 (by Theorem 5.1(i)), we deduce that

x_{2}(v(w_{1} ⊞ w_{2}))x_{3}((w_{1} ⊞ w_{2})v) = [x_{1}(w_{1} ⊞ w_{2}), x_{4}(v)^{−1}]
= [x_{1}(w_{1})x_{1}(w_{2}), x_{4}(v)^{−1}]
= [x_{1}(w_{1}), x_{4}(v)^{−1}]^{x_{1}(w_{2})}[x_{1}(w_{2}), x_{4}(v)^{−1}]
= (x_{2}(vw_{1})x_{3}(w_{1}v))^{x_{1}(w_{2})}x_{2}(vw_{2})x_{3}(w_{2}v)

= x_{2}(vw_{1})x_{2}(H(w_{2}, w_{1}v))x_{3}(w_{1}v)x_{2}(vw_{2})x_{3}(w_{2}v)

= x_{2}(vw_{1} + vw_{2} + H(w_{2}, w_{1}v))x_{3}(w_{1}v ⊞ w_{2}v) ,

for all w_{1}, w_{2} ∈ W and all v_{1}, v_{2} ∈ V. By Theorem 5.1(ii), this implies that

x_{2}(v(w_{1} ⊞ w_{2})) = x_{2}(vw_{1} + vw_{2} + H(w_{2}, w_{1}v)) and
x_{3}((w_{1} ⊞ w_{2})v) = x_{3}(w_{1}v ⊞ w_{2}v) ,

for all w_{1}, w_{2} ∈ W and all v_{1}, v_{2} ∈ V, from which it follows that (Q_{12}) and (Q_{3}) hold.

Similarly, it follows from the identity [a, (bc)^{−1}] = [a, b^{−1}][a, c^{−1}]^{b^{−1}}, the fact that [Im(F), W] = 1 and the fact that [U_{2}, U_{3}] = 1 (because of Theorem 5.1(i)), that (Q_{11}) and (Q_{4}) hold.

Now, we will define a map v ↦ \overline{v} from V to V, by setting

x_{2}(v)^{μ(e_{4})} = x_{2}(−\overline{v}) ,

for all v ∈ V; we will prove later on (see page 31) that \overline{v} = ϵF (ϵ, v) − v. Note that, by Theorem 5.10, we have x_{2}(v)^{μ(e_{4})^{2}} = x_{2}(v), and hence −\overline{(−\overline{v})} = v, for all v ∈ V. If we invert the identity x_{2}(v)^{μ(e_{4})} = x_{2}(−\overline{v}), then we get x_{2}(−v)^{μ(e_{4})} = x_{2}(\overline{v}); it follows that \overline{−v} = −\overline{v}, for all v ∈ V. Combining these two relations, we also get \overline{\overline{v}} = v, for all v ∈ V.

Theorem 5.12. For all w ∈ W and all v ∈ V, we have:

(i) x_{0}(v)^{μ(e_{4})} = x_{4}(v) ; (ii) x_{1}(w)^{μ(e_{4})} = x_{3}(w(−ϵ)) ; (iii) x_{2}(v)^{μ(e_{4})} = x_{2}(−\overline{v}) ; (iv) x_{3}(w)^{μ(e_{4})} = x_{1}(w) ;

(v) x_{4}(v)^{μ(e_{4})} = x_{0}(v) ;

(vi) x_{1}(w)^{μ(e_{1})} = x_{5}(w) ;
(vii) x_{2}(v)^{μ(e_{1})} = x_{4}(−v) ;
(viii) x_{3}(w)^{μ(e_{1})} = x_{3}(w) ;
(ix) x_{4}(v)^{μ(e_{1})} = x_{2}(v) ;
(x) x_{5}(w)^{μ(e_{1})} = x_{1}(w(−ϵ)) .

Proof. First of all, (iii), (v) and (vi) hold by definition. By Theorem 5.10, x_{4}(v)^{μ(e_{4})^{2}} = x_{4}(v). So if we conjugate (v) by μ(e_{4}), we get (i). If we apply Theorem 5.4(i) on the identity

[x_{1}(w), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv) ,

we get that

x_{1}(w) = x_{3}(wv)^{μ(x_{4}(v))} and
x_{2}(vw) = x_{4}(v)^{μ(x_{1}(w))} ,

for all w ∈ W^{∗} and all v ∈ V^{∗}. If we choose v = ϵ in the first equality, we get, by (Q_{1}), that x_{1}(w) = x_{3}(w)^{μ(e_{4})}, which proves (iv). If we choose v = −ϵ in this same equality, we get

x_{1}(w) = x_{3}(w(−ϵ))^{μ(e_{4}^{−1})}
= x_{3}(w(−ϵ))^{μ(e_{4})^{−1}} ,

by Lemma 5.3(i); conjugating by μ(e_{4}) yields (ii).

If we choose w = δ in the second equality, then it follows from (Q_{2}) that x_{2}(v) = x_{4}(v)^{μ(e_{1})}, which proves (ix). By Theorem 5.7(ii), we have that x_{4}(v)^{μ(e_{1})^{2}} = x_{4}(−v). So if we conjugate (ix) by μ(e_{1}), we get (vii).

By Theorem 5.6, we know that μ(e_{1}) ∈ Y_{5}Y_{1}Y_{5}. Since Y_{1} = C_{U_{1}}(U_{3}) and Y_{5} = C_{U_{5}}(U_{3}), it follows that [μ(e_{1}), U_{3}] = 1, which implies (viii).

If we conjugate the identity

[x_{1}(w), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv)

by μ(e_{1})^{2}, we get, using (vi), (vii), (viii) and (ix), that

[x_{5}(w)^{μ(e_{1})}, x_{4}(−v)^{−1}] = x_{2}(−vw)x_{3}(wv) ,

for all w ∈ W and all v ∈ V. If we choose v = −ϵ, then this yields

[x_{5}(w)^{μ(e_{1})}, e_{4}^{−1}] = x_{2}(ϵw)x_{3}(w(−ϵ)) .

It now follows from Theorem 5.4(i) and (iv) that

x_{5}(w)^{μ(e_{1})} = x_{3}(w(−ϵ))^{μ(e_{4})} = x_{1}(w(−ϵ)) ,

for all w ∈ W, which proves (x). □

So far, we have proved (Q_{1}), (Q_{2}), (Q_{3}), (Q_{4}), (Q_{7}), (Q_{9}), (Q_{10}), (Q_{11}) and (Q_{12}). We now continue to prove the other axioms.

If we conjugate the identity

[x_{1}(w_{1}), x_{3}(w_{2})^{−1}] = x_{2}(H(w_{1}, w_{2}))

by μ(e_{4}), we get, by Theorem 5.12, that

[x_{3}(w_{1}(−ϵ)), x_{1}(w_{2})^{−1}] = x_{2}(−\overline{H(w_{1}, w_{2})}) ,

for all w_{1}, w_{2} ∈ W. Using the fact that [b, a] = [a, b]^{−1}, it follows that

[x_{1}(⊟w_{2}), x_{3}(⊟w_{1}(−ϵ))^{−1}] = x_{2}(\overline{H(w_{1}, w_{2})}) ,

hence

x_{2}(H(⊟w_{2}, ⊟w_{1}(−ϵ))) = x_{2}(\overline{H(w_{1}, w_{2})}) ,

for all w_{1}, w_{2} ∈ W. Using the fact that H is additive, it follows from this last equality that H(w_{2}, w_{1}(−ϵ)) = \overline{H(w_{1}, w_{2}))}, for all w_{1}, w_{2} ∈ W. Note that it follows from (Q_{12}) that H(w_{2}, w_{1}(−ϵ)) = −H(w_{2}, w_{1}), for all w_{1}, w_{2} ∈ W, so we have that −H(w_{2}, w_{1}) = \overline{H(w_{1}, w_{2}))}, which proves (Q_{22}).

Completely similarly, we can conjugate the identity

[x_{2}(v_{1}), x_{4}(v_{2})^{−1}] = x_{3}(F (v_{1}, v_{2}))

by μ(e_{1}), and, again by Theorem 5.12, we find after a short calculation that F (v_{1}, v_{2}) = F (v_{2}, v_{1}), for all v_{1}, v_{2} ∈ V, which proves (Q_{21}).

If we conjugate the identity

[x_{1}(w), x_{4}(−v)^{−1}] = x_{2}(−vw)x_{3}(w(−v))

by μ(e_{1})^{2}, then we get, by Theorem 5.12, that

[x_{1}(w(−ϵ)), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(w(−v)) ,

for all w ∈ W and all v ∈ V. But on the other hand, we have that

[x_{1}(w(−ϵ)), x_{4}(v)^{−1}] = x_{2}(v(w(−ϵ)))x_{3}(w(−ϵ)v) ,

for all w ∈ W and all v ∈ V. By Theorem 5.1(ii), this implies that vw = v(w(−ϵ)) and w(−v) = w(−ϵ)v, for all w ∈ W and all v ∈ V. Thus we have proved (Q_{6}) and (Q_{5}).

If we conjugate the same identity

[x_{1}(w), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv)

by μ(e_{4})^{2}, we get, again by Theorem 5.12, that

[x_{1}(w(−ϵ)), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv(−ϵ)) ,

from which it follows immediately (by Theorem 5.1(ii)) that w(−ϵ)v = wv(−ϵ), for all w ∈ W and all v ∈ V. This means that w(−v) = wv(−ϵ) as well.

We will now prove (Q_{8}). We will make use of the identity [a, b] = a^{−1}a^{b} and of the identity [abc, d] = [a, d]^{bc}[b, d]^{c}[c, d].

x_{3}(F (H(w_{2}, w_{1}), v)) = [x_{2}(H(w_{2}, w_{1})), x_{4}(v)^{−1}]
= [[x_{1}(w_{2}), x_{3}(w_{1})^{−1}], x_{4}(v)^{−1}]
= [x_{1}(w_{2}), x_{3}(w_{1})^{−1}]^{−1}[x_{1}(w_{2}), x_{3}(w_{1})^{−1}]^{x_{4}(v)^{−1}}

= [x_{1}(w_{2}), x_{3}(w_{1})^{−1}]^{−1}[x_{1}(w_{2})x_{2}(vw_{2})x_{3}(w_{2}v), x_{3}(w_{1})^{−1}] .

If a_{1} ∈ U_{1}, a_{2} ∈ U_{2} and a_{3}, b_{3} ∈ U_{3}, then [a_{2}, b_{3}] ∈ [U_{2}, U_{3}] = 1 and [a_{1}, b_{3}] ∈ [U_{1}, U_{3}] ≤ U_{2} (by Theorem 5.1(i)), and since [U_{2}, U_{2}U_{3}] = 1, we have that [a_{1}, b_{3}]^{a_{2}a_{3}} = [a_{1}, b_{3}]. Therefore [a_{1}a_{2}a_{3}, b_{3}] = [a_{1}, b_{3}][a_{3}b_{3}]. Hence

x_{3}(F (H(w_{2}, w_{1}), v)) = [x_{1}(w_{2}), x_{3}(w_{1})^{−1}]^{−1}[x_{1}(w_{2}), x_{3}(w_{1})^{−1}][x_{3}(w_{2}v), x_{3}(w_{1})^{−1}]
= [x_{3}(w_{2}v), x_{3}(w_{1})^{−1}]
= x_{3}(⊟w_{2}v ⊞ w_{1} ⊞ w_{2}v ⊟ w_{1}) ,

and since Im(F) ≤ Z(W), we have that

x_{3}(F (H(w_{2}, w_{1}), v)) = x_{3}(⊟w_{1} ⊟ w_{2}v ⊞ w_{1} ⊞ w_{2}v)
= x_{3}([w_{1}, w_{2}v]_{⊞})

as well, for all w_{1}, w_{2} ∈ W and all v ∈ V, which proves (Q_{8}).

We will now apply the Shift Lemma 5.4(ii) on the identity

[x_{1}(w), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv) .

This gives us the identity

[κ(x_{4}(v)), x_{3}(wv)^{−1}] = x_{1}(w)x_{2}(vw) .

Note that, by definition, we have κ(x_{4}(v)) = x_{0}(v^{−1}). If we conjugate this identity by μ(e_{4}), we thus get, by Theorem 5.12, that

[x_{4}(v^{−1}), x_{1}(wv)^{−1}] = x_{3}(w(−ϵ))x_{2}(−\overline{vw}) .

Inverting this identity and replacing w by ⊟w yields

[x_{1}(wv), x_{4}(−(v^{−1}))^{−1}] = x_{2}(\overline{v(⊟w)})x_{3}(w(−ϵ)) ,

for all w ∈ W and all v ∈ V^{∗}. But on the other hand, we have

[x_{1}(wv), x_{4}(−(v^{−1}))^{−1}] = x_{2}(−(v^{−1})(wv))x_{3}(wv(−(v^{−1}))) ,

for all w ∈ W and all v ∈ V^{∗}. By Theorem 5.1(ii), this implies that

\overline{v(⊟w)} = −(v^{−1})(wv) ,
w(−ϵ) = wv(−(v^{−1})) ,

for all w ∈ W and all v ∈ V^{∗}. If we apply the identity w(−v) = wv(−ϵ) on the second equality, we can conclude that this is equivalent to

v^{−1}(wv) = −\overline{v(⊟w)} ,
wvv^{−1} = w ,

for all w ∈ W and all v ∈ V^{∗}. So we have proved (Q_{16}) and (Q_{15}).

If we replace v by v^{−1} and w by wv in (Q_{16}), then we get

(v^{−1})^{−1}(wvv^{−1}) = −\overline{v^{−1}(⊟wv)} ,

for all w ∈ W and all v ∈ V^{∗}. Using (Q_{15}) and (Q_{16}) once again, and using the fact that −\overline{(−\overline{v})} = v for all v ∈ V, we get

(v^{−1})^{−1}w = vw ,

for all w ∈ W and all v ∈ V^{∗}. If we choose w = δ, it follows that (v^{−1})^{−1} = v, for all v ∈ V^{∗}, which proves (Q_{13}).

If we take a_{i} = x_{4}(v) in Lemma 5.3(ii), then we get that λ(x_{4}(−v)) = κ(x_{4}(v))^{−1}, for all v ∈ V^{∗}. By the definition of v^{−1}, this is equivalent to x_{0}((−v)^{−1}) = x_{0}(−(v^{−1})), from which it follows that (−v)^{−1} = −(v^{−1}), for all v ∈ V^{∗}.

Similarly, if we choose a_{i} = x_{1}(w) in Lemma 5.3(ii), then we get that λ(x_{1}(⊟w)) = κ(x_{1}(w))^{−1}, for all w ∈ W^{∗}. By the definition of κ and λ, this is equivalent to x_{5}(λ(⊟w)) = x_{5}(⊟κ(w)), from which it follows that λ(⊟w) = ⊟κ(w), for all w ∈ W^{∗}.

If we apply the Shift Lemma 5.4(iii) on the identity

[x_{1}(w), x_{4}(v)^{−1}] = x_{2}(vw)x_{3}(wv) ,

then we get that

[x_{2}(vw), λ(x_{1}(w))^{−1}] = x_{3}(wv)x_{4}(v) ,

for all w ∈ W^{∗} and all v ∈ V. By definition, we have λ(x_{1}(w)) = x_{5}(λ(w)). If we conjugate this identity by μ(e_{1})^{−1}, we thus get, by Theorem 5.12, that

[x_{4}(vw), x_{1}(λ(w))^{−1}] = x_{3}(wv)x_{2}(−v) ,

for all w ∈ W^{∗} and all v ∈ V. We can rewrite this identity as

[x_{1}(⊟λ(w)), x_{4}(−vw)^{−1}] = x_{2}(v)x_{3}(⊟wv) ,

for all w ∈ W^{∗} and all v ∈ V. On the other hand, we also have that

[x_{1}(⊟λ(w)), x_{4}(−vw)^{−1}] = x_{2}(−vw(⊟λ(w)))x_{3}(⊟λ(w)(−vw)) ,

for all w ∈ W^{∗} and all v ∈ V. It follows from Theorem 5.1(ii) that

v = −vw(⊟λ(w)) ,
wv = λ(w)(−vw) ,

for all w ∈ W^{∗} and all v ∈ V. If we replace w by ⊟λ(w) and v by vw in the second equality, then we get

⊟λ(w)(vw) = λ(⊟λ(w))(−vw(⊟λ(w))) ,

for all w ∈ W^{∗} and all v ∈ V. If we use these same equalities once again, then we can simplify this to

⊟w(−v) = λ(⊟λ(w))v ,

for all w ∈ W^{∗} and all v ∈ V. If we choose v = ϵ, then we get λ(⊟λ(w)) = ⊟w(−ϵ), for all w ∈ W^{∗}. Since λ(⊟w) = ⊟κ(w), for all w ∈ W^{∗}, this is the same as κ(λ(w)) = w(−ϵ), for all w ∈ W^{∗}, so we have proved (Q_{14}). If we replace w by ⊟w, then we get λ(κ(w)) = w(−ϵ) as well. Now, we substitute κ(w) for w in the equations

v = −vw(⊟λ(w)) ,
wv = λ(w)(−vw) ;

this gives us, using the fact that λ(κ(w)) = w(−ϵ), that

v = −vκ(w)(⊟w(−ϵ)) ,
κ(w)v = w(−ϵ)(−vκ(w)) ,

for all w ∈ W^{∗} and all v ∈ V. It suffices to use (Q_{6}) and (Q_{5}) to see that those two equations are equivalent to (Q_{18}) and (Q_{19}), respectively.

If we put a_{2} = x_{2}(v_{2}) and a_{4} = x_{4}(v_{1}) in Theorem 5.9(ii), then we get

[[λ(x_{4}(v_{1})), x_{2}(v_{2})^{−1}], x_{4}(v_{1})]_{3} = [x_{2}(v_{2}), x_{4}(v_{1})]^{−1} ,

for all v_{1} ∈ V^{∗} and all v_{2} ∈ V. First of all, we have that

[λ(x_{4}(v_{1})), x_{2}(v_{2})^{−1}] = [x_{0}(v_{1}^{−1}), x_{2}(v_{2})^{−1}]
= [x_{4}(v_{1}^{−1}), x_{2}(−\overline{v_{2}})^{−1}]^{μ(e_{4})}
= ([x_{2}(\overline{v_{2}}), x_{4}(−v_{1}^{−1})^{−1}]^{−1})^{μ(e_{4})}
= (x_{3}(F (\overline{v_{2}}, −v_{1}^{−1}))^{−1})^{μ(e_{4})}
= x_{3}(F (\overline{v_{2}}, v_{1}^{−1}))^{μ(e_{4})}
= x_{1}(F (\overline{v_{2}}, v_{1}^{−1})) ,

for all v_{1} ∈ V^{∗} and all v_{2} ∈ V. So it follows from this identity that

[x_{1}(F (\overline{v_{2}}, v_{1}^{−1})), x_{4}(−v_{1})^{−1}]_{3} = [x_{2}(v_{2}), x_{4}(−v_{1})^{−1}]^{−1} ,

for all v_{1} ∈ V^{∗} and all v_{2} ∈ V, from which it follows that

F (\overline{v_{2}}, v_{1}^{−1})(−v_{1}) = F (v_{2}, v_{1}) ,

for all v_{1} ∈ V^{∗} and all v_{2} ∈ V. If we now replace v_{1} by −v_{1}, then we get, using the fact that (−v_{1})^{−1} = −(v_{1}^{−1}) and (Q_{21}), that (Q_{17}) holds.

If we choose v_{1} = ϵ in (Q_{17}), then we get that F (ϵ^{−1}, \overline{v}) = F (ϵ, v), for all v ∈ V. If we put a_{2} = x_{2}(v) and a_{4} = e_{4} in Theorem 5.9(i), then we get

x_{2}(v)^{μ(e_{4})}x_{2}(v)^{−1} = [[λ(e_{4}), x_{2}(v)^{−1}], e_{4}]_{2} ,

for all v ∈ V. We have that

[λ(e_{4}), x_{2}(v)^{−1}] = x_{1}(F (\overline{v}, ϵ^{−1}))
= x_{1}(F (ϵ, v)) ,

for all v ∈ V. Thus we have

x_{2}(v)^{μ(e_{4})}x_{2}(v)^{−1} = [[λ(e_{4}), x_{2}(v)^{−1}], e_{4}]_{2}
= [x_{1}(F (ϵ, v)), x_{4}(−ϵ)^{−1}]_{2}
= x_{2}(−ϵF (ϵ, v)) ,

for all v ∈ V. Since x_{2}(v)^{μ(e_{4})}x_{2}(v)^{−1} = x_{2}(−\overline{v} − v), we conclude that

\overline{v} = ϵF (ϵ, v) − v ,

for all v ∈ V; see page 27. If we put a_{1} = x_{1}(w_{1}) and a_{3} = x_{3}(w_{2}) in Theorem 5.9(iii), then we get

[x_{1}(w_{1}), [x_{3}(w_{2}), κ(x_{1}(w_{1}))]^{−1}]_{2} = [x_{1}(w_{1}), x_{3}(w_{2})^{−1}]^{−1} ,

for all w_{1} ∈ W and all w_{2} ∈ W^{∗}. First of all, we have that

[x_{3}(w_{2}), κ(x_{1}(w_{1}))] = [x_{3}(w_{2}), x_{5}(κ(w_{1}))]
= [x_{3}(w_{2}), x_{1}(κ(w_{1}))]^{μ(e_{1})}
= ([x_{1}(κ(w_{1})), x_{3}(⊟w_{2})^{−1}]^{−1})^{μ(e_{1})}
= (x_{2}(−H(κ(w_{1}), w_{2}))^{−1})^{μ(e_{1})}
= x_{2}(H(κ(w_{1}), w_{2}))^{μ(e_{1})}
= x_{4}(−H(κ(w_{1}), w_{2})) ,

for all w_{1} ∈ W and all w_{2} ∈ W^{∗}. So it follows from this identity that

[x_{1}(w_{1}), x_{4}(−H(κ(w_{1}), w_{2}))^{−1}]_{2} = [x_{1}(w_{1}), x_{3}(w_{2})^{−1}]^{−1} ,

for all w_{1} ∈ W and all w_{2} ∈ W^{∗}, from which it follows that

−H(κ(w_{1}), w_{2})w_{1} = −H(w_{1}, w_{2}) ,

for all w_{1} ∈ W and all w_{2} ∈ W^{∗}. So we have proved (Q_{20}).

Since we have shown all of the identities (Q_{1}) – (Q_{20}), we can conclude that every Moufang quadrangle can be obtained from a quadrangular system.

In particular, we are now allowed to use the results of chapter 4 as well. We thus continue to show that the identities (Q_{23}) – (Q_{26}) hold.

In order to show (Q_{23}), we will calculate the expression

x_{2}(v)^{[μ(x_{1}(δ))μ(x_{1}(z))]^{μ(x_{3}(w))μ(x_{3}(δ))}}

with v ∈ V and w, z ∈ W^{∗} in two different ways. We have shown in Theorem 4.3(iii) that

x_{3}(z)^{μ(x_{1}(w))} = x_{3}(Π_{w}(z))

for all w, z ∈ W^{∗}. If we let μ(e_{4}) act on both sides of this equality, then it follows by Lemma 5.3(iii) and Theorem 5.12 that

x_{1}(z)^{μ(x_{3}(w(−ϵ)))} = x_{1}(Π_{w}(z))

and it thus follows by substituting w(−ϵ) for w and by Lemma 5.3(iii) that

μ(x_{1}(z))^{μ(x_{3}(w))} = μ(x_{1}(Π_{w(−ϵ)}(z)))

for all w, z ∈ W^{∗}. By Lemma 3.15(ii), Π_{w(−ϵ)}(z) = Π_{w}(z) for all w, z ∈ W^{∗}. Since δ ∈ Rad(H) by (Q_{9}), it now follows that

μ(x_{1}(z))^{μ(x_{3}(w))μ(x_{3}(δ))} = μ(x_{1}(Π_{w}(z)))

and hence, since Π_{w}(δ) = δ,

[μ(x_{1}(δ))μ(x_{1}(z))]^{μ(x_{3}(w))μ(x_{3}(δ))} = μ(x_{1}(δ))μ(x_{1}(Π_{w}(z)))

for all w, z ∈ W^{∗}. Note that vκ(δ) = v for all v ∈ V. Since we have shown in Theorem 4.3 that

x_{2}(v)^{μ(x_{1}(w))} = x_{4}(−vκ(w)) and
x_{4}(v)^{μ(x_{1}(w))} = x_{2}(vw)

for all v ∈ V and all w ∈ W^{∗}, it thus follows, by Lemma 5.3(i), that

x_{2}(v)^{[μ(x_{1}(δ))μ(x_{1}(z))]^{μ(x_{3}(w))μ(x_{3}(δ))}} = x_{2}(v)^{μ(x_{1}(δ))μ(x_{1}(Π_{w}(z)))}
= x_{4}(−v)^{μ(x_{1}(Π_{w}(z)))}
= x_{2}(−v · Π_{w}(z))

for all w, z ∈ W^{∗}.

On the other hand, if we let μ(e_{4}) act on both sides of the identity x_{4}(v)^{μ(x_{1}(w))} = x_{2}(vw), then we can deduce that

x_{0}(v)^{μ(x_{3}(w))} = x_{2}(−\overline{vw}) and
x_{2}(v)^{μ(x_{3}(w))} = x_{0}(\overline{v}κ(w))

for all v ∈ V and all w ∈ W^{∗}. Hence, by Lemma 5.3(i),

x_{2}(v)^{[μ(x_{1}(δ))μ(x_{1}(z))]^{μ(x_{3}(w))μ(x_{3}(δ))}} = x_{2}(v)^{μ(x_{3}(⊟δ))μ(x_{3}(⊟w))μ(x_{1}(δ))μ(x_{1}(z))μ(x_{3}(w))μ(x_{3}(δ))}
= x_{2}(\overline{\overline{v}(⊟w)})^{μ(x_{1}(δ))μ(x_{1}(z))μ(x_{3}(w))μ(x_{3}(δ))}
= x_{2}(−\overline{\overline{v}(⊟w)}z)^{μ(x_{3}(w))μ(x_{3}(δ))}
= x_{2}(\overline{\overline{\overline{\overline{v}(⊟w)}z}κ(w)})

for all w, z ∈ W^{∗}. Hence we have shown that (Q_{23}) holds.

The proof of (Q_{24}) follows in a completely similar way by calculating the expression

x_{3}(w)^{[μ(x_{4}(ϵ))μ(x_{4}(c))]^{μ(x_{2}(v))μ(x_{2}(ϵ))}}

with w ∈ W and v, c ∈ V^{∗} in two different ways.

We will now show (Q_{25}). Let c ∈ V, v ∈ V^{∗} and w ∈ W^{∗} be arbitrary. This time, we will calculate the expression

x_{2}(c)^{[μ(x_{1}(δ))μ(x_{1}(w))]^{μ(x_{4}(−v))}}

in two different ways. By Theorem 4.2(ii),

x_{1}(w)^{μ(x_{4}(v))} = x_{3}(w(−v))

for all v ∈ V^{∗} and all w ∈ W, and hence, by Lemma 5.3(iii),

[μ(x_{1}(δ))μ(x_{1}(w))]^{μ(x_{4}(−v))} = μ(x_{3}(δv))μ(x_{3}(wv)) .

It follows that

x_{2}(c)^{[μ(x_{1}(δ))μ(x_{1}(w))]^{μ(x_{4}(−v))}} = x_{2}(c)^{μ(x_{3}(δv))μ(x_{3}(wv))}
= x_{2}(\overline{c} · κ(δv))^{μ(x_{3}(wv))}
= x_{2}(−\overline{\overline{c} · κ(δv) · wv}) .

On the other hand, we have shown in Theorem 4.2(iii) that

x_{2}(u)^{μ(x_{4}(v))} = x_{2}(π_{v}(u))

for all u ∈ V and all v ∈ V^{∗}, and hence, by Lemma 5.3(i),

x_{2}(c)^{[μ(x_{1}(δ))μ(x_{1}(w))]^{μ(x_{4}(−v))}} = x_{2}(c)^{μ(x_{4}(v))μ(x_{1}(δ))μ(x_{1}(w))μ(x_{4}(−v))}
= x_{2}(π_{v}(c))^{μ(x_{1}(δ))μ(x_{1}(w))μ(x_{4}(−v))}
= x_{2}(−π_{v}(c)w)^{μ(x_{4}(−v))}
= x_{2}(−π_{−v}(π_{v}(c)w)) .

Since π_{−v}(u) = π_{v}(u) for all u ∈ V and all v ∈ V^{∗}, it follows by comparing these two expressions that

\overline{\overline{c} · κ(δv) · wv} = π_{v}(π_{v}(c)w) .

If we substitute \overline{c · δv} for c in the last identity and apply π_{v} on both sides, then we get, by Lemma 3.23(i), that

π_{v}(\overline{c · δv · κ(δv) · wv}) = π_{v}(\overline{c · δv})w .

Since δv ∈ Rad(H) by (Q_{9}) and Lemma 3.19, it follows by Lemma 3.13(ii) and Lemma 3.2(i) that

c · δv · κ(δv) = −c · (⊟δv) · κ(δv) = c ,

which completes the proof of (Q_{25}).

The proof of (Q_{26}) follows in a completely similar way by calculating the expression

x_{3}(w)^{[μ(x_{4}(ϵ))μ(x_{4}(−v))]^{μ(x_{1}(z))}}

with w ∈ W, z ∈ W^{∗} and v ∈ V^{∗} in two different ways.

This concludes the proof of all of the identities (Q_{1}) – (Q_{26}).

CHAPTER 6 Some Remarks

We start by pointing out that we have really shown that the identities (Q_{23}) – (Q_{26}) follow from the axioms (Q_{1}) – (Q_{20}).

Theorem 6.1. Let Ω := (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then the identities (Q_{23}) – (Q_{26}) hold, for all v, c ∈ V and all w, z ∈ W.

Proof. Let Ω := (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. Then it follows from chapter 4 that we can construct a Moufang quadrangle Γ starting from Ω. In chapter 5, it is shown that every Moufang quadrangle can be constructed from a quadrangular system for which additionally the identities (Q_{23}) – (Q_{26}) hold. In particular, Γ can be constructed from a quadrangular system, which can be chosen to coincide with the quadrangular system Ω that we started with, since the choice of the parametrization of the groups U_{1} and U_{4} and of the elements ϵ and δ was arbitrary. (Note that the parametrization of the groups U_{2} and U_{3} and the definition of the maps F and H then automatically coincide by construction.) This shows that Ω is a quadrangular system for which additionally the identities (Q_{23})

– (Q_{26}) hold. Since Ω was arbitrary, these identities hold for every quadrangular system. □

Remark 6.2. One might wonder why we pay so much attention to these last four identities (Q_{23}), (Q_{24}), (Q_{25}) and (Q_{26}). The reason is that these identities turn out to be essential for the classification of the quadrangular systems, but still, we are not aware of a direct proof for the fact that they follow from the other axioms.

Remark 6.3. Although every quadrangular system gives rise to a Moufang quadrangle and every Moufang quadrangle can be constructed from a quadrangular system, it is not true that there is a bijection between the set of classes of isomorphic quadrangular systems and the set of classes of isomorphic Moufang quadrangles. In particular, two non-isomorphic quadrangular systems can give rise to isomorphic Moufang quadrangles. However, two isomorphic quadrangular systems will always give rise to isomorphic Moufang quadrangles.

Remark 6.4. We could as well have defined a quadrangular system without axiom (Q_{10}). The reason that we added this axiom has to do with the classification of the so-called wide quadrangular systems which are the extension of a quadrangular system of quadratic form type. Without axiom (Q_{10}), one would have to define a translate of a quadrangular system of type F_{4} in order to describe all possible quadrangular systems (up to isomorphism), which is not needed now because of this extra axiom. (See section 8.5 for more details.)

On the other hand, if there are no quadrangles of type F_{4} involved in a certain application, then it can often be more convenient to drop this axiom (Q_{10}), since

36 6. SOME REMARKS

it gives more freedom in the choice of the base point ϵ ∈ V^{∗}. See also Remark 5.11 and Remark A.1.

CHAPTER 7 Examples

We will now present a list of six examples of quadrangular systems. These examples correspond to the six different classes of Moufang quadrangles in [20]. The goal of the next chapter is to prove that, up to isomorphism, this list is complete.

In each case, we will describe a parametrization for the groups V and W, that is, we will describe V and W as groups which are isomorphic to certain other groups Ṽ and W̃, respectively; we will denote the isomorphisms from Ṽ to V and from W̃ to W by square brackets: a ∈ Ṽ ↦ [a] ∈ V and b ∈ W̃ ↦ [b] ∈ W.

7.1. Quadrangular Systems of Quadratic Form Type

Consider a non-trivial anisotropic quadratic space (K, V_{0}, q), that is, a commutative field K (of arbitrary characteristic), a non-trivial vector space V_{0} over K, and a quadratic form q from V_{0} to K such that q(v) = 0 if and only if v = 0. Choose an arbitrary element ϵ ∈ V_{0}, and replace q by q(ϵ)^{−1}q. Then q(ϵ) = 1. Denote the corresponding bilinear form of q by f, i.e. f(u, v) := q(u + v) − q(u) − q(v). We define a map v ↦ \overline{v} by setting \overline{v} := f(ϵ, v)ϵ − v for all v ∈ V. (This map is the negative of the reflection about the base point ϵ.) Let V be parametrized by (V_{0}, +), and let W be parametrized by the additive group of K. We define a map τ_{V} from V × W to V and a map τ_{W} from W × V to W as follows:

τ_{V} ([v], [t]) := [v][t] := [tv] ,
τ_{W} ([t], [v]) := [t][v] := [tq(v)] ,

for all v ∈ V_{0} and all t ∈ K. Then (V, W, τ_{V} , τ_{W} , [ϵ], [1]) is a quadrangular system. One can check that

F ([u], [v]) = [f(u, v)] ,
H([s], [t]) = [0] ,

for all u, v ∈ V_{0} and all s, t ∈ K, and that

[v]^{−1} = [q(v)^{−1}\overline{v}] ,
κ([t]) = [t^{−1}] ,

for all v ∈ V_{0}^{∗} and all t ∈ K^{∗}. Note that \overline{[v]} = ϵF (ϵ, [v])−[v] = [ϵ][f(ϵ, v)]−[v] = [\overline{v}] for all v ∈ V.

These are the quadrangular systems of quadratic form type. They will be denoted by Ω_{Q}(K, V_{0}, q).

7.2. Quadrangular Systems of Involutory Type

Following [20], we define an involutory set as a triple (K, K_{0}, σ), where K is a field or a skew-field, σ is an involution of K, and K_{0} is an additive subgroup of K containing 1 such that

K_{σ} ⊆ K_{0} ⊆ Fix_{K}(σ) and
a^{σ}K_{0}a ⊆ K_{0} for all a ∈ K ,

where K_{σ} := {a + a^{σ} | a ∈ K}. Note that if char(K) ≠ 2, then K_{σ} = Fix_{K}(σ), and hence K_{0} = K_{σ} as well, so the second condition is superfluous in this case.

Let V be parametrized by the additive group of K, and let W be parametrized by K_{0}. We define a map τ_{V} from V × W to V and a map τ_{W} from W × V to W as follows:

τ_{V} ([a], [t]) := [a][t] := [ta] ,
τ_{W} ([t], [a]) := [t][a] := [a^{σ}ta] ,

for all a ∈ K and all t ∈ K_{0}. Then (V, W, τ_{V} , τ_{W} , [1], [1]) is a quadrangular system. One can check that

F ([a], [b]) = [a^{σ}b + b^{σ}a] ,
H([s], [t]) = [0] ,

for all a, b ∈ K and all s, t ∈ K_{0}, and that

[a]^{−1} = [a^{−1}] ,
κ([t]) = [t^{−1}] ,

for all a ∈ K^{∗} and all t ∈ K_{0}^{∗}. Note that \overline{[a]} = ϵF (ϵ, [a])−[a] = [1][a+a^{σ}]−[a] = [a^{σ}] for all a ∈ K.

These are the quadrangular systems of involutory type. They will be denoted by Ω_{I}(K, K_{0}, σ).

7.3. Quadrangular Systems of Indifferent Type

Again following [20], we define an indifferent set as a triple (K, K_{0}, L_{0}), where K is a commutative field of characteristic 2, and K_{0} and L_{0} are additive subgroups of K both containing 1, such that

K_{0}^{2}L_{0} ⊆ L_{0} , L_{0}K_{0} ⊆ K_{0} , K_{0} generates K as a ring.

Let V be parametrized by L_{0}, and let W be parametrized by K_{0}. We define a map τ_{V} from V × W to V and a map τ_{W} from W × V to W as follows:

τ_{V} ([a], [t]) := [a][t] := [t^{2}a] ,
τ_{W} ([t], [a]) := [t][a] := [ta] ,

for all a ∈ L_{0} and all t ∈ K_{0}. Then (V, W, τ_{V} , τ_{W} , [1], [1]) is a quadrangular system. One can check that

F ([a], [b]) = [0] ,
H([s], [t]) = [0] ,

for all a, b ∈ L_{0} and all s, t ∈ K_{0}, and that

[a]^{−1} = [a^{−1}] ,
κ([t]) = [t^{−1}] ,

for all a ∈ K^{∗} and all t ∈ K_{0}^{∗}. Note that \overline{[a]} = [a] for all a ∈ K.

These are the quadrangular systems of indifferent type. They will be denoted by Ω_{D}(K, K_{0}, L_{0}).

7.4. Quadrangular Systems of Pseudo-quadratic Form Type

Let K be an arbitrary field or skew-field, let σ be an involution of K (which may be trivial), and let V_{0} be a right vector space over K. A map h from V_{0} × V_{0} to K is called a sesquilinear form (with respect to σ) if and only if h is additive in both variables, and h(at, bs) = t^{σ}h(a, b)s, for all a, b ∈ V_{0} and all t, s ∈ K. A form h : V_{0} × V_{0} → K is called hermitian, respectively skew-hermitian, (with respect to σ) if and only if h is sesquilinear with respect to σ and h(a, b)^{σ} = h(b, a), respectively h(a, b)^{σ} = −h(b, a), for all a, b ∈ V_{0}.

Let (K, K_{0}, σ) be an involutory set, let V_{0} be a right vector space over K and let p be a map from V_{0} to K. Then p is an anisotropic pseudo-quadratic form on V (with respect to K_{0} and σ) if there is a form h on V_{0} which is skew-hermitian with respect to σ such that

p(a + b) ≡ p(a) + p(b) + h(a, b) (mod K_{0}) ,
p(at) ≡ t^{σ}p(a)t (mod K_{0}) ,
p(a) ∈ K_{0} ⟺ a = 0 ,

for all a, b ∈ V_{0} and all t ∈ K.

As in [20], we define an anisotropic pseudo-quadratic space as a quintuple (K, K_{0}, σ, V_{0}, p) such that (K, K_{0}, σ) is an involutory set, V_{0} is a right vector space over K and p is an anisotropic pseudo-quadratic form on V_{0} with respect to K_{0} and σ.

Let (K, K_{0}, σ, V_{0}, p) be an arbitrary anisotropic pseudo-quadratic space with corresponding skew-hermitian form h. We define a group (T, ⊞) as

T := {(a, t) ∈ V_{0} × K | p(a) − t ∈ K_{0}} ,

where the group action is given by

(a, t) ⊞ (b, s) := (a + b, t + s + h(b, a)) ,

for all (a, t), (b, s) ∈ T. One can check that T is indeed a group with neutral element (0, 0), and with the inverse given by ⊟(a, t) = (−a, −t + h(a, a)), for all (a, t) ∈ T.

Let V be parametrized by the additive group of K, and let W be parametrized by T. We define a map τ_{V} from V × W to V and a map τ_{W} from W × V to W as follows:

τ_{V} ([v], [a, t]) := [v][a, t] := [tv] ,
τ_{W} ([a, t], [v]) := [a, t][v] := [av, v^{σ}tv] ,

for all v ∈ K and all (a, t) ∈ T. Then (V, W, τ_{V} , τ_{W} , [1], [0, 1]) is a quadrangular system. One can check that

F ([u], [v]) = [0, u^{σ}v + v^{σ}u] ,
H([a, t], [b, s]) = [h(a, b)] ,

for all u, v ∈ K and all (a, t), (b, s) ∈ T, and that

[v]^{−1} = [v^{−1}] ,
κ([a, t]) = [at^{−σ}, t^{−σ}] ,

for all v ∈ K^{∗} and all (a, t) ∈ T^{∗}. Note that

\overline{[v]} = [1]F ([1], [v]) − [v] = [1][0, v + v^{σ}] − [v] = [v^{σ}]

for all v ∈ K.

These are the quadrangular systems of pseudo-quadratic form type. They will be denoted by Ω_{P} (K, K_{0}, σ, V_{0}, p).

7.5. Quadrangular Systems of Type E_{6}, E_{7} and E_{8}

We first recall the notion of a norm splitting, which has been introduced in [20, (12.9)]. Let K be an arbitrary commutative field, let V_{0} be a 2d-dimensional vector space over K, and let q be a regular quadratic form from V_{0} to K. First of all, observe that, if E/K is a separable quadratic extension with norm N, then N is a 2-dimensional anisotropic regular quadratic form from E (as a vector space over K) to K. We say that (E, {v_{1}, …, v_{d}}) is a norm splitting of q if and only if V_{0} is a vector space over E, where the scalar multiplication from E × V_{0} to V_{0} is an extension of the scalar multiplication from K × V_{0} to V_{0}, such that {v_{1}, …, v_{d}} is a basis of V_{0} over E for which

q(t_{1}v_{1} + ⋯ + t_{d}v_{d}) = s_{1}N(t_{1}) + ⋯ + s_{d}N(t_{d}) ,

for all t_{1}, …, t_{d} ∈ E. Observe that s_{i} = q(v_{i}), for all i ∈ {1, …, d}. The constants s_{1}, s_{2}, …, s_{d} are called the constants of the norm splitting. Note that these constants are all non-zero, since q is regular. It follows that a regular quadratic form which has a norm splitting is anisotropic.

Remark 7.1. This definition is equivalent to the assumption that q has an orthogonal decomposition q ≃ q_{1} ⊥ q_{2} ⊥ ⋯ ⊥ q_{d}, where each q_{i} is a 2-dimensional regular quadratic form with the same non-trivial discriminant.

Now let (K, V_{0}, q) be an arbitrary anisotropic quadratic space with corresponding bilinear map f. An automorphism T of V_{0} is called a norm splitting map of q if and only if there exist constants α, β ∈ K with α = 0 if char(K) ≠ 2 and α ≠ 0 if char(K) = 2, and with β ≠ 0 in all characteristics, such that

q(T (v)) = βq(v) ,
f(v, T (v)) = αq(v) ,
T (T (v)) + αT (v) + βv = 0 ,

for all v ∈ V_{0}. For each norm splitting map T, we can define a corresponding norm splitting map \overline{T}, defined by the relation \overline{T} (v) := αv − T (v) for all v ∈ V_{0}. It is straightforward to check that \overline{T} is a norm splitting map with the same parameters α and β as the original norm splitting map T.

7.5. QUADRANGULAR SYSTEMS OF TYPE E_{6}, E_{7} AND E_{8} 41

Definition 7.2. Let K be an arbitrary commutative field, let V_{0} be a vector space over K, and let q be an anisotropic quadratic form from V_{0} to K. Then

• q is a quadratic form of type E_{6} if and only if dim_{K}V_{0} = 6 and q has a norm splitting q ≃ s_{1}N ⊥ s_{2}N ⊥ s_{3}N.

• q is a quadratic form of type E_{7} if and only if dim_{K}V_{0} = 8 and q has a norm splitting q ≃ s_{1}N ⊥ s_{2}N ⊥ s_{3}N ⊥ s_{4}N such that s_{1}s_{2}s_{3}s_{4} ∉ N(E).

• q is a quadratic form of type E_{8} if and only if dim_{K}V_{0} = 12 and q has a norm splitting q ≃ s_{1}N ⊥ ⋯ ⊥ s_{6}N such that −s_{1}s_{2}s_{3}s_{4}s_{5}s_{6} ∈ N(E).

An anisotropic quadratic space (K, V_{0}, q) is called of type E_{6}, E_{7} or E_{8} if and only if q is a quadratic form of type E_{6}, E_{7} or E_{8}, respectively.

Theorem 7.3. Let (K, V_{0}, q) be a quadratic space of type E_{k} with k ∈ {6, 7, 8}, and suppose that ϵ is an element of V_{0} such that q(ϵ) = 1. Let T be a norm splitting map of q, and let X_{0} be a vector space over K of dimension 2^{k−3}. Then there exists a unique map (a, v) ↦ av from X_{0} × V_{0} to X_{0} and an element ξ ∈ X_{0}^{∗} such that

at = a(tϵ) ,
(av)\overline{v} = aq(v) ,
ξT (v) = (ξT (ϵ))v ,

for all a ∈ X_{0}, t ∈ K and v ∈ V_{0}.

Proof. This follows from [20, (12.56) and (13.11)]. □

From now on, we let T be a fixed arbitrary norm splitting map of q, and we let X_{0} be a fixed vector space over K of dimension 2^{k−3}. We apply Theorem 7.3 with these choices of T and X_{0}. Note that ξ is not uniquely determined; see [20, (13.12)].

Remark 7.4. The first two conditions of Theorem 7.3 say that X_{0} is a C(q, ϵ)- module, where C(q, ϵ) is the Clifford algebra of q with base point ϵ.

Theorem 7.5. We can choose the norm splitting (E, {v_{1}, …, v_{d}}) in such a way that v_{1} = ϵ (and hence s_{1} = 1). Furthermore, if k = 8, then we can choose it in such a way that ξv_{2}v_{3}v_{4}v_{5}v_{6} = ξ as well.

Proof. This follows from [20, (27.20) and (27.13)]. □

So assume that the norm splitting satisfies the conditions of this Theorem. Then we can now define a subspace M_{0} of X_{0} as follows. If k = 6, then we set

M_{0} := {ξtv_{2}v_{3} | t ∈ E} ;

If k = 7, then we set

M_{0} := {ξt_{1}v_{2}v_{3} + ξt_{2}v_{1}v_{3} + ξt_{3}v_{1}v_{2} + ξtv_{1}v_{2}v_{3} | t_{1}, t_{2}, t_{3}, t ∈ E} ;

If k = 8, then we set

M_{0} := \{ ∑_{\begin{matrix}i,j∈\{2,...,6\} \\ i<j \end{matrix}} ξt_{ij}v_{i}v_{j} | t_{ij} ∈ E \} .

Theorem 7.6. X_{0} = ξV_{0} ⊕ M_{0}.

Proof. See [20, (13.14)]. □

Theorem 7.7. There is a unique map h from X_{0} × X_{0} to V_{0} which is bilinear over K, such that

(i) h(ξ, ξv) = T (v) − \overline{T} (v) , for all v ∈ V_{0} ; (ii) h(ξ, a) = 0 , for all a ∈ M_{0} ; (iii) h(a, b) = −\overline{h(b, a)} , for all a, b ∈ X_{0} ; (iv) h(a, bv) = h(b, av) + f(h(a, b), ϵ)v, for all a, b ∈ X_{0} and all v ∈ V_{0} .

Proof. See [20, (13.15)]. □

We now define an element ζ ∈ V_{0} as follows. Note that, if char(K) = 2, then f(ϵ, T (ϵ)) = α ≠ 0 by the definition of T.

ζ := \begin{cases}ϵ/2 & if char(K) ≠ 2 \\ T (ϵ)/f(ϵ, T (ϵ)) & if char(K) = 2 \end{cases} .

Next, let g be the bilinear form from X_{0} × X_{0} to K given by

g(a, b) := f(h(b, a), ζ)

for all a, b ∈ X_{0}. Set

v^{∗} := \begin{cases}0 & if char(K) ≠ 2 \\ f(v, ζ)ϵ + f(v, ϵ)ζ + v & if char(K) = 2 \end{cases} ,

for all v ∈ V_{0}.

Theorem 7.8. There is a unique map θ from X_{0} × V_{0} to V_{0} satisfying the following conditions, for all a, b ∈ X_{0} and all u, v ∈ V_{0}:

(i) θ(ξ, v) = T (v) ;

(ii) θ(a + b, v) = θ(a, v) + θ(b, v) + h(b, av) − g(a, b)v; (iii) θ(av, w) = \overline{θ(a, w̄)}q(v) − \overline{θ(a, v)}f(w, v̄) + f(θ(a, v), w̄)v̄ + f(θ(a, v^{∗}), v)w.

Proof. See [20, (13.30), (13.31) and (13.37)]. □

Let φ be the map from X_{0} × V_{0} to K defined as

φ(a, v) := f(θ(a, v^{∗}), v) ,

for all a ∈ X_{0} and all v ∈ V_{0}.

Finally, we define a group (S, ⊞) as S := X_{0} × K where the group action is given by

(a, t) ⊞ (b, s) := (a + b, t + s + g(a, b)) ,

for all (a, t), (b, s) ∈ S. One can check that S is indeed a group with neutral element (0, 0), and with the inverse given by ⊟(a, t) = (−a, −t + g(a, a)), for all (a, t) ∈ S.

Let V be parametrized by (V_{0}, +), and let W be parametrized by S. We define a map τ_{V} from V × W to V and a map τ_{W} from W × V to W as follows:

τ_{V} ([v], [a, t]) := [v][a, t] := [θ(a, v) + tv] ,
τ_{W} ([a, t], [v]) := [a, t][v] := [av, tq(v) + φ(a, v)] ,

for all v ∈ V and all (a, t) ∈ S. Then (V, W, τ_{V} , τ_{W} , [ϵ], [0, 1]) is a quadrangular system. One can check that

F ([u], [v]) = [0, f(u, v)] ,
H([a, t], [b, s]) = [h(a, b)] ,

7.6. QUADRANGULAR SYSTEMS OF TYPE F_{4} 43

for all u, v ∈ V and all (a, t), (b, s) ∈ S, and that

[v]^{−1} = [q(v)^{−1}\overline{v}] ,
κ([a, t]) = [\frac{aθ(a, ϵ) + ta}{q(θ(a, ϵ) + tϵ)}, \frac{t}{q(θ(a, ϵ) + tϵ)}] ,

for all v ∈ K^{∗} and all (a, t) ∈ S^{∗}.

Remark 7.9. It is not obvious at all to verify that this is a quadrangular system. Quite a lot of identities involving these functions h, g, θ and φ are needed. We will omit these calculations, since our main interest here is to give a classification, and not to prove existence. However, see [20, Chapter 13 and (32.2)] for more details about these identities.

These are the quadrangular systems of type E_{6}, E_{7} and E_{8}. They will be denoted by Ω_{E}(K, V_{0}, q).

7.6. Quadrangular Systems of Type F_{4}

Consider an anisotropic space (K, V_{0}, q) with char(K) = 2 and with non-trivial radical R := Rad(f) = Def(q) = {v ∈ V_{0} | f(v, V_{0}) = 0} ≠ 0. Suppose that there is an element ϵ ∈ R such that q(ϵ) = 1. Then this quadratic space is said to be of type F_{4} if and only if L := q(R) is a subfield of K, and there is a complement S of R in V_{0} such that the restriction of q to the subspace S has a norm splitting (E, {v_{1}, v_{2}}) with constants s_{1}, s_{2} ∈ K^{∗} such that s_{1}s_{2} ∈ L^{∗}.

From now on, we will assume that (K, V_{0}, q) is of type F_{4}. Since t^{2} = q(tϵ) ∈ q(R) = L for all t ∈ K, we have that K^{2} ⊆ L ⊆ K. Denote the restriction of q to S by q_{1}. Denote the norm of the extension E/K by N, and denote the non-trivial element of Gal(E/K) by u ↦ \overline{u} (not to be confused with the map v ↦ \overline{v} in the definition of a quadrangular system). Set B_{0} := E ⊕E. Then B_{0} is a 4-dimensional vector space over K which can be identified with S by the relation

(u, v) ∈ B_{0} ⟷ uv_{1} + vv_{2} ∈ S .

In particular, we will write q_{1}(u, v) = s_{1}N(u) + s_{2}N(v) for all (u, v) ∈ B_{0}.

Next, we define a commutative field D := E^{2}L = {u^{2}s | u ∈ E, s ∈ L}. Then E^{2} ⊆ D ⊆ E, D/L is a separable quadratic extension, and D ∩ K = L. The non-trivial element of Gal(D/L) is precisely the restriction of the map u ↦ \overline{u} to D; hence we will also denote it by x ↦ \overline{x}. Also, the norm of D is precisely the restriction of N to D, and so we will denote it by N as well. Now set A_{0} := D ⊕ D; then A_{0} is a 4-dimensional vector space over L. Observe that both s_{1}^{−1}s_{2} and s_{1}^{−3}s_{2} are elements of L. We now define a quadratic form q_{2} on A_{0} given by

q_{2}(x, y) := s_{1}^{−1}s_{2}N(x) + s_{1}^{−3}s_{2}N(y)

for all (x, y) ∈ A_{0}. If we set α := s_{1}^{−1}s_{2} ∈ L and β := s_{1}^{−1} ∈ K, then we have

q_{1}(u, v) = β^{−1} · (N(u) + αN(v)) for all (u, v) ∈ B_{0} .
q_{2}(x, y) = α · (N(x) + β^{2}N(y)) for all (x, y) ∈ A_{0} .

We will denote the bilinear forms corresponding to q_{1} and q_{2} by f_{1} and f_{2}, respectively.

Theorem 7.10. For all (u, v) ∈ B_{0} and all (x, y) ∈ A_{0} we have:
(i) q_{1}(u, v) ∈ L ⟺ (u, v) = (0, 0) ;

(ii) q_{2}(x, y) ∈ K^{2} ⟺ (x, y) = (0, 0) ; (iii) α ∈ L \ K^{2} ; (iv) β ∈ K \ L.

Proof. See [20, (14.8)]. □

Note that it follows from (iii) and (iv) of this Theorem that K^{2} ⊂ L ⊂ K. In particular, K is not perfect.

Since L ⊆ K, we can consider K as a (left) vector space over L by the trivial scalar multiplication s · t := st for all s ∈ L and all t ∈ K. Since K^{2} ⊆ L and char(K) = 2, we can also consider L as a (left) vector space over K by the scalar multiplication t ∗ s := t^{2}s for all t ∈ K and all s ∈ L. One can check that in this sense, q is a vector space isomorphism from R to L = q(R).

From now on, we will identify R with L via q, and we still identify S with B_{0} = E ⊕ E. Combining those two identifications, we have actually identified V_{0} with B_{0} ⊕ L. Then ϵ = (0, 1), and we have q(b, s) = q_{1}(b) + s, for all (b, s) ∈ V_{0}.

Now set W_{0} := A_{0} ⊕ K. Then W_{0} is a vector space over L, and we can define a quadratic form q̂ from W_{0} to f given by q̂(a, t) = q_{2}(a) + t^{2} for all (a, t) ∈ W_{0}. It follows from Theorem 7.10(ii) that q̂ is anisotropic as well. One can actually check that (L, W_{0}, q̂) is again a quadratic form of type F_{4}.

Finally, we define a map Θ from A_{0} ⊕ B_{0} to B_{0}, a map Υ from A_{0} ⊕ B_{0} to A_{0}, a map ν from A_{0} ⊕ B_{0} to K, and a map ψ from A_{0} ⊕ B_{0} to L as follows.

Θ((x, y), (u, v)) := (α · (x̄v + βyv̄), xu + βyū) ,
Υ((x, y), (u, v)) := (yū^{2} + αȳv^{2}, β^{−2} · (xu^{2} + αx̄v^{2})) ,
ν((x, y), (u, v)) := α · (β^{−1} · (xuv̄ + x̄ūv) + yūv̄ + ȳuv) ,
ψ((x, y), (u, v)) := α · (xȳu^{2} + x̄yū^{2} + α · (xyv̄^{2} + x̄ȳv^{2})) ,

for all (x, y) ∈ A_{0} = D ⊕ D and all (u, v) ∈ B_{0} = E ⊕ E.

Let V be parametrized by (V_{0}, +), and let W be parametrized by (W_{0}, +). We define a map τ_{V} from V × W to V and a map τ_{W} from W × V to W as follows:

τ_{V} ([b, s], [a, t]) := [b, s][a, t] := [Θ(a, b) + tb, q̂(a, t)s + ψ(a, b)] ,
τ_{W} ([a, t], [b, s]) := [a, t][b, s] := [Υ(a, b) + sa, q(b, s)t + ν(a, b)] ,

for all (b, s) ∈ V_{0} and all (a, t) ∈ W_{0}. Then (V, W, τ_{V} , τ_{W} , [0, 1], [0, 1]) is a quadrangular system. One can check that

F ([b, s], [b′, s′]) = [0, f_{1}(b, b′)] ,
H([a, t], [a′, t′]) = [0, f_{2}(a, a′)] ,

for all (b, s), (b′, s′) ∈ V_{0} and all (a, t), (a′, t′) ∈ W_{0}, and that

[b, s]^{−1} = [q(b, s)^{−1}b, q(b, s)^{−2}s] ,
κ([a, t]) = [q̂(a, t)^{−1}a, q̂(a, t)^{−1}t] ,

for all (b, s) ∈ V_{0}^{∗} and all (a, t) ∈ W_{0}^{∗}.

Remark 7.11. It would be a very tedious job to check that this is indeed a quadrangular system by only using the definitions of the different functions involved. However, it is not very hard to prove the following list of twelve identities, after which the verification of the axioms for the quadrangular systems is straightforward.

Theorem 7.12. For all a, a′ ∈ A_{0} and all b, b′ ∈ B_{0}, we have that

(i) ν(a, b + b′) = ν(a, b) + ν(a, b′) + f_{1}(Θ(a, b), b′) ; (ii) ψ(a + a′, b) = ψ(a, b) + ψ(a′, b) + f_{2}(Υ(a, b), a′) ; (iii) Υ(Υ(a, b), b) = q_{1}(b)^{2}a; (iv) Θ(a, Θ(a, b)) = q_{2}(a)b;

(v) Θ(Υ(a, b), b) + bν(a, b) = q_{1}(b)Θ(a, b) ; (vi) Υ(a, Θ(a, b)) + aψ(a, b) = q_{2}(a)Υ(a, b) ; (vii) ν(Υ(a, b), b) = q_{1}(b)ν(a, b) ; (viii) ψ(a, Θ(a, b)) = q_{2}(a)ψ(a, b) ; (ix) ψ(Υ(a, b), b) = q_{1}(b)^{2}ψ(a, b) ;

(x) ν(a, Θ(a, b)) = q_{2}(a)ν(a, b) ; (xi) q_{1}(Θ(a, b)) = q_{1}(b)q_{2}(a) + ψ(a, b) ; (xii) q_{2}(Υ(a, b)) = q_{1}(b)^{2}q_{2}(a) + ν(a, b)^{2} .

These are the quadrangular systems of type F_{4}. They will be denoted by Ω_{F} (K, V_{0}, q).

Remark 7.13. Although it can be very useful to have these explicit formulas to calculate with, this description is not very clarifying. The description in terms of the quadrangular systems ought to give more insight in the structure of these F_{4}-quadrangles. See section A.3.

CHAPTER 8 The Classification

We will now start the classification of the quadrangular systems. We start with some definitions.

Definition 8.1. A quadrangular system Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is called indifferent if F ≡ 0 and H ≡ 0, reduced if F ≢ 0 and H ≡ 0 and wide if F ≢ 0 and H ≢ 0.

Remark 8.2. We will prove that if Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is a quadrangular system with F ≡ 0 and H ≢ 0, then Ω^{∗} := (W, V, τ_{W} , τ_{V} , δ, ϵ) is a reduced quadrangular system; see Theorem 8.13.

Remark 8.3. Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. If X ⊆ V and Y ⊆ W, then the restriction of τ_{V} to X ×Y and the restriction of τ_{W} to Y ×X will also be denoted by τ_{V} and τ_{W}, respectively.

Definition 8.4. Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a wide quadrangular system. Set Y := Rad(H). We will show below that Γ := (V, Y, τ_{V} , τ_{W} , ϵ, δ) is a reduced quadrangular system; see Theorem 8.14. We then say that Ω is an extension of Γ.

Definition 8.5. Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a reduced quadrangular system. Then Ω is said to be normal if and only if for all w_{1}, w_{2}, …, w_{i} ∈ W, there exists a w ∈ W such that ϵw_{1}w_{2} … w_{i} = ϵw.

Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be an arbitrary quadrangular system. The classification will be divided up into the following five theorems.

Theorem 8.6. If Ω is reduced but not normal, then Ω ≅ Ω_{I}(K, K_{0}, σ) for some involutory set (K, K_{0}, σ) such that σ ≠ 1 and K is generated by K_{0} as a ring.

Theorem 8.7. If Ω is normal, then Ω ≅ Ω_{Q}(K, V_{0}, q) for some anisotropic quadratic space (K, V_{0}, q).

Theorem 8.8. If Ω is indifferent, then Ω ≅ Ω_{D}(K, K_{0}, L_{0}) for some indifferent set (K, K_{0}, L_{0}).

Theorem 8.9. If Ω is an extension of the reduced quadrangular system Γ = Ω_{I}(K, K_{0}, σ) for some involutory set (K, K_{0}, σ) such that σ ≠ 1 and K is generated by K_{0} as a ring, then Ω ≅ Ω_{P} (K, K_{0}, σ, V_{0}, p) for some anisotropic pseudo-quadratic space (K, K_{0}, σ, V_{0}, p).

Theorem 8.10. If Ω is an extension of the reduced quadrangular system Γ = Ω_{Q}(K, V_{0}, q) for some anisotropic quadratic space (K, V_{0}, q), then one of the following holds:

• There exists (a) a multiplication on V_{0} making the vector space V_{0} into an algebra over K such that either V_{0} is a field and V_{0}/K is a separable quadratic extension with norm q or V_{0} is a quaternion division algebra over K with norm q,

(b) an involution σ of V_{0} (which is the unique non-trivial element of Gal(V_{0}/K) if dim_{K} V_{0} = 2 and which is the standard involution of V_{0} if dim_{K} V_{0} = 4) ,

(c) a non-trivial right vector space X over V_{0} ,

(d) a pseudo-quadratic form π on X,

such that (V_{0}, K, σ, X, π) is an anisotropic pseudo-quadratic space, Γ ≅ Ω_{I}(V_{0}, K, σ) and Ω ≅ Ω_{P} (V_{0}, K, σ, X, π).

• (K, V_{0}, q) is a quadratic space of type E_{6}, E_{7} or E_{8}, and Ω ≅ Ω_{E}(K, V_{0}, q).

• (K, V_{0}, q) is a quadratic space of type F_{4}, and Ω ≅ Ω_{F} (K, V_{0}, q).

We will now prove the two theorems which we mentioned in the above remarks. But first, we make an easy but useful observation.

Lemma 8.11. If Rad(F) ≠ 0, then W is abelian, and all elements of V and W have order 1 or 2. Furthermore, π_{r}(v) = v for all r ∈ Rad(F)^{∗} and all v ∈ V, and \overline{v} = v for all v ∈ V.

Proof. Let r be an arbitrary non-zero element of Rad(F). If we substitute r for v in (Q_{8}), then we get that [w_{1}, w_{2}r]_{⊞} = 0 for all w_{1}, w_{2} ∈ W. Substituting w_{2}r^{−1} for w_{2} shows, by (Q_{15}), that W is abelian.

If we substitute r for v in 3.13(i), then we get that w(−r) = F (rw, r) ⊟ wr = ⊟wr for all w ∈ W. By (Q_{5}), this implies that w(−ϵ)r = ⊟wr, hence w(−ϵ) = ⊟w for all w ∈ W. It thus follows from (Q_{6}) that v(⊟w) = vw for all v ∈ V and all w ∈ W. In particular, v(⊟δ) = v. On the other hand, it follows from 3.13(ii) that v(⊟δ) = H(δ, δv) − vδ = −v, for all v ∈ V. Hence v = −v for all v ∈ V, so every element of V has order 1 or 2.

In particular, ϵ = −ϵ, hence w = w(−ϵ) = ⊟w for all w ∈ W, that is, every element of W has order 1 or 2.

Finally, π_{r}(v) = v + \overline{r^{−1}F (r, v)} = v for all r ∈ Rad(F)^{∗} and all v ∈ V. Since it follows from (Q_{10}) that ϵ ∈ Rad(F), we have in particular that \overline{v} = π_{ϵ}(v) = v for all v ∈ V. □

Remark 8.12. Apart from the last statement, we have avoided to use (Q_{10}). We thereby want to stress the fact that this axiom is not essential, and is really only needed to simplify the list of the wide quadrangular systems which have Rad(F) ≠ 0. See Remark 6.4.

Theorem 8.13. Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system with F ≡ 0 and H ≢ 0. Then Ω^{∗} := (W, V, τ_{W} , τ_{V} , δ, ϵ) is a reduced quadrangular system.

Proof. Since F ≡ 0, it follows from Lemma 8.11 that W is abelian and that all elements of V and W have order 1 or 2. In particular, we will write + in place of ⊞ and ⊟. We define \overline{w} := δH(δ, w) + w, for all w ∈ W. Then it follows from (Q_{9}) that \overline{w} = w, for all w ∈ W. We also set w^{−1} := κ(w) for all w ∈ W^{∗} and κ(v) := v^{−1} for all v ∈ V^{∗}. Let F^{∗} ≡ H and H^{∗} ≡ F ≡ 0.

We will denote the axioms that we have to prove for Ω^{∗} by (Q_{i})^{∗}.

Since all elements of V and W have order 1 or 2, the axioms (Q_{5})^{∗} and (Q_{6})^{∗} are trivial. Note that \overline{v} = v for all v ∈ V and \overline{w} = w for all w ∈ W. We now prove the remaining axioms. We observe that (Q_{1})^{∗} ≡ (Q_{2}), (Q_{2})^{∗} ≡ (Q_{1}), (Q_{3})^{∗} ≡ (Q_{4}), (Q_{4})^{∗} ≡ (Q_{3}), (Q_{13})^{∗} ≡ (Q_{14}), (Q_{14})^{∗} ≡ (Q_{13}), (Q_{17})^{∗} ≡ (Q_{20}) and (Q_{20})^{∗} ≡ (Q_{17}). It follows from F ≡ 0 that Rad(F) = V and hence Im(H) ⊆ Rad(F) ∋ ϵ; this shows (Q_{7})^{∗} and (Q_{9})^{∗}. Now (Q_{8})^{∗} follows from the fact that V is abelian and that F ≡ 0; (Q_{10})^{∗} follows from (Q_{9}). By (Q_{22}), H(w_{1}, w_{2}) = H(w_{2}, w_{1}) for all w_{1}, w_{2} ∈ W. Since W is abelian, it follows from (Q_{12}) that v(w_{1} + w_{2}) = v(w_{2} + w_{1}) = vw_{2} + vw_{1} + H(w_{1}, w_{2}v) = vw_{1} + vw_{2} + H(w_{2}v, w_{1}) for all v ∈ V and all w_{1}, w_{2} ∈ W. This proves (Q_{11})^{∗}. Vice versa, it follows from (Q_{11}) that w(v_{1} + v_{2}) = wv_{1} + wv_{2} for all w ∈ W and all v_{1}, v_{2} ∈ V, which proves (Q_{12})^{∗}. By (Q_{13}), we have that (Q_{15}) is equivalent to wv^{−1}v = w. It follows that (Q_{15})^{∗} ≡ (Q_{18}) and (Q_{18})^{∗} ≡ (Q_{15}). Again by (Q_{13}), we have that (Q_{16}) is equivalent to v(wv^{−1}) = v^{−1}w. Hence (Q_{16})^{∗} ≡ (Q_{19}) and (Q_{19})^{∗} ≡ (Q_{16}). It follows that Ω^{∗} is a quadrangular system, which is reduced since H^{∗} ≡ 0 and F^{∗} ≢ 0. □

Theorem 8.14. Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a wide quadrangular system. Set Y := Rad(H). Then Γ := (V, Y, τ_{V} , τ_{W} , ϵ, δ) is a reduced quadrangular system; see Remark 8.3.

Proof. First of all, we observe that Y is a subgroup of W: if w_{1}, w_{2} ∈ Y, then H(w_{1} ⊞ w_{2}, w) = H(w_{1}, w) + H(w_{2}, w) = 0 for all w ∈ W, so w_{1} ⊞ w_{2} ∈ Y as well. It only remains to show that τ_{W} (Y × V) ⊆ Y, F (V, V) ⊆ Y and κ(Y^{∗}) ⊆ Y. So let w be an arbitrary element of Y, and let v be an arbitrary element of V. Then [w, w_{2}]_{⊞} = 0 for all w_{2} ∈ W by (Q_{8}), and therefore v(w ⊞ w_{2}) = v(w_{2} ⊞ w). It follows from (Q_{12}) that H(w_{2}, wv) = H(w, w_{2}v) = 0 for all w_{2} ∈ W. By (Q_{22}), this implies that H(wv, w_{2}) = 0 for all w_{2} ∈ W, hence wv ∈ Y. So we have proved that τ_{W} (Y × V) = Y · V ⊆ Y.

It follows from (Q_{7}) that F (V, V) ⊆ Y. Now let w be an arbitrary element of Y^{∗}. Substituting −ϵ for v in 3.2(ii) yields κ(w)(−ϵ(⊟w)) = w, hence κ(w) = w(−ϵ(⊟w))^{−1} ∈ Y · V ⊆ Y. So Γ is a quadrangular system, which is reduced since H restricted to Y × Y is identically zero. □

8.1. Quadrangular Systems of Involutory Type

Our goal in this section is to classify the quadrangular systems which are reduced but not normal.

Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system. For the moment, we only assume that H ≡ 0, so Ω is reduced or indifferent.

Since H ≡ 0, it follows from (Q_{8}) that W is abelian. We will write + and − in place of ⊞ and ⊟, respectively. It follows from 3.13(ii) that v(−w) = −vw for all v ∈ V and all w ∈ W. If we replace w_{2} by −w_{2} in (Q_{12}), we get v(w_{1} − w_{2}) = vw_{1}−vw_{2}, for all v ∈ V and all w_{1}, w_{2} ∈ W. In particular, it follows from ϵw_{1} = ϵw_{2} that w_{1} = w_{2}, by 3.3(ii).

By (Q_{18}), we have ϵκ(w)w = ϵ for all w ∈ W. If we replace w by −w, we get that ϵ(−κ(−w))w = ϵ for all w ∈ W. Hence ϵκ(w)w = ϵ(−κ(−w))w, so ϵκ(w) = ϵ(−κ(−w)) by 3.5(ii) and hence κ(w) = −κ(−w) for all w ∈ W by the previous paragraph. Moreover, by (Q_{6}) and the result of the previous paragraph, we have that w(−ϵ) = w for all w ∈ W. It now follows from (Q_{14}) that κ(κ(w)) = w, for all w ∈ W. Hence we will write w^{−1} in place of κ(w), for all w ∈ W. Note that it follows from 3.2(i) that vww^{−1} = v, for all v ∈ V and all w ∈ W^{∗}.

For all w_{1}, w_{2}, …, w_{n} ∈ W^{∗}, let m = w_{1}•w_{2}•⋯•w_{n} be the automorphism of V which maps v to vw_{1}w_{2} … w_{n} for all v ∈ V; see Theorem 3.6(i). Let M be the set of all such automorphisms. Then (M, •) is a group with neutral element δ. We denote the action of an element m ∈ M by right juxtaposition, i.e. vm = vw_{1}w_{2} … w_{n}. Let K be the set of homomorphisms from V to itself (additively) generated by a finite number of elements of M. We write k = m_{1} + ⋯ + m_{ℓ}, where m_{1}, …, m_{ℓ} ∈ M. Again, we denote the action of an element k ∈ K by right juxtaposition, so we have vk = vm_{1}+⋯+vm_{ℓ}. Then K with this + as addition and with • as multiplication is a ring with multiplicative identity δ. Note that, by (Q_{12}), v(w_{1} + w_{2}) = vw_{1} + vw_{2}, hence the notation w_{1} + w_{2} is unambiguous. Let σ be the automorphism of M which maps m = w_{1} • ⋯ • w_{n} to m^{σ} := w_{n} • ⋯ • w_{1}. We extend σ to K by setting k^{σ} := m_{1}^{σ} + ⋯ + m_{ℓ}^{σ} for all k = m_{1} + ⋯ + m_{ℓ} ∈ K. Let E := ϵK.

Lemma 8.15. For all v ∈ V, w ∈ W and m ∈ M, we have that w·ϵm·v = w·vm.

Proof. Let m = w_{1} • ⋯ • w_{n} be an arbitrary element of M, so w_{1}, …, w_{n} are elements of W^{∗}. We will show the lemma by induction on n.

Note that Π_{⊟z} ≡ id_{W} since H ≡ 0; hence by (Q_{26}), the lemma holds for n = 1. Assume that w · ϵw_{1} … w_{n−1} · v = w · vw_{1} … w_{n−1} for all w ∈ W and all v ∈ V. Then, by repeated use of (Q_{26}),

w · ϵw_{1} … w_{n} · v = w · (ϵw_{1} … w_{n−1} · w_{n}) · v
= w · (ϵw_{n}) · ϵw_{1} … w_{n−1} · v
= w · (ϵw_{n}) · vw_{1} … w_{n−1}
= w · (vw_{1} … w_{n−1} · w_{n})
= w · vw_{1} … w_{n}

for all w ∈ W and all v ∈ V, and we are done. □

Lemma 8.16. For all v ∈ V and all w, z ∈ W, we have that \overline{\overline{v}w}z = \overline{\overline{vz}w}.

Proof. Let v ∈ V and w, z ∈ W. We may assume that w ≠ 0. Note that Π_{w}(z) = z since H ≡ 0. Since v(−w) = −vw, it follows from (Q_{23}) that vz =

\overline{\overline{\overline{\overline{v}w}z}w^{−1}}. Since (w^{−1})^{−1} = w and by 3.12, it follows from this that \overline{\overline{vz}w} = \overline{\overline{v}w}z. □

Lemma 8.17. For all v ∈ V, all w ∈ W and all m ∈ M, we have that \overline{\overline{v}w}m = \overline{\overline{vm}w} and \overline{\overline{v}m}w = \overline{\overline{vw}m}.

Proof. Let m = w_{1} • ⋯ • w_{n} be an arbitrary element of M, so w_{1}, …, w_{n} are elements of W^{∗}. We will prove by induction on n that \overline{\overline{v}w}w_{1} … w_{n} = \overline{\overline{vw_{1} … w_{n}}w}. For n = 1, this was shown in Lemma 8.16. Assume that we have proved the current lemma for n − 1. Then

\overline{\overline{v}w}w_{1} … w_{n} = \overline{\overline{v}w}w_{1} … w_{n−1} · w_{n}
= \overline{\overline{vw_{1} … w_{n−1}}w}w_{n}
= \overline{\overline{vw_{1} … w_{n}}w} ,

again by Lemma 8.16 with vw_{1} … w_{n−1} in place of v. This proves the first identity; the second then follows from the first by substituting \overline{v} for v. □

Lemma 8.18. For all v ∈ V and all k_{1}, k_{2} ∈ m, we have that \overline{\overline{v}k_{1}}k_{2} = \overline{\overline{vk_{2}}k_{1}}.

Proof. Since both sides are additive in k_{1} and k_{2}, it suffices to show this for k_{1} = m_{1} ∈ M and k_{2} = m_{2} ∈ M. Let m_{2} = z_{1} • ⋯ • z_{n} for some z_{1}, …, z_{n} ∈ W^{∗}. By repeated use of Lemma 8.17, we have that

\overline{\overline{v}m_{1}}m_{2} = \overline{\overline{v}m_{1}}z_{1} … z_{n}
= \overline{\overline{vz_{1}}m_{1}}z_{2} … z_{n}
= …
= \overline{\overline{vz_{1} … z_{n}}m_{1}}
= \overline{\overline{vm_{2}}m_{1}} ,

which proves the lemma. □

Theorem 8.19. \overline{ϵk} = ϵk^{σ}, for all k ∈ K.

Proof. It is sufficient to prove this for k ∈ M. Let k = w_{1} • ⋯ • w_{n} be an arbitrary element of M, so w_{1}, …, w_{n} ∈ W^{∗}. We will prove by induction on n that \overline{ϵw_{1} … w_{n}} = ϵw_{n} … w_{1}.

First assume that n = 1. Since v(−w) = −vw for all v ∈ V and all w ∈ W, it follows from 3.21 that \overline{ϵw_{1}} = −π_{ϵ}(ϵw_{1}) = −ϵ(−w_{1}) = ϵw_{1}.

Now assume that we have proved that \overline{ϵw_{1} … w_{n−1}} = ϵw_{n−1} … w_{1}, for all w_{1}, …, w_{n−1} ∈ W^{∗}. Then it follows from 3.12 and Lemma 8.17 that

\overline{ϵw_{1} … w_{n}} = \overline{\overline{\overline{ϵw_{1} … w_{n−1}}}w_{n}}
= \overline{\overline{ϵw_{n−1} … w_{1}}w_{n}}
= \overline{\overline{ϵ}w_{n}}w_{n−1} … w_{1}
= ϵw_{n}w_{n−1} … w_{1} ,

since \overline{\overline{ϵ}w_{n}} = \overline{ϵw_{n}} = ϵw_{n}. □

Theorem 8.20. Δ := (E, W, τ_{V} , τ_{W} , ϵ, δ) is a quadrangular system; see Remark 8.3.

Proof. First of all, we observe that E = ϵK is a subgroup of V, since ϵk_{1} + ϵk_{2} = ϵ(k_{1} + k_{2}) for all k_{1}, k_{2} ∈ K. It only remains to show that τ_{V} (E × W) ⊆ E, H(W, W) ⊆ E and (E^{∗})^{−1} ⊆ E. Since K • W = K, we have that τ_{V} (E × W) = ϵK · W = ϵ(K • W) = ϵK = E. Since H ≡ 0, it is obvious that H(W, W) ⊆ E. Finally, if we substitute δ for w in (Q_{16}) and apply the fact that vww^{−1} = v, we get that v^{−1} = \overline{v}(δv)^{−1} for all v ∈ V^{∗}. In particular, we get that (ϵk)^{−1} = \overline{ϵk}(δ · ϵk)^{−1} = ϵ(k^{σ} • (δ · ϵk)^{−1}) ∈ ϵK for all k ∈ K^{∗}, where we have used Theorem 8.19. Thus Δ := (E, W, τ_{V} , τ_{W} , ϵ, δ) is a quadrangular system. □

Lemma 8.21. If vk = 0 for some v ∈ V and some k ∈ K, then vk_{2}k = 0 for all k_{2} ∈ K.

Proof. We may assume that v ≠ 0. We will first show the lemma for k_{2} = w ∈ W. Since v(−w) = −vw, it follows from 3.20 that vw = \overline{\overline{v}m} where m = δ\overline{v^{−1}} •wv ∈ M. By Lemma 8.18, it follows from vk = 0 that vwk = \overline{\overline{v}m}k = \overline{\overline{vk}m} = 0, which proves the lemma in this case.

Now let k_{2} = m = w_{1} • ⋯ • w_{n} be an arbitrary element of M, then it follows by induction on n that vmk = 0.

Finally, let k_{2} = m_{1} + ⋯ + m_{ℓ} be an arbitrary element of K, then it follows from the previous paragraph that vk_{2}k = vm_{1}k + ⋯ + vm_{ℓ}k = 0, which completes the proof of this lemma. □

Lemma 8.22. For all v_{1}, v_{2} ∈ V and all k ∈ K, we have that F (v_{1}k, v_{2}) = F (v_{1}, v_{2}k^{σ}).

Proof. Since both V and W are abelian, it follows from (Q_{21}) and (Q_{11}) that F (v_{1}w, v_{2}) = F (v_{1}, v_{2}w) for all v_{1}, v_{2} ∈ V and all w ∈ W. For m ∈ M, say m = w_{1} • ⋯ • w_{n}, it follows by induction on n that F (v_{1}m, v_{2}) = F (v_{1}, v_{2}m^{σ}). Since F is additive in both variables, it now follows that F (v_{1}k, v_{2}) = F (v_{1}, v_{2}k^{σ}) for any k = m_{1} + ⋯ + m_{ℓ} ∈ K. □

Our first goal is to prove that Δ is a quadrangular system of involutory type. Assume from now on that V = E, that is, that Ω = Δ.

Lemma 8.23. (i) Let k ∈ K. If ϵk = 0, then k = 0.

(ii) Let k_{1}, k_{2} ∈ K. If ϵk_{1} = ϵk_{2}, then k_{1} = k_{2}.

Proof. Let k ∈ K be such that ϵk = 0. By Lemma 8.21, it follows that ϵk_{2}k = 0 for all k_{2} ∈ K, and hence V k = Ek = ϵKk = 0, which implies that k = 0 (remember that the elements of K are endomorphisms of V). This proves (i); (ii) now follows from (i) by substituting k_{1} − k_{2} for k. □

Definition 8.24. For each k ∈ K^{∗}, we define k′ as the (unique!) element in K such that (ϵk)^{−1} = ϵk′.

Lemma 8.25. For all k ∈ K^{∗} and all w ∈ W, we have that k•w = w(ϵk)•(k′)^{σ}.

Proof. By (Q_{16}), we have that

(ϵk)w = \overline{(ϵk)^{−1} · w(ϵk)}
= \overline{ϵ · k′ · w(ϵk)}
= \overline{ϵ · (k′ • w(ϵk))}
= ϵ · (w(ϵk) • (k′)^{σ}) ,

where we have used Theorem 8.19. It follows by Lemma 8.23(ii) that k • w = w(ϵk) • (k′)^{σ}. □

Lemma 8.26. For all k ∈ K^{∗} and all w ∈ W^{∗}, we have that (k •w)′ = w^{−1} •k′.

Proof. By Lemma 8.25 with k • w in place of k and w^{−1} in place of w and by (Q_{19}), we have that

k • w = ((k • w) • w^{−1}) • w
= w^{−1}(ϵ(k • w)) • (k • w)^{′σ} • w
= w(ϵk) • (k • w)^{′σ} • w ,

which, together with Lemma 8.25, implies that

w(ϵk) • (k′)^{σ} • w^{−1} = w(ϵk) • (k • w)^{′σ} .

If we apply σ to both sides, we get that

w^{−1} • k′ • w(ϵk) = (k • w)′ • w(ϵk) ,

from which it follows that w^{−1} • k′ = (k • w)′ since w(ϵk) is invertible in K. □

Lemma 8.27. For all w_{1}, …, w_{n} ∈ W^{∗}, we have that

(w_{1} • ⋯ • w_{n})′ = w_{n}^{−1} • ⋯ • w_{1}^{−1} .

Proof. By Lemma 8.26 with k = δ, we have that w′_{1} = w_{1}^{−1}. Again by Lemma 8.26, it now follows by induction on n that

(w_{1} • ⋯ • w_{n})′ = w_{n}^{−1} • (w_{1} • ⋯ • w_{n−1})′
= w_{n}^{−1} • w_{n−1}^{−1} • ⋯ • w_{1}^{−1} ,

which is what we wanted to show. □

Since it follows from this lemma that m • m′ = m′ • m = δ, we will from now on write m^{−1} in place of m′ for all m ∈ M.

Lemma 8.28. For all m ∈ M and all w ∈ W, we have that w(ϵm) = m•w•m^{σ}.

Proof. First of all, observe that it follows from Lemma 8.27 that (m^{−1})^{σ} = (m^{σ})^{−1}. By Lemma 8.25, we have that m • w = w(ϵm) • (m^{−1})^{σ}. It follows that m • w • m^{σ} = w(ϵm). □

Lemma 8.29. For all k_{1}, k_{2} ∈ K, we have that F (ϵk_{1}, ϵk_{2}) = k_{1} • k_{2}^{σ} + k_{2} • k_{1}^{σ}.

Proof. Since v +\overline{v} = ϵF (ϵ, v) for all v ∈ V, we have that ϵF (ϵ, ϵk) = ϵk +\overline{ϵk} = ϵk + ϵk^{σ} and hence F (ϵ, ϵk) = k + k^{σ}, for all k ∈ K. It now follows from Lemma 8.22 that

F (ϵk_{1}, ϵk_{2}) = F (ϵ, ϵk_{2}k_{1}^{σ})
= F (ϵ, ϵ(k_{2} • k_{1}^{σ}))
= k_{2} • k_{1}^{σ} + k_{1} • k_{2}^{σ} ,

which proves the lemma. □

Theorem 8.30. For all k ∈ K and all w ∈ W, we have that w(ϵk) = k •w •k^{σ}.

Proof. In Lemma 8.28, we have shown this theorem for all k ∈ M. Now suppose that the theorem holds for k_{1}, k_{2} ∈ K. We will show that it then holds for k_{1} + k_{2} as well, which will prove the theorem for all k ∈ K.

It follows from (Q_{11}) and Lemma 8.29 that

w(ϵ(k_{1} + k_{2})) = w(ϵk_{1} + ϵk_{2})
= w(ϵk_{1}) + w(ϵk_{2}) + F (ϵk_{2}w, ϵk_{1})
= k_{1} • w • k_{1}^{σ} + k_{2} • w • k_{2}^{σ} + (k_{2} • w) • k_{1}^{σ} + k_{1} • (k_{2} • w)^{σ}
= k_{1} • w • k_{1}^{σ} + k_{2} • w • k_{2}^{σ} + k_{2} • w • k_{1}^{σ} + k_{1} • w • k_{2}^{σ}
= (k_{1} + k_{2}) • w • (k_{1} + k_{2})^{σ} ,

which is what we had to show. □

Theorem 8.31. K_{+,•} is a field or a skew-field.

Proof. We already know that K_{+,•} is a ring. Let k be an arbitrary element of K^{∗}. We will show that k′ • k = k • k′ = δ.

By (Q_{15}), Theorem 8.30 and Lemma 8.25, we have that

δ = δ(ϵk)(ϵk)^{−1}
= δ(ϵk)(ϵk′)
= k′ • δ(ϵk) • (k′)^{σ}
= k′ • (k • δ)
= k′ • k ,

and if we substitute k′ for k, then we get that δ = k • k′ as well, since it follows from the definition of k′ that k′′ = k.

Hence every non-zero element k ∈ K^{∗} is invertible with inverse k^{−1} = k′. It follows that K_{+,•} is a field or a skew-field. □

For technical reasons which will be clear in a moment, we let K_{+,·} := K_{+,•}^{op}, that is, we set k_{1}k_{2} := k_{2} • k_{1} for all k_{1}, k_{2} ∈ K.

Theorem 8.32. (K, W, σ) is an involutory set. Furthermore, K is generated by W as a ring.

Proof. We have just shown that K is a field or a skew-field. It is obvious from the definition that σ^{2} = 1 and that (k_{1}k_{2})^{σ} = (k_{2} • k_{1})^{σ} = k_{1}^{σ} • k_{2}^{σ} = k_{2}^{σ}k_{1}^{σ} for all k_{1}, k_{2} ∈ K, so σ is an involution of K. W is an additive subgroup of K containing δ. By Lemma 8.29, k +k^{σ} = F (ϵ, ϵk) ∈ Im(F) ⊆ W for all k ∈ K, hence K_{σ} ⊆ W, and by the definition of σ, all elements of W are fixed by σ. Finally, it follows from Theorem 8.30 that k^{σ}W k = k • W • k^{σ} ⊆ W (ϵk) ⊆ W for all k ∈ K. Thus (K, W, σ) is an involutory set.

The fact that K is generated by W as a ring follows immediately from the definition of the ring K. □

Theorem 8.33. (E, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{I}(K, W, σ).

Proof. Let ϕ be the isomorphism from [K] to E which maps [k] to ϵk for all k ∈ K, and let ψ be the isomorphism from [W] to W which maps [w] to w for all w ∈ W. Then ϕ([δ]) = ϵδ = ϵ and ψ([δ]) = δ. Furthermore,

ϕ([k][w]) = ϕ([wk]) = ϵ(wk) = ϵ(k • w) = ϵk · w = ϕ([k])ψ([w]) , and
ψ([w][k]) = ψ(k^{σ}wk) = k^{σ}wk = k • w • k^{σ} = w(ϵk) = ψ([w])ϕ([k]) ,

for all w ∈ W and all k ∈ K. Hence (ϕ, ψ) is an isomorphism from Ω_{I}(K, W, σ) to (E, W, τ_{V} , τ_{W} , ϵ, δ). □

The next lemma shows the “σ ≠ 1” part of Theorem 8.6.

Lemma 8.34. If (K, K_{0}, σ) is an involutory set with σ = 1, then Ω_{I}(K, K_{0}, σ) is normal or indifferent.

Proof. Since σ is an involution, ab = (ab)^{σ} = b^{σ}a^{σ} = ba for all a, b ∈ K, hence K is abelian. It follows that F ([a], [b]) = [2ab] for all a, b ∈ K. If char(K) = 2, then F ≡ 0, hence Ω_{I}(K, K_{0}, σ) is indifferent. So we can assume that char(K) ≠ 2. But then K_{σ} = {2a | a ∈ K} = K, and hence K_{0} = K. It follows that for all elements t_{1}, t_{2}, …, t_{n} ∈ K_{0}, the product t_{1}t_{2} … t_{n} lies in K_{0} as well, and hence [1][t_{1}][t_{2}] … [t_{n}] = [1][t_{1}t_{2} … t_{n}], which implies that Ω_{I}(K, K_{0}, σ) is normal. □

From now on, we drop our assumption that V = E (but we still assume that H ≡ 0). Our next goal is to show that if Ω is reduced but not normal, then V = E after all.

We start with a generalization of Lemma 8.18:

Lemma 8.35. For all v ∈ V^{∗}, all c ∈ V and all k_{1}, k_{2} ∈ K, we have that

π_{v}(π_{v}(c)k_{1})k_{2} = π_{v}(π_{v}(ck_{2})k_{1}) .

Proof. If we substitute \overline{c}(δv)^{−1} for c in (Q_{25}), then we get that π_{v}(c)w = π_{v}(\overline{\overline{c}(δv)^{−1}(wv)}), and hence π_{v}(π_{v}(c)w) = \overline{\overline{c}(δv)^{−1}(wv)} for all c, v ∈ V and all w ∈ W. It thus follows by 3.23(i) and Lemma 8.18 that

π_{v}(π_{v}(c)w_{1})w_{2} = \overline{\overline{c}(δv)^{−1}(w_{1}v)}w_{2}
= \overline{\overline{cw_{2}}(δv)^{−1}(w_{1}v)}
= π_{v}(π_{v}(cw_{2})w_{1}) ,

which shows the lemma for all k_{1}, k_{2} ∈ W. In the same way as in Lemma 8.17 and Lemma 8.18, we can use induction to deduce from this that the lemma holds for all k_{1}, k_{2} ∈ M. Since π_{v} is additive, it then follows that the lemma holds for all k_{1}, k_{2} ∈ K. □

Lemma 8.36. For all v ∈ V^{∗} and all k ∈ K, we have that π_{v}(vk) = −vk^{σ}.

Proof. Since π_{v} is additive, it suffices to show that π_{v}(vm) = −vm^{σ} for all m = w_{1} • ⋯ • w_{n} ∈ M, which we will do by induction on n.

It already follows from 3.21 that π_{v}(vw_{1}) = −vw_{1}, which shows the statement for n = 1. Now assume that π_{v}(vw_{1} … w_{n−1}) = −vw_{n−1} … w_{1} for all w_{1}, …, w_{n−1} ∈ W. Then it follows by Lemma 8.35 that

π_{v}(vw_{1} … w_{n}) = −π_{v}(π_{v}(vw_{n−1} … w_{1})w_{n})
= −π_{v}(π_{v}(v)w_{n})w_{n−1} … w_{1}
= −vw_{n}w_{n−1} … w_{1} ,

since π_{v}(π_{v}(v)w_{n}) = −π_{v}(vw_{n}) = vw_{n}. □

Lemma 8.37. For all v ∈ V and all w ∈ W, we have that vww = v · δ(ϵw).

Proof. We may assume that v ≠ 0. Since H ≡ 0, it follows from (Q_{26}) that δ(ϵw)v = δ(vw), and hence δ(ϵw) = δ(vw)v^{−1}. By (Q_{16}), it follows that

v · δ(ϵw) = v · (δ(vw)v^{−1}) = \overline{v^{−1} · δ(vw)} .

If we substitute vw for v, v^{−1} for c and w^{−1} for w in (Q_{25}), then we get, by (Q_{19}), (Q_{16}) and 3.21, that

π_{vw}(\overline{v^{−1} · δ(vw)})w^{−1} = π_{vw}(\overline{v^{−1} · w^{−1}(vw)})
= π_{vw}(\overline{v^{−1} · wv})
= π_{vw}(vw)
= −vw ,

and hence π_{vw}(v ·δ(ϵw)) = −vww, from which it follows, by 3.23(i), that v ·δ(ϵw) = −π_{vw}(vww) = vww by Lemma 8.36, which is what we had to show. □

Lemma 8.38. For all v ∈ V and all w, z ∈ W, we have that vwzw = v · z(ϵw).

Proof. We may assume that v ≠ 0. Since H ≡ 0, it follows from (Q_{26}) that z(ϵw)v = z(vw). By (Q_{16}), it follows that

v · z(ϵw) = v · (z(vw)v^{−1}) = \overline{v^{−1} · z(vw)} .

If we substitute vw for v, v^{−1} for c and z for w in (Q_{25}), then we get that

π_{vw}(\overline{v^{−1} · δ(vw)})z = π_{vw}(\overline{v^{−1} · z(vw)}) ,

and hence

π_{vw}(v · δ(ϵw))z = π_{vw}(v · z(ϵw)) .

It now follows from Lemma 8.37 and Lemma 8.36 that

v · z(ϵw) = π_{vw}(π_{vw}(vww)z)
= π_{vw}(−vwwz)
= vwzw ,

which proves the lemma. □

Lemma 8.39. For all v ∈ V, all z ∈ W and all m ∈ M, we have that vmzm^{σ} = v · z(ϵm). In particular, vmm^{σ} = v · δ(ϵm).

Proof. Let m = w_{1} •⋯•w_{n} with w_{1}, …, w_{n} ∈ W^{∗}. We will prove the lemma by induction on n.

We have already shown in Lemma 8.38 that the current lemma holds for n = 1. Now assume that vw_{1} … w_{n−1}yw_{n−1} … w_{1} = v · y(ϵw_{1} … w_{n−1}) for all y ∈ W. Then by Lemma 8.38 and (Q_{26}), we have that

vw_{1} … w_{n}zw_{n} … w_{1} = ((vw_{1} … w_{n−1})w_{n}zw_{n})w_{n−1} … w_{1}
= vw_{1} … w_{n−1} · z(ϵw_{n}) · w_{n−1} … w_{1}
= v · z(ϵw_{n})(ϵw_{1} … w_{n−1})
= v · z(ϵw_{1} … w_{n−1}w_{n}) ,

and we are done. □

Lemma 8.40. For all v ∈ V and all w_{1}, w_{2}, w_{3} ∈ W, we have that
(i) F (vw_{1}, vw_{2}w_{3}) = F (ϵw_{1}, ϵw_{2}w_{3})v;
(ii) vw_{1}w_{2}w_{3} + vw_{3}w_{2}w_{1} = vF (ϵw_{1}, ϵw_{3}w_{2}) .
Proof. By (Q_{21}), (Q_{11}) and (Q_{26}), we have that
F (vw_{1}, vw_{2}w_{3}) = w_{3}(vw_{1} + vw_{2}) − w_{3}(vw_{1}) − w_{3}(vw_{2})
= w_{3} · v(w_{1} + w_{2}) − w_{3} · vw_{1} − w_{3} · vw_{2}
= w_{3} · ϵ(w_{1} + w_{2}) · v − w_{3} · ϵw_{1} · v − w_{3} · ϵw_{2} · v
= (w_{3} · (ϵw_{1} + ϵw_{2}) − w_{3} · ϵw_{1} − w_{3} · ϵw_{2}) · v
= F (ϵw_{1}, ϵw_{2}w_{3})v ,

which proves (i). By Lemma 8.36, the definition of π_{v}, Lemma 8.22, (i) and (Q_{16}), we have that

vw_{1}w_{2}w_{3} + vw_{3}w_{2}w_{1} = vw_{3}w_{2}w_{1} − π_{v}(vw_{3}w_{2}w_{1})
= \overline{v^{−1}F (v, vw_{3}w_{2}w_{1})}
= \overline{v^{−1}F (vw_{1}, vw_{3}w_{2})}
= \overline{v^{−1} · F (ϵw_{1}, ϵw_{3}w_{2})v}
= vF (ϵw_{1}, ϵw_{3}w_{2}) ,

which proves (ii). □

Lemma 8.41. Let w_{1}, w_{2}, w_{3} ∈ W be arbitrary. Let k = w_{1} + w_{2} • w_{3} ∈ K. Then

(i) vkk^{σ} = v · δ(ϵk) for all v ∈ V ; (ii) If ϵk = 0, then k = 0 .

Proof. By Lemma 8.39, Lemma 8.40(ii) and (Q_{11}), we have that

vkk^{σ} = v(w_{1} + w_{2} • w_{3})(w_{1} + w_{3} • w_{2})
= vw_{1}w_{1} + vw_{2}w_{3}w_{3}w_{2} + vw_{1}w_{3}w_{2} + vw_{2}w_{3}w_{1}
= v · δ(ϵw_{1}) + v · δ(ϵw_{2}w_{3}) + vF (ϵw_{1}, ϵw_{2}w_{3})
= v · (δ(ϵw_{1}) + δ(ϵw_{2}w_{3}) + F (ϵw_{1}, ϵw_{2}w_{3}))
= v · δ(ϵw_{1} + ϵw_{2}w_{3})
= v · δ(ϵk)

for all v ∈ V, which proves (i). Now suppose that ϵk = ϵw_{1} + ϵw_{2}w_{3} = 0. Then it follows, by (Q_{11}), Lemma 8.15 and Lemma 8.40(i), that

0 = δ · (ϵw_{1} + ϵw_{2}w_{3}) · v
= δ · ϵw_{1} · v + δ · ϵw_{2}w_{3} · v + F (ϵw_{1}, ϵw_{2}w_{3}) · v
= δ · vw_{1} + δ · vw_{2}w_{3} + F (vw_{1}, vw_{2}w_{3})
= δ · (vw_{1} + vw_{2}w_{3})
= δ · vk ,

and hence vk = 0, for all v ∈ V. So k = 0. □

Remark 8.42. It will follow from the classification that the statements in Lemma 8.40 and Lemma 8.41 actually hold in a much broader generality, for all reduced quadrangular systems. More precisely, we have that

(i) F (vk_{1}, vk_{2}) = F (ϵk_{1}, ϵk_{2})v; (ii) vk + vk^{σ} = vF (ϵk, ϵ) ; (iii) vkk^{σ} = v · δ(ϵk) ; (iv) if ϵk = 0, then k = 0 ;

for all v ∈ V and all k, k_{1}, k_{2} ∈ K. However, we are not aware of a simple proof of these facts at this step of the classification.

By definition, Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is normal if and only if ϵM = ϵW. To complete the proof of Theorem 8.6, it will thus suffice to prove the following theorem:

Theorem 8.43. If V ≠ E and F ≢ 0, then ϵM = ϵW. Furthermore, vm = vm^{σ} and \overline{vm} = \overline{v}m for all v ∈ V \ E and all m ∈ M.

Proof. We start by showing that \overline{vw} = \overline{v}w for all v ∈ V \ E and all w ∈ W. So let v ∈ V \ E and w ∈ W be arbitrary. By 3.20, \overline{\overline{v}w} = v · δv^{−1} · w\overline{v}, and hence vw − \overline{\overline{v}w} = vk for k = w − δv^{−1} • w\overline{v}. On the other hand,

vw − \overline{\overline{v}w} = vw − \overline{ϵF (v, ϵ)w − vw}
= vw − ϵF (ϵF (v, ϵ)w − vw, ϵ) + ϵF (v, ϵ)w − vw
= ϵF (v, ϵ)w − ϵF (ϵF (v, ϵ)w − vw, ϵ)
∈ ϵK ,

hence vk ∈ ϵK. Suppose that vk ≠ 0. Then it would follow from Lemma 8.41 that ϵk ≠ 0 and that vkk^{σ} · (δ(ϵk))^{−1} = v. Hence v ∈ ϵK · k^{σ} · (δ(ϵk))^{−1} ⊆ ϵK = E, which contradicts the choice of v. So we must have vk = 0, and thus \overline{\overline{v}w} = vw, which shows that \overline{vw} = \overline{v}w.

We will now show by induction on n that \overline{vw_{1} … w_{n}} = \overline{v}w_{1} … w_{n} for all v ∈ V \ E and all w_{1}, …, w_{n} ∈ W. We have already shown this for n = 1, so suppose that it holds for n − 1. We may assume that w_{i} ≠ 0 for all i ∈ {1, …, n}. If v ∈ V \ E, then also vw_{1} … w_{n−1} ∈ V \ E, since v = vw_{1} … w_{n−1}w_{n−1}^{−1} … w_{1}^{−1}. Hence we can substitute vw_{1} … w_{n−1} for v in the result of the previous paragraph, and we get that

\overline{vw_{1} … w_{n}} = \overline{vw_{1} … w_{n−1} · w_{n}}
= \overline{vw_{1} … w_{n−1}} · w_{n}
= \overline{v}w_{1} … w_{n} ;

the statement thus holds for n as well. This shows that \overline{vm} = \overline{v}m for all v ∈ V \ E and all m ∈ M.

We will now prove by induction on n that \overline{vw_{1} … w_{n}} = \overline{v}w_{n} … w_{1} for all v ∈ V and all w_{1}, …, w_{n} ∈ W. Again, we have already shown this for n = 1, so suppose that it holds for n − 1. Then, by Lemma 8.17,

\overline{vw_{1} … w_{n}} = \overline{\overline{\overline{v}w_{n−1} … w_{1}}w_{n}}
= \overline{vw_{n}}w_{n−1} … w_{1}
= \overline{v}w_{n} … w_{1} ,

so it holds for n as well. We have thus proved that \overline{vm} = \overline{v}m^{σ} for all m ∈ M.

Since v ∈ V \ E if and only if \overline{v} ∈ V \ E, it follows from the previous two paragraphs that vm = vm^{σ} for all v ∈ V \ E and all m ∈ M.

Now, we first assume that ϵ ∉ Rad(F). Since E is a proper subgroup of V, V is generated by V \ E. Since ϵ ∉ Rad(F), this implies that F (ϵ, V \ E) ≠ 0, so there exists an element v ∈ V \ E such that F (ϵ, v) ≠ 0. Let m be an arbitrary element of M, and let m_{2} := F (ϵ, v)^{−1} • m ∈ M. Then it follows from \overline{vm_{2}} = \overline{v}m_{2} that

ϵF (ϵ, vm_{2}) − vm_{2} = ϵF (ϵ, v)m_{2} − vm_{2}

and hence

ϵF (ϵ, vF (ϵ, v)^{−1}m) = ϵF (ϵ, v)F (ϵ, v)^{−1}m

from which it follows that

ϵm = ϵF (ϵ, vF (ϵ, v)^{−1}m) ∈ ϵW ,

for all m ∈ M. So we have shown that ϵM = ϵW in this case.

Now assume that ϵ ∈ Rad(F). By Lemma 8.11, all elements of V and W have order at most 2, and \overline{v} = v for all v ∈ V.

Since F ≢ 0, there exists an element η ∈ V \ Rad(F). By Lemma 8.22, Rad(F) · K = Rad(F), so E = ϵK ⊆ Rad(F), and hence η ∈ V \ E. It follows that ηm = ηm^{σ} for all m ∈ M, which implies, by Lemma 8.36, that π_{η}(ηm) = ηm^{σ} = ηm. By the definition of π_{η}, this implies that η^{−1}F (η, ηm) = 0, and hence F (η, ηm) = 0 for all m ∈ M. Since F is additive, this in turn implies that F (η, ηK) = 0. Since we chose η ∉ Rad(F), we conclude that V ≠ ηK.

We now show that π_{η}(vw) = π_{η}(v)w for all v ∈ V \ ηK and all w ∈ W. So let v ∈ V \ ηK and w ∈ W be arbitrary. If we substitute η for v and v(δη)^{−1} for c in (Q_{25}), then we get that π_{η}(v)w = π_{η}(v(δη)^{−1}(wη)), and hence π_{η}(π_{η}(v)w) = v(δη)^{−1}(wη). Hence vw + π_{η}(π_{η}(v)w) = vk where k = w + (δη)^{−1} • (wη) ∈ K. On the other hand,

vw + π_{η}(π_{η}(v)w) = vw + π_{η}(vw + ηF (η^{−1}, v)w)
= vw + vw + ηF (η^{−1}, v)w + ηF (η^{−1}, vw + ηF (η^{−1}, v)w)
= ηF (η^{−1}, v)w + ηF (η^{−1}, vw + ηF (η^{−1}, v)w)
∈ ηK,

hence vk ∈ ηK. In a similar way as in the first paragraph, it would follow from vk ≠ 0 that v ∈ ηK, which would contradict the choice of v. Hence vk = vw + π_{η}(π_{η}(v)w) = 0, and thus π_{η}(vw) = π_{η}(v)w.

Again, it follows by induction on n that π_{η}(vw_{1} … w_{n}) = π_{η}(v)w_{1} … w_{n} for all v ∈ V \ ηK and all w_{1}, … w_{n} ∈ W, that is, π_{η}(vm) = π_{η}(v)m for all v ∈ V \ ηK and all m ∈ M.

Since ηK is a proper subgroup of V, V is generated by V \ ηK. Since η ∉ Rad(F), this implies that F (η, V \ ηK) ≠ 0, so there exists an element v ∈ V \ ηK such that F (η, v) ≠ 0. It follows from (Q_{17}) that F (η^{−1}, v) = F (η, v)η^{−1} ≠ 0 as well.

Let m be an arbitrary element of M, and let m_{2} := F (η^{−1}, v)^{−1} • m ∈ M. Then it follows from π_{η}(vm_{2}) = π_{η}(v)m_{2} that

vm_{2} + ηF (η^{−1}, vm_{2}) = vm_{2} + ηF (η^{−1}, v)m_{2}

and hence

ηF (η^{−1}, vF (η^{−1}, v)^{−1}m) = ηF (η^{−1}, v)F (η^{−1}, v)^{−1}m ,

from which it follows that

ηm = ηF (η^{−1}, vF (η^{−1}, v)^{−1}m) ,

for all m ∈ M. So we have shown that ηM = ηW.

Since η ∈ V \ Rad(F) and ϵ ∈ Rad(F), we have that η + ϵ ∈ V \ Rad(F) as well. So the conclusion of the previous paragraph is also valid for η + ϵ, that is, (η + ϵ)M = (η + ϵ)W. Now let m be an arbitrary element of M. Then ηm = ηw_{1} and (η + ϵ)m = (η + ϵ)w_{2} for some w_{1}, w_{2} ∈ W. It follows that

ϵm = ηm + (η + ϵ)m
= ηw_{1} + (η + ϵ)w_{2}
= η(w_{1} + w_{2}) + ϵw_{2} .

If w_{1} + w_{2} ≠ 0, then it would follow from this that

η = (ϵm + ϵw_{2}) · (w_{1} + w_{2})^{−1} ∈ ϵK = E ⊆ Rad(F) ,

which contradicts the choice of η. Hence we must have w_{1} + w_{2} = 0, and it follows that ϵm = ϵw_{2}. Since m was arbitrary, we have shown that ϵM = ϵK also in this case.

This completes the proof of this theorem, and thereby also the proof of Theorem 8.6. □

8.2. Quadrangular Systems of Quadratic Form Type

Our goal in this section is to classify the quadrangular systems which are normal.

Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system which is normal. In particular, Ω is reduced, so H ≡ 0.

Lemma 8.44. K is abelian, i.e. vk_{1}k_{2} = vk_{2}k_{1} for all v ∈ V and all k_{1}, k_{2} ∈ K. Equivalently, σ = 1.

Proof. Note that by the definition of σ, K is abelian if and only if σ = 1. It suffices to show that vw_{1}w_{2} = vw_{2}w_{1} for all v ∈ V and all w_{1}, w_{2} ∈ W. If

v ∈ V \ E, then this follows by substituting w_{1} • w_{2} for m in Theorem 8.43. If v = ϵ, then ϵw_{1}w_{2} = ϵw_{3} for some w_{3} ∈ W since Ω is normal. Hence, by Theorem 8.19, ϵw_{1}w_{2} = ϵw_{3} = \overline{ϵw_{3}} = \overline{ϵw_{1}w_{2}} = ϵw_{2}w_{1}. Finally, assume that v = ϵk for some k ∈ K. Then vw_{1}w_{2} = ϵkw_{1}w_{2} = \overline{ϵw_{2}w_{1}k^{σ}} = \overline{ϵw_{1}w_{2}k^{σ}} = ϵkw_{2}w_{1} = vw_{2}w_{1}, again by Theorem 8.19. This shows the lemma in all possible cases. □

Lemma 8.45. For all v ∈ V, we have that vK = vW.

Proof. It suffices to show that vM = vW.

First, assume that v ∈ E, say v = ϵk for some k ∈ K. Let m ∈ M be arbitrary. Then ϵm = ϵw for some w ∈ W, since Ω is normal. Since K is abelian by Lemma 8.44, it follows that vm = ϵkm = ϵmk = ϵwk = ϵkw = vw, which shows that vM = vW in this case.

Now, assume that v ∈ V \E. Note that it is sufficient to show that vw_{1}w_{2} ∈ vW for all w_{1}, w_{2} ∈ W; it then follows by induction that vM = vW. Choose two arbitrary elements w_{1}, w_{2} ∈ W. By Lemma 8.40(ii) and Lemma 8.44, we have that vw_{1}w_{2}w_{3} + vw_{1}w_{2}w_{3} = vF (ϵw_{1}, ϵw_{2}w_{3}), or equivalently, vw_{1}w_{2}w_{3}(δ + δ) = vF (ϵw_{1}, ϵw_{2}w_{3}), for all w_{3} ∈ W.

We now distinguish two cases. First, assume that δ + δ = 0. It then follows, by (Q_{12}), that all elements of V and W have order at most 2. Since Ω is normal, there exists a w ∈ W such that ϵw_{1}w_{2} = ϵw. By the previous paragraph, F (ϵw_{1}, ϵw_{2}w) = 0, and hence, by Lemma 8.22 and Lemma 8.40(i), F (vw_{1}w_{2}, vw) = F (vw_{1}, vw_{2}w) = F (ϵw_{1}, ϵw_{2}w)v = 0 as well. By Lemma 8.15, it follows that

δ(vw_{1}w_{2} + vw) = δ · vw_{1}w_{2} + δ · vw + F (vw_{1}w_{2}, vw)
= δ · ϵw_{1}w_{2} · v + δ · ϵw · v
= δ · ϵw · v + δ · ϵw · v
= 0 ,

which implies that vw_{1}w_{2} + vw = 0, hence vw_{1}w_{2} = vw ∈ vW, which is what we had to show.

Now, assume that δ + δ ≠ 0. Then we set w_{3} = (δ + δ)^{−1} in the identity vw_{1}w_{2}w_{3}(δ + δ) = vF (ϵw_{1}, ϵw_{2}w_{3}), which yields

vw_{1}w_{2} = vF (ϵw_{1}, ϵw_{2}(δ + δ)^{−1}) ∈ vW ,

which proves the lemma in this case as well. □

Lemma 8.46. Let w ∈ W and k ∈ K be such that vk = vw for some v ∈ V^{∗}. Then k = w.

Proof. We will show that uk = uw for all u ∈ V. We distinguish two cases. First, assume that u ∈ vK. Since vK = vW by Lemma 8.45, there exists a

z ∈ W such that u = vz. Then uk = vzk = vkz = vwz = vzw = uw, since K is abelian.

Now, assume that u ∉ vK. By Lemma 8.45, there exists a w_{2} ∈ W such that uk = uw_{2}, and there exists a w_{3} ∈ W such that (v + u)k = (v + u)w_{3}. We have to show that w = w_{2}. We have that vw_{3} + uw_{3} = (v + u)w_{3} = (v + u)k = vk + uk = vw + uw_{2}, from which it follows that u(w_{3} − w_{2}) = v(w − w_{3}). Since u ∉ vK, this can only occur if w_{3} − w_{2} = 0, and then w − w_{3} = 0 as well. Hence w = w_{3} = w_{2}, which is what we had to show. □

Remark 8.47. It follows from this lemma that if k_{1}, k_{2} ∈ K are such that vk_{1} = vk_{2} for some v ∈ V^{∗}, then k_{1} = k_{2}, since, by Lemma 8.45, there exists a w ∈ W such that vk_{1} = vw = vk_{2}.

Theorem 8.48. K_{+,•} is a commutative field.

Proof. We have already shown in Lemma 8.44 that K is a commutative ring. It only remains to show that every element of K^{∗} is invertible. Let k be an arbitrary non-zero element of K. Since Ω is normal, ϵk = ϵw for some w ∈ W; hence by Lemma 8.46, k = w. It follows that k is invertible with inverse k^{−1} = w^{−1}, since w • w^{−1} = w^{−1} • w = δ. □

By Lemma 8.46, (K, +) ≅ W as additive groups. We will denote the isomorphism by square brackets, that is, for every t ∈ K, we will denote the corresponding element of W by [t]. Since K is a commutative field, V is a (left) vector space over K, with scalar multiplication given by tv := v[t], for all t ∈ K and all v ∈ V. From now on, we will denote the multiplicative identity of K by 1 in place of δ. Then δ = [1] ∈ W. If there is no danger of confusion, we will also write st in place of s • t for s, t ∈ K, and t^{2} in place of t • t for t ∈ K.

Definition 8.49. We define a map q from V to K by setting [q(v)] = δv = [1]v for all v ∈ V. Furthermore, we define a map f from V × V to K by setting [f(v_{1}, v_{2})] = F (v_{1}, v_{2}) for all v_{1}, v_{2} ∈ V.

Lemma 8.50. For all v ∈ V, all w ∈ W and all t ∈ K, we have that

(i) \overline{vw} = \overline{v}w; (ii) \overline{tv} = t\overline{v}.

Proof. We first show (i). If v ∈ ϵW, then vw ∈ ϵW as well, and it follows from Theorem 8.19 and Lemma 8.44 that \overline{vw} = vw = \overline{v}w (remember that σ = 1). If v ∉ ϵW = ϵK, then we have already shown this in Theorem 8.43.

Identity (ii) now follows by substituting [t] for w in (i). □

Lemma 8.51. For all v ∈ V and all t ∈ K, we have that [t]v = [tq(v)].

Proof. Let w := [t] ∈ W. We have to show that wv = w • δv. By (Q_{16}) and Lemma 8.50(i), v^{−1} · wv = \overline{vw} = \overline{v}w. On the other hand, v^{−1} · (w • δv) = v^{−1} · δv · w = \overline{v}w, again by (Q_{16}). Hence v^{−1} · wv = v^{−1} · (w • δv), which implies by Lemma 8.46 that wv = w • δv. □

Lemma 8.52. For all u, v ∈ V and all t ∈ K, we have that π_{u}(tv) = tπ_{u}(v).

Proof. Let w := [t] ∈ W. If we substitute u for v and \overline{v}(δu)^{−1} for c in (Q_{25}), then we get that π_{u}(v)w = π_{u}(\overline{\overline{v}(δu)^{−1}(wu)}), and hence, by Lemma 8.50(i) and Lemma 8.51, that

tπ_{u}(v) = π_{u}(v)w
= π_{u}(v(δu)^{−1}(wu))
= π_{u}(v · [q(u)^{−1}] · [tq(u)])
= π_{u}(tq(u)q(u)^{−1}v)
= π_{u}(tv) ,

which is what we had to show. □

Theorem 8.53. q is an anisotropic quadratic form from V to K with corresponding bilinear form f.

Proof. Let v ∈ V and t ∈ K be arbitrary, and let w := [t] ∈ W. Then, by (Q_{26}),

[q(tv)] = [q(vw)] = δ · vw = δ · ϵw · v = δv · ϵw .

By Lemma 8.46, it follows from Lemma 8.38 that w • w • z = z · ϵw for all z ∈ W. Hence

[q(tv)] = δv · ϵw = w • w • δv = [t] • [t] • [q(v)] = [t^{2}q(v)] .

Next, it follows from (Q_{11}) that for all u, v ∈ V, [q(u + v)] = δ(u + v) = δu + δv + F (u, v) = [q(u)] + [q(v)] + [f(u, v)] = [q(u) + q(v) + f(u, v)]. We now show that f is bilinear over K. Let u, v ∈ V^{∗} and t ∈ K be arbitrary. By Lemma 8.52, we have that π_{u}(tv) = tπ_{u}(v). By the definition of π_{u}, this yields

tv − \overline{u^{−1}F (u, tv)} = tv − t\overline{u^{−1}F (u, v)} .

By Lemma 8.50(ii), it follows that u^{−1}F (u, tv) = tu^{−1}F (u, v), hence

u^{−1} · [f(u, tv)] = u^{−1} · [f(u, v)] · [t] = u^{−1} · [tf(u, v)] .

By Lemma 8.46, this implies that f(u, tv) = tf(u, v). Since f is symmetric, it follows from this that f is bilinear over K.

Finally, q is anisotropic, since q(v) = 0 implies that δv = 0 and hence v = 0. □

Lemma 8.54. For all u, v ∈ V, we have that q(\overline{v}) = q(v) and f(\overline{u}, \overline{v}) = f(u, v).

Proof. We have that

q(\overline{v}) = q(f(ϵ, v)ϵ − v)
= q(f(ϵ, v)ϵ) + q(v) − f(f(ϵ, v)ϵ, v)
= f(ϵ, v)^{2}q(ϵ) + q(v) − f(ϵ, v)f(ϵ, v)
= q(v) ,

and hence

f(\overline{u}, \overline{v}) = q(\overline{u} + \overline{v}) − q(\overline{u}) − q(\overline{v})
= q(u + v) − q(u) − q(v)
= f(u, v)

as well. □

Theorem 8.55. (V, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{Q}(K, V, q).

Proof. Observe that q(ϵ) = 1, since [q(ϵ)] = δϵ = δ = [1].

Let ϕ be the isomorphism from [V] to V which maps [v] to v for all v ∈ V, and let ψ be the isomorphism from [K] to W which maps [t] to [t] for all t ∈ W. Then ϕ([ϵ]) = ϵ and ψ([1]) = [1] = δ. Furthermore,

ϕ([v][t]) = ϕ([tv]) = tv = v[t] = ϕ([v])ψ([t]) , and
ψ([t][v]) = ψ([tq(v)]) = [tq(v)] = [t]v = ψ([t])ϕ([v]) ,

for all t ∈ K and all v ∈ V. Hence (ϕ, ψ) is an isomorphism from Ω_{Q}(K, V, q) to (V, W, τ_{V} , τ_{W} , ϵ, δ). □

This completes the proof of Theorem 8.7.

8.3. Quadrangular Systems of Indifferent Type

Our goal in this section is to classify the quadrangular systems which are indifferent.

Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a quadrangular system which is indifferent. Then F ≡ 0 and H ≡ 0. By Lemma 8.11, all elements of V and W have order 1 or 2, and for all v ∈ V, we have \overline{v} = v. Furthermore, we have π_{v}(c) = c for all v, c ∈ V.

Lemma 8.56. K is abelian, i.e. vk_{1}k_{2} = vk_{2}k_{1} for all v ∈ V and all k_{1}, k_{2} ∈ K.

Proof. By Lemma 8.36, we have that π_{v}(vk) = vk^{σ} and hence vk = vk^{σ} for all v ∈ V and all k ∈ K. It follows that vk_{1}k_{2} = v(k_{1} • k_{2}) = v(k_{1} • k_{2})^{σ} = v(k_{2}^{σ} • k_{1}^{σ}) = vk_{2}^{σ}k_{1}^{σ} = vk_{2}k_{1} for all v ∈ V and all k_{1}, k_{2} ∈ K. □

Lemma 8.57. For all v ∈ V and all k ∈ K, we have that vkk = v · δ(ϵk).

Proof. It already follows from Lemma 8.39 that vmm = v · δ(ϵm) for all m ∈ M. Now suppose that vk_{1}k_{1} = v · δ(ϵk_{1}) and vk_{2}k_{2} = v · δ(ϵk_{2}) for some k_{1}, k_{2} ∈ K. We will then show that v(k_{1} + k_{2})(k_{1} + k_{2}) = v · δ(ϵ(k_{1} + k_{2})), which will prove the lemma.

By (Q_{11}), (Q_{12}) and Lemma 8.56, we have that

v(k_{1} + k_{2})(k_{1} + k_{2}) = vk_{1}k_{1} + vk_{2}k_{2} + vk_{1}k_{2} + vk_{2}k_{1}
= v · δ(ϵk_{1}) + v · δ(ϵk_{2}) + vk_{1}k_{2} + vk_{1}k_{2}
= v · (δ(ϵk_{1}) + δ(ϵk_{2}))
= v · δ(ϵk_{1} + ϵk_{2})
= v · δ(ϵ(k_{1} + k_{2})) ,

and we are done. □

Lemma 8.58. For all v ∈ V and all k ∈ K, we have that δ · vk = δ · ϵk · v.

Proof. In Lemma 8.15, we have already shown this for all k ∈ M. Now suppose that δ · vk_{1} = δ · ϵk_{1} · v and δ · vk_{2} = δ · ϵk_{2} · v for some k_{1}, k_{2} ∈ K. We will then show that δ · v(k_{1} + k_{2}) = δ · ϵ(k_{1} + k_{2}) · v, which will prove the lemma.

By (Q_{11}), we have that

δ · v(k_{1} + k_{2}) = δ · (vk_{1} + vk_{2})
= δ · vk_{1} + δ · vk_{2}
= δ · ϵk_{1} · v + δ · ϵk_{2} · v
= δ · (ϵk_{1} + ϵk_{2}) · v
= δ · ϵ(k_{1} + k_{2}) · v ,

and we are done. □

Theorem 8.59. K_{+,•} is a commutative field of characteristic 2 with multiplicative identity δ.

Proof. We have already shown in Lemma 8.56 that K_{+,•} is a commutative ring. Let k ∈ K be arbitrary. If δ(ϵk) = 0, then it would follow from Lemma 8.58 that vk = 0 for all v ∈ V and thus k = 0. Hence δ(ϵk) is invertible for all k ≠ 0, and it then follows from Lemma 8.57 that vkk(δ(ϵk))^{−1} = v for all v ∈ V. This implies that k is invertible with inverse k^{−1} := k • (δ(ϵk))^{−1}.

Furthermore, for all v ∈ V and all k ∈ K, we have that v(k + k) = vk + vk = 0, hence k + k = 0, so char(K) = 2. □

Lemma 8.60. If vk_{1} = vk_{2} for some v ∈ V^{∗} and some k_{1}, k_{2} ∈ K, then k_{1} = k_{2}.

Proof. If vk_{1} = vk_{2} for some v ∈ V^{∗} and some k_{1}, k_{2} ∈ K, then v(k_{1}+k_{2}) = 0. If we would have that k_{1} ≠ k_{2}, then k_{1} + k_{2} would be invertible, and it would then follow that v = v(k_{1} +k_{2})(k_{1} +k_{2})^{−1} = 0, a clear contradiction. Hence k_{1} = k_{2}. □

Theorem 8.61. (K, W, δV) is an indifferent set. Moreover, δv • w = wv and w • w • δv = δ · vw for all v ∈ V and all w ∈ W.

Proof. It is obvious that W is a subgroup of (K, +). Since δv_{1} + δv_{2} = δ(v_{1} + v_{2}) by (Q_{11}), δV is a subgroup of (K, +) as well. Furthermore, both W and δV contain the multiplicative identity δ.

By (Q_{25}), ϵ · δv · w = ϵ · wv for all v ∈ V and all w ∈ W. It follows by Lemma 8.60 that δv • w = wv, and hence δV • W ⊆ W.

By (Q_{26}), we have that δ · vw = δ · ϵw · v = δv · ϵw, for all v ∈ V and all w ∈ W. By Lemma 8.60, it follows from Lemma 8.38 that w • w • z = z · ϵw for all z ∈ W. Hence δv · ϵw = w • w • δv. It follows that w • w • δv = δ · vw, and hence W^{2} • δV ⊆ δV.

Finally, it follows from the definition of K that K is generated by W as a ring. This shows that (K, W, δV) is an indifferent set. □

Theorem 8.62. (V, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{D}(K, W, δV).

Proof. First, we observe that v is uniquely determined by δv, since δ(v_{1} + v_{2}) = δv_{1} + δv_{2} by (Q_{11}).

Let ϕ be the isomorphism from [δV] to V which maps [δv] to v for all v ∈ V, and let ψ be the isomorphism from [W] to W which maps [w] to w for all w ∈ W.

Then ϕ([δ]) = ϕ([δϵ]) = ϵ and ψ([δ]) = δ. Furthermore, it follows from Theorem 8.61 that

ϕ([δv][w]) = ϕ([w^{2} • δv]) = ϕ([δ · vw]) = vw = ϕ([δv])ψ([w]) , and
ψ([w][δv]) = ψ([δv • w]) = ψ([wv]) = wv = ψ([w])ϕ([δv]) ,

for all v ∈ V and all w ∈ W. Hence (ϕ, ψ) is an isomorphism from Ω_{D}(K, W, δV) to (V, W, τ_{V} , τ_{W} , ϵ, δ). □

This completes the proof of Theorem 8.8, and thereby the classification of all reduced quadrangular systems.

8.4. Quadrangular Systems of Pseudo-quadratic Form Type, I

Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a wide quadrangular system which is the extension of a reduced quadrangular system Λ of proper involutory type; more precisely, let Λ = (V, Rad(H), τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{I}(K, K_{0}, σ) with K = ⟨K_{0}⟩ and σ ≠ 1, where τ_{V} and τ_{W} are as in Remark 8.3. In particular, V = [K].

Definition 8.63. Let Y := Rad(H). Note that Y is a normal subgroup of W since Y ⊆ Z(W) by (Q_{8}); let W̃ := W/Y. Let ι be the canonical surjection from W to W̃. We will also write w̃ in place of ι(w), for all w ∈ W. Then w̃_{1} = w̃_{2} if and only if w_{1} ⊟ w_{2} ∈ Y.

By (Q_{8}) and (Q_{7}), [w_{1}, w_{2}] ∈ Im(F) ⊆ Y, hence W̃ is abelian; we will use the additive notations + and − for W̃. We can define a map τ̃_{W} from W̃ × V to W̃, which we will denote by · or by juxtaposition, by setting

τ̃_{W} (w̃, v) := w̃ · v := w̃v

for all v ∈ V and all w ∈ W. This is well defined: let w̃_{1} = w̃_{2}, then w_{1} ⊟ w_{2} ∈ Y, and hence w_{1}v ⊟ w_{2}v = (w_{1} ⊟ w_{2})v ∈ Y since Y · V = Y; it then follows that w̃_{1}v = w̃_{2}v.

Remark 8.64. If s ∈ K_{0}, then the notation [s] is ambiguous. If we want to make clear whether we mean [s] ∈ V or [s] ∈ W, we will write [s]_{V} and [s]_{W}, respectively. Note that [s]_{W} ∈ Y for all s ∈ K_{0}, and that ϵ[s]_{W} = [s]_{V} for all s ∈ K_{0}.

Theorem 8.65. W̃ is a right vector space over K, with scalar multiplication given by w̃t := w̃ · [t], for all t ∈ K and all w̃ ∈ W̃.

Proof. We have that

(w̃_{1} + w̃_{2})t = (w̃_{1} + w̃_{2}) · [t] = ι((w_{1} + w_{2}) · [t])
= ι(w_{1} · [t]) + ι(w_{2} · [t]) = w̃_{1} · [t] + w̃_{2} · [t] = w̃_{1}t + w̃_{2}t

for all t ∈ K and all w_{1}, w_{2} ∈ W. By (Q_{11}) and (Q_{7}),

w̃(t_{1} + t_{2}) = w̃ · [t_{1} + t_{2}] = w̃ · ([t_{1}] + [t_{2}]) = ι(w · ([t_{1}] + [t_{2}]))
= ι(w · [t_{1}] ⊞ w · [t_{2}] ⊞ F ([t_{2}]w, [t_{1}]))
= ι(w · [t_{1}]) + ι(w · [t_{2}]) = w̃ · [t_{1}] + w̃ · [t_{2}] = w̃t_{1} + w̃t_{2}

for all t_{1}, t_{2} ∈ K and all w ∈ W.

It only remains to show that w̃(t_{1}t_{2}) = (w̃t_{1})t_{2} for all t_{1}, t_{2} ∈ K and all w ∈ W. (The other axioms for a vector space are obvious.) We thus have to check that ι(w · [t_{1}t_{2}]) = ι(w · [t_{1}] · [t_{2}]). Since K = ⟨K_{0}⟩, it suffices to show this for t_{1} ∈ K_{0}; the result for t_{1} = s_{1} … s_{n} with s_{1}, …, s_{n} ∈ K_{0} will then follow by induction on n, and since we have already shown that w̃ ·[t_{3} +t_{4}] = w̃ ·[t_{3}]+w̃ ·[t_{4}] for all t_{3}, t_{4} ∈ K, the result then follows for all t_{1} ∈ K.

By Remark 8.64, (Q_{26}) and the definition of Ω_{I}(K, K_{0}, σ), we have that

w · [s]_{V} · [t] = w · ϵ[s]_{W} · [t]
= w · [t][s]_{W}
= w · [st]_{V}

and hence ι(w · [s]_{V} · [t]) = ι(w · [st]_{V}) for all s ∈ K_{0} and all t ∈ K, which is what we had to show. □

Definition 8.66. Let π be the map from W̃ to V/[K_{0}]_{V} which maps w̃ to ϵw (mod [K_{0}]_{V}). This map is well defined: let w_{1}, w_{2} ∈ W be such that w̃_{1} = w̃_{2}. Then w_{1} ⊟ w_{2} ∈ Y, hence

ϵw_{1} − ϵw_{2} = ϵ((w_{1} ⊟ w_{2}) ⊞ w_{2}) − ϵw_{2}
= ϵ(w_{1} ⊟ w_{2}) + ϵw_{2} − ϵw_{2}
= ϵ(w_{1} ⊟ w_{2}) ∈ ϵY = ϵ[K_{0}]_{W} = [K_{0}]_{V}

by (Q_{12}).

Lemma 8.67. For all w̃ ∈ W̃, we have that π(w̃) = 0 if and only if w̃ = 0.

Proof. Let w ∈ W be such that π(w̃) = 0. Then ϵw ∈ [K_{0}]_{V}, say ϵw = [s]_{V} with s ∈ K_{0}. By 3.13(ii), ϵ(⊟[s]_{W}) = −ϵ[s]_{W} since [s]_{W} ∈ Y. It follows that ϵ(w ⊟ [s]_{W}) = ϵw + ϵ(⊟[s]_{W}) = ϵw − ϵ[s]_{W} = ϵw − [s]_{V} = 0, and hence w = [s]_{W} ∈ Y. It follows that w̃ = 0. □

Definition 8.68. Let h be the map from W̃ ×W̃ to K, defined by the identity [h(w̃_{1}, w̃_{2})] := H(w_{1}, w_{2}) for all w̃_{1}, w̃_{2} ∈ W̃. Since W̃ = W/ Rad(H), the map h is well defined.

Lemma 8.69. For all v ∈ V, all w ∈ W and all y_{1}, …, y_{n} ∈ Y, we have that \overline{\overline{v}w}y_{1} … y_{n} = \overline{\overline{vy_{1} … y_{n}}w}.

Proof. We will first prove the lemma for n = 1. Let v ∈ V, w ∈ W and y ∈ Y. We may assume that w ≠ 0. Note that Π_{w}(y) = y since y ∈ Rad(H). It follows from (Q_{23}) that \overline{vy} = −\overline{\overline{\overline{v}(⊟w)}y}κ(w), and hence, by (Q_{18}), \overline{\overline{vy}(⊟w)} = \overline{\overline{v}(⊟w)}y. Substituting ⊟w for w now yields the result for n = 1.

We advance to general n by induction. Let v ∈ V, w ∈ W and y_{1}, …, y_{n} ∈ Y, and suppose that \overline{\overline{v}w}y_{1} … y_{n−1} = \overline{\overline{vy_{1} … y_{n−1}}w}. Then

\overline{\overline{v}w}y_{1} … y_{n} = \overline{\overline{vy_{1} … y_{n−1}}w} · y_{n}
= \overline{\overline{vy_{1} … y_{n−1}y_{n}}w} ,

where we have used the lemma for n = 1 in the last equality. This completes the proof of this lemma. □

Lemma 8.70. For all w ∈ W and all y_{1}, …, y_{n} ∈ Y, we have that \overline{ϵw}y_{1} … y_{n} = \overline{ϵy_{n} … y_{1}w}. In particular, for all w ∈ W and all s ∈ K_{0}, we have that \overline{[s]_{V} w} = \overline{ϵw}[s]_{W}.

Proof. By Theorem 8.19, \overline{ϵy_{1} … y_{n}} = ϵy_{n} … y_{1}. The first result follows by substituting ϵ for v in Lemma 8.69.

In the particular case n = 1 and y_{1} = [s]_{W}, we get that \overline{ϵw}[s]_{W} = \overline{ϵ[s]_{W} w} = \overline{[s]_{V} w}. □

Theorem 8.71. The map h is a skew-hermitian form over K with respect to σ.

Proof. Since \overline{H(w_{1}, w_{2})} = −H(w_{2}, w_{1}) for all w_{1}, w_{2} ∈ W by (Q_{22}), and since \overline{[t]} = [t^{σ}] for all t ∈ K, it follows that h(w̃_{1}, w̃_{2}) = −h(w̃_{2}, w̃_{1})^{σ}, for all w̃_{1}, w̃_{2} ∈ W̃.

By (Q_{22}), (Q_{12}) and Lemma 8.70,

H(w_{1} · [s]_{V} , w_{2}) = −\overline{H(w_{2}, w_{1} · [s]_{V})}
= −\overline{[s]_{V} · (w_{1} ⊞ w_{2})} + \overline{[s]_{V} · w_{1}} + \overline{[s]_{V} · w_{2}}
= −\overline{ϵ(w_{1} ⊞ w_{2})}[s]_{W} + \overline{ϵw_{1}}[s]_{W} + \overline{ϵw_{2}}[s]_{W}
= −\overline{H(w_{2}, w_{1})} · [s]_{W}
= H(w_{1}, w_{2}) · [s]_{W} ,

for all w_{1}, w_{2} ∈ W and all s ∈ K_{0}. Hence

[h(w̃_{1}s, w̃_{2})] = [h(w̃_{1}, w̃_{2})] · [s]_{W} = [sh(w̃_{1}, w̃_{2})] ,

from which it follows that h(w̃_{1}s, w̃_{2}) = sh(w̃_{1}, w̃_{2}) for all w_{1}, w_{2} ∈ W and all s ∈ K_{0}. Since K = ⟨K_{0}⟩ and since h is additive in both variables, it follows from this that h(w̃_{1}t, w̃_{2}) = t^{σ}h(w̃_{1}, w̃_{2}) for all w_{1}, w_{2} ∈ W and all t ∈ K.

Finally, h(w̃_{1}, w̃_{2}t) = −h(w̃_{2}t, w̃_{1})^{σ} = −(t^{σ}h(w̃_{2}, w̃_{1}))^{σ} = −h(w̃_{2}, w̃_{1})^{σ}t = h(w̃_{1}, w̃_{2})t for all w_{1}, w_{2} ∈ W and all t ∈ K. This shows that h is a skew-hermitian form over K with respect to σ. □

Definition 8.72. For all t ∈ K, let k_{t} be the homomorphism from V to itself which maps [t′] to [tt′] for all [t′] ∈ [K] = V. We denote the action of k_{t} by right juxtaposition, i.e. [t′]k_{t} = [tt′] for all t, t′ ∈ K. In particular, we can identify k_{s} and [s]_{W} for all s ∈ K_{0}. Moreover, we set k_{t}^{σ} := k_{t^{σ}} for all t ∈ K. Note that the set {k_{t} | t ∈ K} coincides with the set K that we defined in the beginning of section 8.1. In particular, we can apply the lemmas and theorems of that section on the sub-quadrangular system Λ = (V, Y, τ_{V} , τ_{W} , ϵ, δ).

Lemma 8.73. For all w ∈ W and all t_{1}, t_{2} ∈ K, we have that
ϵk_{1}wk_{2}^{σ} + ϵk_{2}wk_{1}^{σ} = ϵF ([t_{2}]w, [t_{1}]) + H(w[t_{2}], w[t_{1}]) ,

where k_{1} := k_{t_{1}} and k_{2} := k_{t_{2}}.

Proof. By Lemma 8.22 and the definition of the map v ↦ \overline{v}, we have that

ϵF ([t_{2}]w, [t_{1}]) = ϵF (ϵk_{2}w, ϵk_{1})
= ϵF (ϵk_{2}wk_{1}^{σ}, ϵ)
= ϵk_{2}wk_{1}^{σ} + \overline{ϵk_{2}wk_{1}^{σ}} ,

and by Theorem 8.71 and 3.13(ii), we have that

H(w[t_{2}], w[t_{1}]) = [h(w̃t_{2}, w̃t_{1})]
= [t_{2}^{σ}h(w̃, w̃t_{1})]
= [h(w̃, w̃t_{1})]k_{2}^{σ}
= H(w, w[t_{1}])k_{2}^{σ}
= [t_{1}](⊟w)k_{2}^{σ} + [t_{1}]wk_{2}^{σ}
= ϵk_{1}(⊟w)k_{2}^{σ} + ϵk_{1}wk_{2}^{σ} .

It only remains to show that \overline{ϵk_{2}wk_{1}^{σ}} = −ϵk_{1}(⊟w)k_{2}^{σ}. By Lemma 8.70, we have that ϵk_{t}^{σ}w = \overline{\overline{ϵw}k_{t}} for all w ∈ W and all t ∈ K. By (Q_{16}) with ϵ in place of v and Lemma 8.18 with ϵ(⊟w) in place of v, it follows that

\overline{ϵk_{2}wk_{1}^{σ}} = \overline{\overline{\overline{ϵw}k_{2}^{σ}}k_{1}^{σ}}
= −\overline{\overline{ϵ(⊟w)k_{2}^{σ}}k_{1}^{σ}}
= −\overline{\overline{ϵ(⊟w)}k_{1}^{σ}}k_{2}^{σ}
= −ϵk_{1}(⊟w)k_{2}^{σ} ,

which completes the proof of this lemma. □

Theorem 8.74. For all w ∈ W and all t ∈ K, we have that ϵ · w[t] = ϵk_{t}wk_{t}^{σ}.

Proof. First assume that t = s_{1} … s_{n} with s_{1}, …, s_{n} ∈ K_{0}. Let y_{i} := [s_{i}]_{W} ∈ Y for all i ∈ {1, …, n}. Then we have to show that ϵ · w(ϵy_{1} … y_{n}) = ϵy_{1} … y_{n}wy_{n} … y_{1} for all w ∈ W. By (Q_{16}), Lemma 8.27 and Lemma 8.70,

ϵy_{1} … y_{n}w = −\overline{(ϵy_{1} … y_{n})^{−1}(⊟w · ϵy_{1} … y_{n})}
= −\overline{ϵy_{n}^{−1} … y_{1}^{−1}(⊟w · ϵy_{1} … y_{n})}
= −\overline{ϵ(⊟w · ϵy_{1} … y_{n})}y_{1}^{−1} … y_{n}^{−1}
= ϵ(w · ϵy_{1} … y_{n})y_{1}^{−1} … y_{n}^{−1} ,

from which it follows that ϵ(w · ϵy_{1} … y_{n}) = ϵy_{1} … y_{n}wy_{n} … y_{1}.

Now suppose that ϵ · w[t_{1}] = ϵk_{1}wk_{1}^{σ} and ϵ · w[t_{2}] = ϵk_{2}wk_{2}^{σ} for some t_{1}, t_{2} ∈ K, where where k_{1} := k_{t_{1}} and k_{2} := k_{t_{2}}. We will show that

ϵ · w[t_{1} + t_{2}] = ϵ(k_{1} + k_{2})w(k_{1} + k_{2})^{σ} ,

which will prove the theorem for all t ∈ K, since K = ⟨K_{0}⟩.

By (Q_{11}), (Q_{12}) with v = ϵ, (Q_{7}) and Lemma 8.73, we have that

ϵ · w[t_{1} + t_{2}] = ϵ · w([t_{1}] + [t_{2}])
= ϵ · (w[t_{1}] ⊞ w[t_{2}] ⊞ F ([t_{2}]w, t_{1}))
= ϵ · w[t_{1}] + ϵ · w[t_{2}] + ϵF ([t_{2}]w, t_{1}) + H(w[t_{2}], w[t_{1}])
= ϵk_{1}wk_{1}^{σ} + ϵk_{2}wk_{2}^{σ} + ϵk_{1}wk_{2}^{σ} + ϵk_{2}wk_{1}^{σ}
= ϵ(k_{1} + k_{2})w(k_{1} + k_{2})^{σ} ,

which completes the proof of this theorem. □

Lemma 8.75. Let w ∈ W and t ∈ K be arbitrary. Let x ∈ K be such that ϵw = [x]. Then [t]w = [xt], and ϵk_{t}wk_{t}^{σ} = [t^{σ}xt].

Proof. By Lemma 8.70, we have that

[t]w = ϵk_{t}w = \overline{\overline{ϵw}k_{t}^{σ}} = \overline{[x^{σ}]k_{t}^{σ}} = \overline{[t^{σ}x^{σ}]} = [xt] ;

it then follows that

ϵk_{t}wk_{t}^{σ} = [xt]k_{t}^{σ} = [t^{σ}xt] ,

and we are done. □

Definition 8.76. For all w̃ ∈ W̃, let p(w̃) be any element t ∈ K such that [t] is contained in the coset π(w̃) ∈ V/[K_{0}]_{V}. Hence p is a map from W̃ to K such that [p(w̃)] ≡ ϵw (mod [K_{0}]_{V}).

Theorem 8.77. (K, K_{0}, σ, W̃, p) is an anisotropic pseudo-quadratic space with corresponding skew-hermitian form h.

Proof. It only remains to show that p is a pseudo-quadratic form. All equivalences will be modulo [K_{0}]_{V}. By (Q_{12}), we have that

[p(w̃_{1} + w̃_{2})] ≡ [p(w̃_{2} + w̃_{1})]
≡ ϵ(w_{2} ⊞ w_{1})
≡ ϵw_{2} + ϵw_{1} + H(w_{1}, w_{2})
≡ [p(w̃_{1})] + [p(w̃_{2})] + [h(w̃_{1}, w̃_{2})]

for all w_{1}, w_{2} ∈ W, which shows the first property.

Next, let w be an arbitrary element of W, and let x ∈ K be such that ϵw = [x]. Then [p(w̃)] ≡ [x], and hence [t^{σ}p(w̃)t] ≡ [t^{σ}xt], since t^{σ}K_{0}t ⊆ K_{0}. It follows from Theorem 8.74 and Lemma 8.75 that

[p(w̃t)] ≡ ϵ · w[t] ≡ ϵk_{t}wk_{t}^{σ} ≡ [t^{σ}xt] ≡ [t^{σ}p(w̃)t] ,

which shows the second property.

Finally, if [p(w̃)] ≡ 0 for some w̃ ∈ W̃, then π(w̃) = 0 and hence w̃ = 0 by Lemma 8.67. □

Lemma 8.78. Let the group T be as in section 7.4. For each element (a, x) ∈ T, there is a unique element w ∈ W such that w ∈ a and ϵw = [x]. If we denote this element by χ(a, x), then χ is an isomorphism from T to W.

Proof. Let (a, x) ∈ T be arbitrary. Choose an arbitrary element z ∈ a. Then a = z̃, and ϵz ≡ [x] (mod [K_{0}]_{V}) by the definition of T. Hence ϵz − [x] ∈ [K_{0}]_{V} = ϵY, say ϵz − [x] = ϵy with y ∈ Y. Set w = z ⊟ y, then w̃ = z̃ = a, and ϵw = ϵ(z ⊟ y) = ϵz − ϵy = [x] by (Q_{12}) and 3.13(ii). This shows the existence of w.

Now suppose that w_{1}, w_{2} ∈ W are such that w̃_{1} = w̃_{2} and ϵw_{1} = ϵw_{2}. Then w_{1} ⊟ w_{2} ∈ Y, and hence, by (Q_{12}),

0 = ϵw_{1} − ϵw_{2}
= ϵ((w_{1} ⊟ w_{2}) ⊞ w_{2}) − ϵw_{2}
= ϵ(w_{1} ⊟ w_{2}) + ϵw_{2} − ϵw_{2}
= ϵ(w_{1} ⊟ w_{2}) ,

from which it follows that w_{1} = w_{2}.

Hence χ : T → W is a well defined map, which is bijective, with the inverse map given by χ^{−1}(w) = (w̃, x) ∈ T, where [x] = ϵw. In order to show that χ is an isomorphism, it now suffices to show that χ^{−1}(w_{1} ⊞ w_{2}) = χ^{−1}(w_{1}) ⊞ χ^{−1}(w_{2}).

Let x_{1}, x_{2} ∈ K be such that [x_{1}] = ϵw_{1} and [x_{2}] = ϵw_{2}. Then ϵ(w_{1} ⊞ w_{2}) = ϵw_{1} + ϵw_{2} + H(w_{2}, w_{1}) = [x_{1}] + [x_{2}] + [h(w̃_{2}, w̃_{1})], and hence

χ^{−1}(w_{1} ⊞ w_{2}) = (w̃_{1} + w̃_{2}, x_{1} + x_{2} + h(w̃_{2}, w̃_{1}))
= (w̃_{1}, x_{1}) ⊞ (w̃_{2}, x_{2})
= χ^{−1}(w_{1}) ⊞ χ^{−1}(w_{2}) ,

which completes the proof of this lemma. □

Theorem 8.79. (V, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{P} (K, K_{0}, σ, W̃, p).

Proof. Let ϕ be the isomorphism from [K] to V which maps [t] to [t] for all t ∈ K, and let ψ be the isomorphism from [T] to W which maps [a, x] to χ(a, x) for all (a, x) ∈ T. Then ϕ([1]) = [1] = ϵ and ψ([0, 1]) = δ since δ ∈ Y (hence δ̃ = 0) and ϵδ = [1].

Now, let t ∈ K and (a, x) ∈ T be arbitrary. Let w = ψ([a, x]) = χ(a, x), then a = w̃ and ϵw = [x]. By Lemma 8.75, [xt] = [t]w, hence

ϕ([t][a, x]) = ϕ([xt]) = [xt] = [t]w = ϕ([t])ψ([a, x]) , and
ψ([a, x][t]) = ψ([at, t^{σ}xt]) = w[t] = ψ([a, x])ϕ([t]) ,

since w̃[t] = w̃t = at and ϵ · w[t] = [t^{σ}xt] by Theorem 8.74 and Lemma 8.75. Hence (ϕ, ψ) is an isomorphism from Ω_{P} (K, K_{0}, σ, W̃, p) to (V, W, τ_{V} , τ_{W} , ϵ, δ). □

This completes the proof of Theorem 8.9.

8.5. Quadrangular Systems of Type F_{4}

Let Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) be a wide quadrangular system which is the extension of a reduced quadrangular system Λ of quadratic form type; more precisely, let Λ = (V, Rad(H), τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{Q}(K, V_{0}, q), where τ_{V} and τ_{W} are as in Remark 8.3.

Our goal in this section is to classify these quadrangular systems in the case that Rad(F) ≠ 0.

So assume that Rad(F) ≠ 0. It then follows from (Q_{10}) that ϵ ∈ Rad(F). Note that \overline{v} = v for all v ∈ V by Lemma 8.11.

Remark 8.80. We will identify V and V_{0} in the sequel if there is no danger of confusion, which will allow us to use notations like tv with t ∈ K and v ∈ V.

Observe that the axiom system is very symmetrical now. (See section A.3.1 in the appendix.) In particular, every identity will have a “dual identity”, which is obtained by switching the roles of V and W.

Lemma 8.81. For all v, v_{1}, v_{2} ∈ V and all w, w_{1}, w_{2} ∈ W, we have that

(i) F (v_{1}w, v_{2}) = F (v_{1}, v_{2}w) ; (ii) H(w_{1}v, w_{2}) = H(w_{1}, w_{2}v) .

Proof. Since both V and W are abelian, it follows from (Q_{21}) and (Q_{11}) that F (v_{1}w, v_{2}) = w(v_{2} + v_{1}) + wv_{2} + wv_{1} = w(v_{1} + v_{2}) + wv_{1} + wv_{2} = F (v_{2}w, v_{1}) = F (v_{1}, v_{2}w), which proves (i). Similarly, (ii) follows from (Q_{22}) and (Q_{12}). (Identity (ii) is the “dual” of identity (i).) □

Definition 8.82. Let R := Rad(F). Then ϵ ∈ R, and R ≠ V since F ≢ 0. Moreover, let L := q(R) ⊆ K.

Lemma 8.83. Σ := (R, W, τ_{V} , τ_{W} , ϵ, δ) is a quadrangular system with F_{Σ} ≡ 0 and H_{Σ} ≢ 0; see Remark 8.3.

Proof. First of all, we observe that R = Rad(F) is a subgroup of V, since F is additive in both variables. We have that Im(F_{Σ}) = F (R, R) = 0, and Im(H_{Σ}) = H(W, W) = Im(H) ≠ 0. It now only remains to show that τ_{V} (R × W) ⊆ R, H(W, W) ⊆ R and (R^{∗})^{−1} ⊆ R.

If v ∈ R, then F (v, V) = 0, hence F (vw, V) = F (v, V w) ⊆ F (v, V) = 0 as well for all w ∈ W, by Lemma 8.81(i). Hence τ_{V} (R × W) = R · W ⊆ R. Since W is abelian, it follows from (Q_{8}) that H(W, W) ⊆ Rad(F) = R. Finally, if v ∈ R^{∗}, then F (v^{−1}, V) = F (v, V)v^{−1} = 0 by (Q_{17}), and hence v^{−1} ∈ R. Thus Σ := (R, W, τ_{V} , τ_{W} , ϵ, δ) is a quadrangular system. □

By Theorem 8.13, Σ^{∗} = (W, R, τ_{W} , τ_{V} , δ, ϵ) is a reduced quadrangular system. Suppose that Σ^{∗} were of involutory type, say Σ^{∗} ≅ Q_{I}(J, J_{0}, σ) for some involutory set (J, J_{0}, σ). Then δ ∈ Rad(H) would imply that 0 = F_{Σ^{∗}}([1], [a]) = [a + a^{σ}] and hence a + a^{σ} = 0 for all a ∈ J, from which it would follow that F_{Σ^{∗}}([a], [b]) = [(a^{σ}b) + (a^{σ}b)^{σ}] = 0 for all a, b ∈ J. Hence H_{Σ} ≡ F_{Σ^{∗}} = 0, a contradiction.

It follows that Σ^{∗} must be of quadratic form type. In particular, R has the structure of a field, W is a (right) vector space over R, and the map p : W → R : w ↦ ϵw is a quadratic form. If we denote the multiplication in R by •, then we have that w(r_{1} • r_{2}) = (wr_{1})r_{2} for all w ∈ W and all r_{1}, r_{2} ∈ R.

Lemma 8.84. L is a subfield of K, with K^{2} ⊆ L ⊆ K. Moreover, q is a field isomorphism from R to L.

Proof. We will first prove that q is an isomorphism (both additive and multiplicative) from R to L. Since L = q(R), q is surjective. For all r_{1}, r_{2} ∈ R, we have that [q(r_{1} + r_{2})] = δ(r_{1} + r_{2}) = δr_{1} + δr_{2} = [q(r_{1}) + q(r_{2})], by (Q_{11}); hence q is additive. In particular, if q(r_{1}) = q(r_{2}), then q(r_{1} + r_{2}) = 0 and hence r_{1} = r_{2}, since q is anisotropic. Hence q is injective. Furthermore, for all r_{1}, r_{2} ∈ R, we have that [q(r_{1} • r_{2})] = δ(r_{1} • r_{2}) = (δr_{1})r_{2} = [q(r_{1})]r_{2} = [q(r_{1})q(r_{2})] by Lemma 8.51, hence q is multiplicative.

It follows that L = q(R) is a commutative field which is isomorphic to R. Finally, for all t ∈ K, we have that q(tϵ) = t^{2}q(ϵ) = t^{2}, hence K^{2} ⊆ q(R) = L since tϵ ∈ Rad(F). □

Definition 8.85. For all s ∈ L, we let [s] := q^{−1}(s) ∈ V. If we want to make clear whether we mean [s] ∈ V or [s] ∈ W, we will write [s]_{V} and [s]_{W}, respectively. By Lemma 8.84, we can consider W as a (left) vector space over L via the scalar multiplication sw := w[s] for all w ∈ W and all s ∈ L.

Definition 8.86. Let q̂ be the map from W to L given by q̂(w) := q(p(w)) = q(ϵw) for all w ∈ W, and let f̂ be the map from W × W to L given by f̂(w_{1}, w_{2}) := q(H(w_{1}, w_{2})) for all w_{1}, w_{2} ∈ W. In particular, ϵw = [q̂(w)]_{V} and H(w_{1}, w_{2}) = [f̂(w_{1}, w_{2})]_{V} for all w, w_{1}, w_{2} ∈ W.

Lemma 8.87. q̂ is a quadratic form from W to L with corresponding bilinear form f̂.

Proof. Since p is a quadratic form from W to R with corresponding bilinear form H, it follows by Lemma 8.84 that q̂ = q ∘ p is a quadratic form from W to L with corresponding bilinear form f̂ = q ∘ H. □

Remark 8.88. For s ∈ L and t ∈ K, we will write q[s] and q̂[t] in place of q([s]) and q̂([t]), respectively.

Lemma 8.89. For all v, v_{1}, v_{2} ∈ V and all w, w_{1}, w_{2} ∈ W, we have that
(i) F (v_{1}, v_{2}) = 0 ⟹ wv_{1}v_{2} = wv_{2}v_{1} ;
(ii) H(w_{1}, w_{2}) = 0 ⟹ vw_{1}w_{2} = vw_{2}w_{1} .

Proof. Observe that (Q_{23}) can be rewritten as “vΠ_{w}(z) = vwzw^{−1}”, and that (Q_{24}) can be rewritten as “wπ_{v}(c) = wvcv^{−1}” since π_{v}(ϵ) = ϵ.

Let v_{1}, v_{2} ∈ V be such that F (v_{1}, v_{2}) = 0, and assume that v_{1} ≠ 0. Then π_{v_{1}}(v_{2}) = v_{2}. It then follows from (Q_{24}) that wv_{2} = wv_{1}v_{2}v_{1}^{−1} for all w ∈ W, hence (i). Identity (ii) is the dual of (i). □

In particular, s(wv) = wv[s] = w[s]v = (sw)v and t(vw) = vw[t] = v[t]w = (tv)w for all v ∈ V, w ∈ W, s ∈ L and t ∈ K. It follows that the notations swv and tvw are unambiguous.

Lemma 8.90. For all v ∈ V^{∗} and all w ∈ W^{∗}, we have that

(i) v^{−1} = q(v)^{−1}v; (ii) w^{−1} = q̂(w)^{−1}w.

Proof. If we substitute δ for w in (Q_{16}), then we get that v = v^{−1} · δv = v^{−1}[q(v)]_{W} = q(v)v^{−1}, which proves (i). Similarly for (ii). □

Lemma 8.91. For all v ∈ V^{∗}, w ∈ W^{∗}, t ∈ K and s ∈ L, we have that

(i) w · tv = q̂[t]wv; (ii) v · sw = q[s]vw; (iii) wv = q̂[q(v)]wv^{−1} ; (iv) vw = q[q̂(w)]vw^{−1} .

Proof. We only prove (i) and (iii). By (Q_{26}),

w · tv = w · v[t] = w · ϵ[t] · v = w · [q̂[t]]_{V} · v = q̂[t]wv ,

which proves (i). It follows from Lemma 8.90(i) and (i) that wv = w · q(v)v^{−1} = q̂[q(v)]wv^{−1}, which proves (iii). □

Lemma 8.92. For all v, c ∈ V, w, z ∈ W, we have that

(i) wvcv = w(f(v, c)v + q(v)c) ; (ii) vwzw = v(f̂(w, z)w + q̂(w)z) .

Proof. We only prove (i). We may assume that v ≠ 0. By (Q_{24}) and by Lemma 8.90(i),

wvcv^{−1} = wπ_{v}(c) = w(c + f(v, c)v^{−1}) = w(c + f(v, c)q(v)^{−1}v) .

It follows by Lemma 8.91(iii) and Lemma 8.91(i) that

wvcv = q̂[q(v)]wvcv^{−1} = q̂[q(v)]w(c + f(v, c)q(v)^{−1}v) = w(q(v)c + f(v, c)v) ,

which is what we had to show. □

Lemma 8.93. For all v, c ∈ V, w, z ∈ W, we have that (i) wvc + wcv = q̂[f(v, c)]w + [f(v, c)f(vw, c)] ; (ii) vwz + vzw = q[f̂(w, z)]v + [f̂(w, z)f̂(wv, z)] .

Proof. Again, we only prove (i). We may assume that v ≠ 0. By (Q_{24}), Lemma 8.91(i) and Lemma 8.90(i),

wvcv^{−1} = wπ_{v}(c)
= w(c + f(v, c)v^{−1})
= wc + w · f(v, c)v^{−1} + F (f(v, c)v^{−1}, cw)
= wc + q̂[f(v, c)]wv^{−1} + [f(f(v, c)q(v)^{−1}v, cw)] ,

and hence, by Lemma 8.51,

wvc = wcv + q̂[f(v, c)]w + [f(v, c)q(v)^{−1}f(v, cw)]v
= wcv + q̂[f(v, c)]w + [f(v, c)f(v, cw)]
= wcv + q̂[f(v, c)]w + [f(v, c)f(vw, c)] ,

which is what we had to show. □

Lemma 8.94. For all v ∈ V, c ∈ V^{∗}, w ∈ W and z ∈ W^{∗}, we have that

(i) z · vz = q̂(z)zv; (ii) c · wc = q(c)cw; (iii) f̂(z, w · vz)z^{−1} = f̂(zv, w)z; (iv) f(c, v · wc)c^{−1} = f(cw, v)c.

Proof. We will only prove (i) and (iii). First of all, observe that it follows from (Q_{12}) that H(w, wv) = 0 for all v ∈ V and all w ∈ W. In particular, Π_{z}(z·ϵz) = z·ϵz and Π_{z}(z · vz) = z · vz. It thus follows from (Q_{26}) that z · vz = z · ϵz · v = q̂(z)zv, which shows (i). By Lemma 8.81(ii) and Lemma 8.90(ii), it now follows that

f̂(z, w · vz)z^{−1} = f̂(z · vz, w)q̂(z)^{−1}z
= f̂(q̂(z)zv, w)q̂(z)^{−1}z
= f̂(zv, w)z ,

which shows (iii). □

Lemma 8.95. For all v ∈ V and all w ∈ W, we have that

(i) [q̂(wv)] = q(v)[q̂(w)] ; (ii) [q(vw)] = q̂(w)[q(v)] .

Proof. We will only prove (i). We may again assume that v ≠ 0. Since ϵ ∈ Rad(F), it follows by Lemma 8.81(i) that F (v, ϵc) = F (vc, ϵ) = 0 and hence π_{v}(ϵc) = ϵc for all c ∈ V. If we set c = ϵ in (Q_{25}), we thus get that ϵ·δv ·w = ϵ·wv, and hence [q̂(wv)] = ϵ · wv = ϵ · δv · w = ϵ[q(v)]w = q(v)ϵw = q(v)[q̂(w)]. □

Lemma 8.96. For all v, c ∈ V and all w, z ∈ W, we have that

(i) w · vz + q̂(z)wv = f̂(w, zv)z + f̂(w, z)zv; (ii) v · wc + q(c)vw = f(v, cw)c + f(v, c)cw.

Proof. We will only prove (i). We may assume that z ≠ 0. By (Q_{26}), we have that

w · ϵz · v + z^{−1}H(z, w · ϵz) · v = w · vz + z^{−1}H(z, w · vz) ,

hence

q̂(z)wv + f̂(z, w · ϵz)z^{−1}v = w · vz + f̂(z, w · vz)z^{−1} ,

and it follows from Lemma 8.94(iii) that

q̂(z)wv + f̂(z, w)zv = w · vz + f̂(zv, w)z ,

which is what we had to show. □

At this point, we will break the symmetry. We cannot avoid this, since L is a subfield of K, but not vice versa.

Lemma 8.97. For all t ∈ K and all s ∈ L, we have that

(i) s[t] = [st] ; (ii) t[s] = [t^{2}s] .

Proof. By Lemma 8.51, we have that s[t] = [t][s] = [tq[s]]. Since [s] = q^{−1}(s) by definition, it follows that s[t] = [st]. On the other hand, t[s] = tq^{−1}(s) = q^{−1}(t^{2}s) = [t^{2}s]. □

Now choose fixed arbitrary elements ξ ∈ W \ Y and d ∈ V \ R.

Theorem 8.98. There exists an element e ∈ V such that f(d, e) = 1 and f(d, eξ) = 0. Moreover, f(dξ, eξ) = q̂(ξ) ∈ L \ K^{2}.

Proof. We will first show the last statement. So let a, b ∈ V be arbitrary elements such that f(a, b) = 1. Then, by Lemma 8.81(i) and Lemma 8.91(iv), we have that f(aξ, bξ) = f(a, bξξ) = f(a, q[q̂(ξ)]b) = q[q̂(ξ)]f(a, b) = q[q̂(ξ)] = q̂(ξ) (note that [s] = q^{−1}(s) for all s ∈ L by definition). Let α := q̂(ξ). Suppose that α ∈ K^{2}, say α = t^{2} for some t ∈ K. Then q(tϵ) = t^{2} = α = q̂(ξ) = q(ϵξ), hence ϵ[t] = tϵ = ϵξ. Since [t] ∈ Y, this implies that ϵ(ξ + [t]) = ϵξ + ϵ[t] = 0, and hence ξ = [t] ∈ Y, which contradicts the choice of ξ. Hence α ∉ K^{2}.

Since d ∉ R = Rad(F), there exist an elements u ∈ V such that F (d, u) ≠ 0. Let v := f(d, u)^{−1}u, then

f(d, v) = f(d, f(d, u)^{−1}u)
= f(d, u)^{−1}f(d, u)
= 1 .

In particular, f(dξ, vξ) = q̂(ξ) = α. Since α ∉ K^{2}, we have that α^{−1}f(d, vξ)^{2} ≠ 1. Now let

e := (1 + α^{−1}f(d, vξ)^{2})^{−1}(v + α^{−1}f(d, vξ)vξ) .

Then, by Lemma 8.81(i),

f(d, e) = (1 + α^{−1}f(d, vξ)^{2})^{−1}f(d, v + α^{−1}f(d, vξ)vξ)
= (1 + α^{−1}f(d, vξ)^{2})^{−1}(f(d, v) + α^{−1}f(d, vξ)f(d, vξ))
= (1 + α^{−1}f(d, vξ)^{2})^{−1}(1 + α^{−1}f(d, vξ)^{2})
= 1 ,

and

f(d, eξ) = (1 + α^{−1}f(d, vξ)^{2})^{−1}f(d, vξ + α^{−1}f(d, vξ)vξξ)
= (1 + α^{−1}f(d, vξ)^{2})^{−1}(f(d, vξ) + α^{−1}f(d, vξ)f(d, vξξ))
= (1 + α^{−1}f(d, vξ)^{2})^{−1}(f(d, vξ) + α^{−1}f(d, vξ)α)
= 0 ,

which shows that e fulfills the required properties. □

From now on, let e ∈ V be as in Theorem 8.98, and let α := f(dξ, eξ) = q̂(ξ). By Theorem 8.98, α ∈ L \ K^{2}.

Theorem 8.99. Let B := ⟨d, e, dξ, eξ⟩. Then dim_{K} B = 4 and B ∩ R = 0.

Proof. Let v = t_{1}d + t_{2}e + t_{3}dξ + t_{4}eξ with t_{1}, t_{2}, t_{3}, t_{4} ∈ K be an arbitrary element of B. Suppose that v ∈ R = Rad(F) = Rad(f). Then f(v, d) = f(v, e) = f(v, dξ) = f(v, eξ) = 0. Note that f(d, d) = f(d, dξ) = f(e, e) = f(e, eξ) = f(dξ, dξ) = f(eξ, eξ) = 0 by (Q_{11}), and that f(d, eξ) = f(e, dξ) = 0 by Theorem 8.98 and Lemma 8.81(i). Moreover, f(d, e) = 1 and f(dξ, eξ) = α ≠ 0. It now follows from f(v, d) = 0 that t_{2} = 0, from f(v, e) = 0 that t_{1} = 0, from f(v, dξ) = 0 that t_{4} = 0 and from f(v, eξ) = 0 that t_{3} = 0. Hence v = 0. This shows that B ∩ R = 0.

Since 0 ∈ R, the previous paragraph also shows that it follows from v = 0 that t_{1} = t_{2} = t_{3} = t_{4} = 0, hence d, e, dξ and eξ are linearly independent. It follows that dim_{K} B = 4. □

Theorem 8.100. B^{⊥} = R, where B^{⊥} := {v ∈ V | f(v, B) = 0}.

Proof. It is obvious that R ⊆ B^{⊥}. So let g be an arbitrary element of B^{⊥}. Then f(g, d) = f(g, e) = f(g, dξ) = f(g, eξ) = 0. If we substitute ξ for z, ξde for w and g for v in Lemma 8.96(i), then we get that

ξde · gξ + q̂(ξ)ξdeg = f̂(ξde, ξg)ξ + f̂(ξde, ξ)ξg .

Since f(e, gξ) = 0 and f(d, gξ) = 0, it follows from Lemma 8.89(i) that ξde · gξ = ξd · gξ · e = ξ · gξ · d · e, and hence ξde · gξ = q̂(ξ)ξgde by Lemma 8.94(i). On the other hand, since f(e, g) = 0 and f(d, g) = 0, it follows from Lemma 8.89(i) that q̂(ξ)ξdeg = q̂(ξ)ξdge = q̂(ξ)ξgde.

Hence ξde · gξ = q̂(ξ)ξdeg, and it follows that f̂(ξde, ξg)ξ = f̂(ξde, ξ)ξg. By Lemma 8.81(i) and (Q_{11}), we have that

[f̂(ξde, ξ)] = [f̂(ξd, ξe)]
= [q̂(ξd + ξe)] + [q̂(ξd)] + [q̂(ξe)]
= [q̂(ξ(d + e))] + [q̂(ξd)] + [q̂(ξe)]

since F (dξ, e) = 0. It follows from Lemma 8.95(i) that

[f̂(ξde, ξ)] = q(d + e)[q̂(ξ)] + q(d)[q̂(ξ)] + q(e)[q̂(ξ)]
= (q(d + e) + q(d) + q(e))[α]
= f(d, e)[α]
= [α] ,

and hence f̂(ξde, ξ) = α ≠ 0. It follows that ξg = α^{−1}f̂(ξde, ξg)ξ = ξr with r = [α^{−1}f̂(ξde, ξg)] ∈ R. Since ξ(g + r) = ξg + ξr = 0 by (Q_{11}), we conclude that g = r ∈ R, which completes the proof of this theorem. □

Since dim_{K} B = 4 is finite by Theorem 8.99, we have V = B + B^{⊥}. Since B ∩ R = 0 by Theorem 8.99 and B^{⊥} = R by Theorem 8.100, it follows that V has a decomposition V = B ⊕ R. In particular, every complement of R in V has dimension 4 over K. By symmetry, it also follows that every complement of Y in W has dimension 4 over L.

Let β := q(d)^{−1}. Then β ∈ K \ L, since β ∈ L would imply that q(d) = β^{−1} = q[β^{−1}] and hence d = [β^{−1}] ∈ [L] = R = Rad(f), which contradicts the fact that f(d, e) = 1.

Theorem 8.101. Let A := ⟨ξ, ξed^{−1}, ξd^{−1}, β^{2}ξe⟩. Then W = A ⊕ Y.

Proof. Let w = s_{1}ξ + s_{2}ξed^{−1} + s_{3}ξd^{−1} + s_{4}β^{2}ξe with s_{1}, s_{2}, s_{3}, s_{4} ∈ L be an arbitrary element of A. Suppose that w ∈ Y = Rad(H) = Rad(f̂). Observe that q̂[q(d)] = q(ϵ[q(d)]) = q(q(d)ϵ) = q(d)^{2} = β^{−2} and hence, by Lemma 8.91(iii), f̂(ξed^{−1}, ξ) = f̂(ξe, ξd^{−1}) = q̂[q(d)]^{−1}f̂(ξe, ξd) = αβ^{2} ≠ 0.

By (Q_{12}), f̂(ξ, ξ) = f̂(ξd^{−1}, ξ) = f̂(ξe, ξ) = 0. It thus follows from f̂(w, ξ) = 0 that f̂(s_{2}ξed^{−1}, ξ) = 0 and hence s_{2} = 0. We now have w = s_{1}ξ + s_{3}ξd^{−1} + s_{4}β^{2}ξe.

Since f̂(ξ, ξe) = f̂(ξe, ξe) = 0 and f̂(ξd^{−1}, ξe) = αβ^{2} ≠ 0, it follows from f̂(w, ξe) = 0 that s_{3} = 0, and hence w = s_{1}ξ + s_{4}β^{2}ξe.

Since f̂(ξ, ξd^{−1}) = 0 and f̂(ξe, ξd^{−1}) = αβ^{2} ≠ 0, it follows from f̂(w, ξd^{−1}) = 0 that s_{4} = 0. Hence w = s_{1}ξ.

Finally, it now follows from f̂(w, ξed^{−1}) = 0 that f̂(s_{1}ξ, ξed^{−1}) = s_{1}αβ^{2} = 0 and hence s_{1} = 0.

So we have shown that w ∈ Y implies w = 0, and at the same time, we have shown that ξ, ξed^{−1}, ξd^{−1} and β^{2}ξe are linearly independent. Hence dim_{L} A = 4 and A ∩ Y = 0, from which it follows that A is contained in a complement of Y in W. Since every complement of Y in W is 4-dimensional, this implies that A itself is a complement of Y, i.e. W = A ⊕ Y. □

Let E be the splitting field of the polynomial ϕ(x) ≡ q(d)x^{2} + x + q(e) over K.

Lemma 8.102. E/K is a separable quadratic extension.

Proof. Suppose that t ∈ K would be a root of ϕ. Then

q(td + e) = q(td) + f(td, e) + q(e) = t^{2}q(d) + tf(d, e) + q(e) = ϕ(t) ,

since f(d, e) = 1, hence q(td + e) = 0. Since q is anisotropic, this implies that td + e = 0, which contradicts the fact that d and e are linearly independent. Hence ϕ has no roots in K, so E/K is a quadratic extension. Since the coefficient of x of ϕ is non-zero, the two roots of ϕ are distinct, hence the extension is separable. □

Let ω ∈ E \ K be one of the roots of ϕ. Let D := E^{2}L = L(ω^{2}). Then D is the splitting field of the polynomial ϕ′(x) ≡ q(d)^{2}x^{2} + x + q(e)^{2} over L. For both extensions E/K and D/L, we will denote the norm by N and the non-trivial element of the Galois group by x ↦ \overline{x}.

We can consider E as a 2-dimensional vector space over K, and D as a 2dimensional vector space over L. Let B_{0} := E ⊕ E, and let A_{0} := D ⊕ D. Then B_{0} is a 4-dimensional vector space over K, and A_{0} is a 4-dimensional vector space over L. We can identify B and A with B_{0} and A_{0}, respectively, by the following relations.

t_{1}d + t_{2}e + t_{3}dξ + t_{4}eξ ⟷ (t_{1} + t_{2}ω, t_{3} + t_{4}ω)
s_{1}ξ + s_{2}ξed^{−1} + s_{3}ξd^{−1} + s_{4}β^{2}ξe ⟷ (s_{1} + s_{2}ω^{2}, s_{3} + s_{4}ω^{2})

Since R = [L] and Y = [K], we have actually identified V and W with B_{0} ⊕ L and A_{0} ⊕ K, respectively:

t_{1}d + t_{2}e + t_{3}dξ + t_{4}eξ + [s] ⟷ (t_{1} + t_{2}ω, t_{3} + t_{4}ω, s)
s_{1}ξ + s_{2}ξed^{−1} + s_{3}ξd^{−1} + s_{4}β^{2}ξe + [t] ⟷ (s_{1} + s_{2}ω^{2}, s_{3} + s_{4}ω^{2}, t)

For all (b, s) ∈ B_{0} ⊕ L and all (a, t) ∈ A_{0} ⊕ K, we will denote the corresponding elements of V and W by [b, s] and [a, t], respectively.

We can now describe the quadratic forms q and q̂ on B_{0} ⊕ L and A_{0} ⊕ K, respectively, via this identification.

Theorem 8.103. For all u, v ∈ E, s ∈ L, x, y ∈ D and t ∈ K, we have that

(i) q[u, v, s] = β^{−1}(N(u) + αN(v)) + s;
(ii) q̂[x, y, t] = α(N(x) + β^{2}N(y)) + t^{2} .

Proof. Let u = t_{1} + t_{2}ω and v = t_{3} + t_{4}ω be arbitrary elements of E, and let s be an arbitrary element of L. Then we have that

q[u, v, s] = q[t_{1} + t_{2}ω, t_{3} + t_{4}ω, s]
= q(t_{1}d + t_{2}e + t_{3}dξ + t_{4}eξ + [s])
= q(t_{1}d + t_{2}e) + q(t_{3}dξ + t_{4}eξ) + q[s] ,

since f(t_{1}d + t_{2}e, t_{3}dξ + t_{4}eξ) = 0 and [s] ∈ Rad(f). By Lemma 8.95(ii) and Lemma 8.97(i), [q(vξ)] = α[q(v)] = [αq(v)], and hence q(vξ) = αq(v) for all v ∈ V. It follows that

q[u, v, s] = q(t_{1}d + t_{2}e) + αq(t_{3}d + t_{4}e) + q[s]

= q(t_{1}d) + f(t_{1}d, t_{2}e) + q(t_{2}e) + α(q(t_{3}d) + f(t_{3}d, t_{4}e) + q(t_{4}e)) + q[s]

= t_{1}^{2}q(d) + t_{1}t_{2} + t_{2}^{2}q(e) + α(t_{3}^{2}q(d) + t_{3}t_{4} + t_{4}^{2}q(e)) + s
= q(d)N(t_{1} + t_{2}ω) + αq(d)N(t_{3} + t_{4}ω) + s
= β^{−1}(N(u) + αN(v)) + s ,

which proves (i). Similarly, let x = s_{1} + s_{2}ω^{2} and y = s_{3} + s_{4}ω^{2} be arbitrary elements of D, and let t be an arbitrary element of K. Then we have that

q̂[x, y, t] = q̂[s_{1} + s_{2}ω^{2}, s_{3} + s_{4}ω^{2}, t]
= q̂(s_{1}ξ + s_{2}ξed^{−1} + s_{3}ξd^{−1} + s_{4}β^{2}ξe + [t]) .

Note that f̂(ξed^{−1}, ξd^{−1}) = f̂(ξe, ξd^{−1}d^{−1}) = q̂[q(d)]^{−1}f̂(ξe, ξ) = 0 by Lemma 8.91(iii), hence f̂(s_{1}ξ + s_{2}ξed^{−1}, s_{3}ξd^{−1} + s_{4}β^{2}ξe) = 0. Since [t] ∈ Rad(f̂), it thus follows that

q̂[x, y, t] = q̂(s_{1}ξ + s_{2}ξed^{−1}) + q̂(s_{3}ξd^{−1} + s_{4}β^{2}ξe) + q̂[t]
= q̂(s_{1}ξ) + f̂(s_{1}ξ, s_{2}ξed^{−1}) + q̂(s_{2}ξed^{−1})
+ q̂(s_{3}ξd^{−1}) + f̂(s_{3}ξd^{−1}, s_{4}β^{2}ξe) + q̂(s_{4}β^{2}ξe) + q̂[t]
= s_{1}^{2}q̂(ξ) + s_{1}s_{2}f̂(ξ, ξed^{−1}) + s_{2}^{2}q̂(ξed^{−1})
+ s_{3}^{2}q̂(ξd^{−1}) + s_{3}s_{4}β^{2}f̂(ξ, ξed^{−1}) + s_{4}^{2}β^{4}q̂(ξe) + q̂[t] .

By Lemma 8.95(i) and Lemma 8.97(ii), [q̂(wv)] = q(v)[q̂(w)] = [q(v)^{2}q̂(w)], and hence q̂(wv) = q(v)^{2}q̂(w) for all v ∈ V and all w ∈ W. Remember that f̂(ξ, ξed^{−1}) = αβ^{2} and that q(d^{−1}) = q(d)^{−1} = β. Since q̂[t] = q(ϵ[t]) = q(tϵ) = t^{2}, it thus follows that

q̂[x, y, t] = s_{1}^{2}α + s_{1}s_{2}αβ^{2} + s_{2}^{2}q(e)^{2}q(d)^{−2}α
+ s_{3}^{2}q(d)^{−2}α + s_{3}s_{4}β^{2}αβ^{2} + s_{4}^{2}β^{4}q(e)^{2}α + t^{2}
= α(s_{1}^{2} + s_{1}s_{2}q(d)^{−2} + s_{2}^{2}q(e)^{2}q(d)^{−2}
+ β^{2}(s_{3}^{2} + s_{3}s_{4}q(d)^{−2} + s_{4}^{2}q(e)^{2}q(d)^{−2})) + t^{2}
= α(N(s_{1} + s_{2}ω^{2}) + β^{2}N(s_{3} + s_{4}ω^{2})) + t^{2}
= α(N(x) + β^{2}N(y)) + t^{2} ,

which proves (ii). □

For all a ∈ A_{0} and all b ∈ B_{0}, we let q_{1}(b) := q[b, 0] and q_{2}(a) := q̂[a, 0]. Denote the corresponding bilinear forms by f_{1} and f_{2}, respectively. We now define maps Υ̃, ν̃, Θ̃ and ψ̃ from A_{0} × B_{0} to A_{0}, K, B_{0} and L, respectively, by setting

[a, 0][b, 0] = [Υ̃(a, b), ν̃(a, b)] ,
[b, 0][a, 0] = [Θ̃(a, b), ψ̃(a, b)] ,

for all a ∈ A_{0} and all b ∈ B_{0}. We will show that these maps coincide with the maps Υ, ν, Θ and ψ defined on page 44.

Lemma 8.104. Υ̃ ≡ Υ.

Proof. All the equivalences in the proof of this lemma are modulo Y. Let

a_{1} := ξ ,  b_{1} := d ,
a_{2} := ξed^{−1} ,  b_{2} := e ,
a_{3} := ξd^{−1} ,  b_{3} := dξ ,
a_{4} := β^{2}ξe ,  b_{4} := eξ ,

and let a_{ij} := a_{i}b_{j} for all i, j ∈ {1, 2, 3, 4}. We first observe that ξde + ξed ≡ f(d, e)^{2}ξ ≡ ξ and that ξd^{−1}e + ξed^{−1} ≡ f(d^{−1}, e)^{2}ξ ≡ β^{2}ξ by Lemma 8.93(i). Then

a_{11} ≡ ξ · d ≡ β^{−2}a_{3} ;
a_{12} ≡ ξ · e ≡ β^{−2}a_{4} ;
a_{13} ≡ ξ · dξ ≡ αξd ≡ αβ^{−2}a_{3} ;
a_{14} ≡ ξ · eξ ≡ αξe ≡ αβ^{−2}a_{4} ;

8.5. QUADRANGULAR SYSTEMS OF TYPE F_{4} 79

a_{21} ≡ ξed^{−1} · d ≡ ξe ≡ β^{−2}a_{4} ; a_{22} ≡ ξed^{−1} · e ≡ (ξd^{−1}e + β^{2}ξ) · e ≡ q(e)^{2}a_{3} + a_{4} ; a_{23} ≡ ξed^{−1} · dξ ≡ ξe · dξ · d^{−1} ≡ ξ · dξ · ed^{−1} ≡ αξded^{−1} ≡ α(ξ + ξed)d^{−1} ≡ αa_{3} + αβ^{−2}a_{4} ; a_{24} ≡ ξed^{−1} · eξ ≡ ξe · eξ · d^{−1} ≡ ξ · eξ · ed^{−1} ≡ αξeed^{−1} ≡ αq(e)^{2}a_{3}; a_{31} ≡ ξd^{−1} · d ≡ a_{1} ; a_{32} ≡ ξd^{−1} · e ≡ β^{2}ξ + ξed^{−1} ≡ β^{2}a_{1} + a_{2} ; a_{33} ≡ ξd^{−1} · dξ ≡ ξ · dξ · d^{−1} ≡ αξdd^{−1} ≡ αa_{1} ; a_{34} ≡ ξd^{−1} · eξ ≡ ξ · eξ · d^{−1} ≡ αξed^{−1} ≡ αa_{2} ; a_{41} ≡ β^{2}ξe · d ≡ q(d)^{2}β^{2}ξed^{−1} ≡ a_{2} ; a_{42} ≡ β^{2}ξe · e ≡ β^{2}q(e)^{2}a_{1} ; a_{43} ≡ β^{2}ξe · dξ ≡ β^{2}ξ · dξ · e ≡ β^{2}αξde ≡ αβ^{2}(ξ + ξed) ≡ αβ^{2}a_{1} + αa_{2} ; a_{44} ≡ β^{2}ξe · eξ ≡ β^{2}ξ · eξ · e ≡ β^{2}αξee ≡ αβ^{2}q(e)^{2}a_{1} . Hence

Υ̃((1, 0), (1, 0)) = (0, β^{−2}) ;  Υ̃((0, 1), (1, 0)) = (1, 0) ;
Υ̃((1, 0), (ω, 0)) = (0, β^{−2}ω^{2}) ;  Υ̃((0, 1), (ω, 0)) = (β^{2} + ω^{2}, 0) ;
Υ̃((1, 0), (0, 1)) = (0, αβ^{−2}) ;  Υ̃((0, 1), (0, 1)) = (α, 0) ;
Υ̃((1, 0), (0, ω)) = (0, αβ^{−2}ω^{2}) ;  Υ̃((0, 1), (0, ω)) = (αω^{2}, 0) ;
Υ̃((ω^{2}, 0), (1, 0)) = (0, β^{−2}ω^{2}) ;  Υ̃((0, ω^{2}), (1, 0)) = (ω^{2}, 0) ;
Υ̃((ω^{2}, 0), (ω, 0)) = (0, q(e)^{2} + ω^{2}) ;  Υ̃((0, ω^{2}), (ω, 0)) = (β^{2}q(e)^{2}, 0) ;
Υ̃((ω^{2}, 0), (0, 1)) = (0, α + αβ^{−2}ω^{2}) ; Υ̃((0, ω^{2}), (0, 1)) = (αβ^{2} + αω^{2}, 0) ;
Υ̃((ω^{2}, 0), (0, ω)) = (0, αq(e)^{2}) ; Υ̃((0, ω^{2}), (0, ω)) = (αβ^{2}q(e)^{2}, 0) .

Since ω^{2} = βω + βq(e)^{2} and \overline{ω} = ω + β, it is now straightforward to check that Υ̃ coincides with the map Υ defined on page 44 on the set

{(1, 0), (ω^{2}, 0), (0, 1), (0, ω^{2})} × {(1, 0), (ω, 0), (0, 1), (0, ω)} .

By (Q_{3}) and (Q_{11}), the map Υ̃ is additive in both variables. Since (sw)v = s(wv) for all s ∈ L, v ∈ V and w ∈ W, it follows that Υ̃(sa, b) = sΥ̃(a, b) for all s ∈ L, a ∈ A_{0} and b ∈ B_{0}. By Lemma 8.91(i), we have that w(tv) = q̂[t]wv = t^{2}wv for all t ∈ K, v ∈ V and w ∈ W, and hence Υ̃(a, tb) = t^{2}Υ̃(a, b) for all t ∈ K, a ∈ A_{0} and b ∈ B_{0}. Since the same properties hold for Υ, and since A_{0} = ⟨(1, 0), (ω^{2}, 0), (0, 1), (0, ω^{2})⟩ and B_{0} = ⟨(1, 0), (ω, 0), (0, 1), (0, ω)⟩, it thus follows that Υ̃ ≡ Υ. □

Lemma 8.105. For all a ∈ A_{0} and all b, b′ ∈ B_{0}, we have that

(i) q_{2}(Υ̃(a, b)) = q_{1}(b)^{2}q_{2}(a) + ν̃(a, b)^{2} ;
(ii) ν̃(a, b + b′) = ν̃(a, b) + ν̃(a, b′) + f_{1}(Θ̃(a, b), b′) ;
(iii) q_{1}(Θ̃(a, b)) = q_{2}(a)q_{1}(b) + ψ̃(a, b) .

Proof. By Lemma 8.95(i) and Lemma 8.97(ii), [q̂(wv)] = q(v)[q̂(w)] = [q(v)^{2}q̂(w)], and hence q̂(wv) = q(v)^{2}q̂(w) for all v ∈ V and all w ∈ W. If we choose v = [b, 0] and w = [a, 0], then we get that

q̂[Υ̃(a, b), ν̃(a, b)] = q[b, 0]^{2}q̂[a, 0] .

Hence q_{2}(Υ̃(a, b)) + ν̃(a, b)^{2} = q_{1}(b)^{2}q_{2}(a), which proves (i).

Similarly, it follows from Lemma 8.95(ii) and Lemma 8.97(i) that q(vw) = q̂(w)q(v) for all v ∈ V and all w ∈ W. It follows that

q(Θ̃(a, b), ψ̃(a, b)) = q̂[a, 0]q[b, 0] .

Hence q_{1}(Θ̃(a, b)) + ψ̃(a, b) = q_{2}(a)q_{1}(b), which proves (iii).

Finally, it follows from (Q_{11}) that

[a, 0] · [b + b′, 0] = [a, 0] · [b, 0] + [a, 0] · [b′, 0] + F ([b, 0] · [a, 0], [b′, 0]) .

Projecting this identity on Y = [0, K] yields

ν(a, b + b′) = ν(a, b) + ν(a, b′) + f([Θ̃(a, b), ψ̃(a, b)], [b′, 0])
= ν(a, b) + ν(a, b′) + f_{1}(Θ̃(a, b), b′) ,

which proves (ii). □

Theorem 8.106. Υ̃ ≡ Υ, ν̃ ≡ ν, Θ̃ ≡ Θ and ψ̃ ≡ ψ.

Proof. We have already shown in Lemma 8.104 that Υ̃ ≡ Υ. It then follows from Lemma 8.105(i) and Theorem 7.12(xii) that ν̃ ≡ ν. Hence, by Lemma 8.105(ii) and Theorem 7.12(i), we have that f_{1}(Θ̃(a, b) − Θ(a, b), b′) = 0 for all a ∈ A_{0} and all b, b′ ∈ B_{0}, from which it follows that Θ̃(a, b) − Θ(a, b) ∈ Rad(f_{1}) for all a ∈ A_{0} and all b ∈ B_{0}. Since B ∩ Rad(f) = B ∩ R = 0 by Theorem 8.99, we have that B_{0} ∩ Rad(f_{1}) = 0 as well, and hence Θ̃ ≡ Θ. Finally, it then follows from Lemma 8.105(iii) and Theorem 7.12(xi) that ψ̃ ≡ ψ. □

Theorem 8.107. (V, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{F} (K, V_{0}, q).

Proof. First of all, observe that q is indeed a quadratic form of type F_{4}, since its regular component q_{1} has a norm splitting q_{1}(u, v) = β^{−1}N(u)+β^{−1}αN(v), and the product of the coefficients of the norm splitting is β^{−1} · β^{−1}α = β^{−2}α, which is an element of L.

Let ϕ be the isomorphism from [V_{0}] = [B_{0} ⊕ L] to V which maps [b, s] to [b, s] for all b ∈ B_{0} and all s ∈ L, and let ψ be the isomorphism from [W_{0}] = [A_{0} ⊕ K] to W which maps [a, t] to [a, t] for all a ∈ A_{0} and all t ∈ K. Then ϕ([0, 1]) = [0, 1] = [1]_{V} = ϵ and ψ([0, 1]) = [0, 1] = [1]_{W} = δ.

Since ϕ and ψ are identity maps, it now only remains to show that it follows from the relations

[a, 0][b, 0] = [Υ̃(a, b), ν̃(a, b)] ,
[b, 0][a, 0] = [Θ̃(a, b), ψ̃(a, b)] ,

for all a ∈ A_{0} and all b ∈ B_{0} that

[a, t][b, s] = [Υ̃(a, b) + sa, ν̃(a, b) + q[b, s]t] ,
[b, s][a, t] = [Θ̃(a, b) + tb, ψ̃(a, b) + q̂[a, t]s] ,

for all a ∈ A_{0} and all b ∈ B_{0}. We will only show the first identity, the second one being completely similar. Since [0, s] ∈ Rad(F), it follows from (Q_{11}) and Lemma 8.51 that

[a, t][b, s] = [a, 0][b, s] + [0, t][b, s]
= [a, 0][b, 0] + [a, 0][0, s] + [0, t][b, s]
= [Υ̃(a, b), ν̃(a, b)] + [sa, 0] + [0, tq[b, s]]
= [Υ̃(a, b) + sa, ν̃(a, b) + q[b, s]t] .

Since ϕ and ψ are identity maps, it is now obvious that ϕ([b, s][a, t]) = ϕ([b, s])ψ([a, t]) and ψ([a, t][b, s]) = ψ([a, t])ϕ([b, s]) for all (a, t) ∈ W_{0} and all (b, s) ∈ V_{0}; hence (ϕ, ψ) is an isomorphism from Ω_{F} (K, V_{0}, q) to (V, W, τ_{V} , τ_{W} , ϵ, δ). □

8.6. Quadrangular Systems of Pseudo-quadratic Form Type, II

In this section, we continue to assume that Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is a wide quadrangular system which is the extension of a quadrangular system Λ of quadratic form type, i.e. Λ = (V, Rad(H), τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{Q}(K, V_{0}, q), where τ_{V} and τ_{W} are as in Remark 8.3.

Our goal in this and the next section is to classify these quadrangular systems if Rad(F) = 0. So assume that Rad(F) = 0. We continue to identify V and V_{0} if there is no danger of confusion.

Lemma 8.108. For all v ∈ V, all w ∈ W and all t ∈ K, we have that (tv)w = t(vw). It follows that the notation tvw is unambiguous.

Proof. If we substitute [t] for z, \overline{v} for v and ⊟w for w in (Q_{23}), then we get, since Π_{w}([t]) = [t], that

t\overline{v} = −\overline{\overline{(t · \overline{vw})}κ(⊟w)} ,

and hence, by Lemma 8.50(ii), that

\overline{tv} = −\overline{(t · vw)κ(⊟w)} .

It follows that

tv · w = −(t · vw)κ(⊟w) · w ,

and hence, by (Q_{18}), that tv · w = t · vw, which is what we had to show. □

Definition 8.109. If char(K) ≠ 2, let ζ := ϵ/2. If char(K) = 2, define S_{1} := {v ∈ V | F (ϵ, v) ≠ 0} and S_{2} := {ϵw | w ∈ W }. If S_{1} ∩ S_{2} ≠ ∅, choose a fixed element z ∈ S_{1} ∩ S_{2}; if S_{1} ∩ S_{2} = ∅, choose a fixed element z ∈ S_{1}. Observe that S_{1} ≠ ∅ since ϵ ∉ Rad(F). In both cases, we let ζ := f(ϵ, z)^{−1}z.

It follows that f(ϵ, ζ) = 1, independent of the characteristic.

Remark 8.110. This somewhat strange definition will become clear in section 8.7.

Definition 8.111. An element w ∈ W is called ζ-orthogonal if and only if f(ζ, ϵw) = 0.

Lemma 8.112. Each coset of Y in W contains a unique ζ-orthogonal element.

Proof. Consider an arbitrary coset w ⊞ Y of Y in W (where w ∈ W). An arbitrary element of this coset, say w ⊞ [t] with t ∈ K, is ζ-orthogonal if and only if f(ζ, ϵ(w⊞[t])) = 0. By (Q_{12}), f(ζ, ϵ(w⊞[t])) = f(ζ, ϵw+ϵ[t]) = f(ζ, ϵw)+f(ζ, tϵ) = f(ζ, ϵw) + t, hence w ⊞ [t] is ζ-orthogonal if and only if t = −f(ζ, ϵw). □

Since Y = Rad(H) is a normal subgroup of W, we can define the quotient group X := W/Y. Since [W, W]_{⊞} ≤ Y, the group X is abelian. We will use the additive notations + and − for X.

Definition 8.113. We define a map ρ : X → W as follows. For each element w ⊞ Y ∈ X, we let ρ(w ⊞ Y) be the unique element w ⊞ y ∈ w ⊞ Y ⊆ W which is ζ-orthogonal; see Lemma 8.112. Moreover, for all x ∈ X and all t ∈ K, we let (x, t) := ρ(x) ⊞ [t] ∈ W. Note that ρ(x) ∈ x for all x ∈ X, and hence (x, t) ∈ x for all x ∈ X and all t ∈ K as well.

Lemma 8.114. For all w ∈ W, there exist unique elements x ∈ X and t ∈ K such that w = (x, t).

Proof. Let w ∈ W be arbitrary. Let x := w ⊞ Y ∈ X, and let

y := ⊟ρ(x) ⊞ w ∈ ⊟(w ⊞ Y) ⊞ w = ⊟Y ⊟ w ⊞ w = Y .

Hence y = [t] for some t ∈ K, and we thus have that (x, t) = ρ(x) ⊞ [t] = ρ(x) ⊞ y = w.

Now suppose that (x_{1}, t_{1}) = (x_{2}, t_{2}) for some x_{1}, x_{2} ∈ X and some t_{1}, t_{2} ∈ K. Since (x_{1}, t_{1}) ∈ x_{1} and (x_{2}, t_{2}) ∈ x_{2}, it follows that the cosets x_{1} and x_{2} have an element in common, and hence they are equal, i.e. x_{1} = x_{2}. It then follows from (x_{1}, t_{1}) = (x_{2}, t_{2}) that t_{1} = t_{2} as well. □

Definition 8.115. We define a map G : X × X → W by setting

G(a, b) := ⊟ρ(a + b) ⊞ ρ(a) ⊞ ρ(b)

for all a, b ∈ X. Note that a and b are cosets of Y in W. It follows that G(a, b) ∈ −(a + b) + a + b = Y. Hence we can define a map g : X × X → K by setting G(a, b) = [g(a, b)] for all a, b ∈ X.

Lemma 8.116. (a, t) ⊞ (b, s) = (a + b, t + s + g(a, b)) for all a, b ∈ X and all t, s ∈ K.

Proof. Since Y ⊆ Z(W) by (Q_{8}), we have that

(a, t) ⊞ (b, s) = ρ(a) ⊞ [t] ⊞ ρ(b) ⊞ [s]
= ρ(a) ⊞ ρ(b) ⊞ [t + s]
= ρ(a + b) ⊞ [g(a, b)] ⊞ [t + s]
= (a + b, g(a, b) + t + s) ,

which is what we had to show. □

Definition 8.117. We define a map θ from X × V to V, a map π from X to V and a map h from X × X to V by setting

θ(a, v) := v · (a, 0) ,
π(a) := θ(a, ϵ) = ϵ · (a, 0) ,
h(a, b) := H((a, 0), (b, 0)) ,

for all a, b ∈ X and all v ∈ V.

By definition, (a, 0) is ζ-orthogonal for all a ∈ X, hence f(π(a), ζ) = f(ϵ · (a, 0), ζ) = 0 for all a ∈ X. Furthermore, it follows from (Q_{12}) that v · (a, t) = v · (a, 0) + v · [t] = θ(a, v) + tv for all v ∈ V, all a ∈ X and all t ∈ K.

Definition 8.118. We define a map (a, v) ↦ av from X × V to X and a map φ from X × V to K by the relation

(a, 0) · v = (av, φ(a, v)) .

Since we did not define a multiplication yet between elements of X and elements of V, this will not cause confusion.

Note that it follows from (a, 0) · ϵ = (a, 0) that aϵ = a and φ(a, ϵ) = 0 for all a ∈ X. Furthermore, we have that (a, t) · v = (av, tq(v) + φ(a, v)) by Lemma 8.51, and that H((a, t), (b, s)) = h(a, b), for all a, b ∈ X, all v ∈ V and all t, s ∈ K.

Lemma 8.119. For all a ∈ X, we have that

g(a, −a) = g(−a, a) = f(π(a), ϵ) = f(π(−a), ϵ) .

Proof. Let w := (a, 0) ∈ W. Then w is ζ-orthogonal. By 3.13(i), we have that w(−ϵ) = [f(ϵw, ϵ)] ⊟ w. By (Q_{6}), f(ϵ · w(−ϵ), ζ) = f(ϵw, ζ) = 0, and hence w(−ϵ) is ζ-orthogonal as well. It follows that w(−ϵ) = (b, 0) for some b ∈ X. Since [f(ϵw, ϵ)] ∈ Y, we now have that b = w(−ϵ)⊞Y = [f(ϵw, ϵ)]⊟w⊞Y = ⊟w⊞Y = −a. It follows by Lemma 8.116 that

[f(π(a), ϵ)] = [f(ϵw, ϵ)]
= w(−ϵ) ⊞ w
= (−a, 0) ⊞ (a, 0)
= (−a + a, 0 + 0 + g(−a, a))
= [g(−a, a)] .

Hence f(π(a), ϵ) = g(−a, a), and since it follows from (Q_{6}) that

f(π(−a), ϵ) = f(ϵ · (−a, 0), ϵ)
= f(ϵ · (a, 0)(−ϵ), ϵ)
= f(ϵ · (a, 0), ϵ)
= f(π(a), ϵ) ,

we conclude that g(a, −a) = g(−a, a) = f(π(a), ϵ) = f(π(−a), ϵ). □

Definition 8.120. We define a map (t, a) ↦ ta from K × X to X by setting ta := a · tϵ for all t ∈ K and all a ∈ X. We will prove later on (see Theorem 8.123) that this makes X into a vector space over K.

Lemma 8.121. For all a ∈ X and all t ∈ K, we have that φ(a, tϵ) = 0 . Moreover, for all a ∈ X, all v ∈ V and all t ∈ K, we have that

(i) ta · v = a · tv = t · av;

(ii) φ(ta, v) = φ(a, tv) = t^{2}φ(a, v) .

Proof. Let w := (a, 0) ∈ W and let y := [t] ∈ Y. Since Π_{⊟y}(z) = z for all z ∈ W, it follows from (Q_{26}) that w · ϵy · v = w · vy, for all v ∈ V. It thus follows from (Q_{11}) that

F (ϵ · w(ϵy), ζ) = ⊟w · ϵy · ϵ ⊟ w · ϵy · ζ ⊞ w · ϵy · (ζ + ϵ)
= ⊟w · ϵy ⊟ w · ζy ⊞ w · (ζ + ϵ)y
= F (ϵy · w, ζy)
= F (tϵw, tζ)
= [t^{2}f(ϵw, ζ)]
= 0 ,

since w is ζ-orthogonal. It follows that w · ϵy is ζ-orthogonal as well. Since

w · ϵy = w · tϵ = (a, 0) · tϵ = (a · tϵ, φ(a, tϵ)) ,

it follows that φ(a, tϵ) = 0.

It now follows from w · ϵy · v = w · vy that (a · tϵ, 0) · v = (a, 0) · tv for all v ∈ V, and hence

(ta · v, φ(ta, v)) = (ta, 0) · v = (a · tϵ, 0) · v = (a, 0) · tv = (a · tv, φ(a, tv))

for all v ∈ V. This implies that ta · v = a · tv and φ(ta, v) = φ(a, tv).

Now observe that π_{tϵ}(c) = π_{ϵ}(c) = −\overline{c}, for all c ∈ V. If we substitute tϵ for v and v for c in (Q_{24}), we thus get that wv = w · tϵ · v · (tϵ)^{−1}, and hence

(a, 0) · v · tϵ = (a, 0) · tϵ · v .

It follows that

(av, φ(a, v)) · tϵ = (ta, 0) · v ,

and finally, since q(tϵ) = t^{2}, that

(t · av, t^{2}φ(a, v)) = (ta · v, φ(ta, v)) ,

and we are done. □

Lemma 8.122. The map (a, v) ↦ av is additive in both variables. Moreover, the following hold for all a, b ∈ X and all u, v ∈ V:

(i) φ(a + b, v) + g(a, b)q(v) = φ(a, v) + φ(b, v) + g(av, bv) ; (ii) φ(a, u + v) = φ(a, u) + φ(a, v) + g(av, au) + f(θ(a, u), v) .

Proof. It follows from (Q_{3}) that

((a, 0) ⊞ (b, 0))v = (a, 0) · v ⊞ (b, 0) · v ,

and hence, by Lemma 8.116, that

(a + b, g(a, b)) · v = (av, φ(a, v)) ⊞ (bv, φ(b, v)) ,

from which it follows that

((a + b)v, g(a, b)q(v) + φ(a + b, v)) = (av + bv, φ(a, v) + φ(b, v) + g(av, bv)) .

So we have shown that (a + b)v = av + bv and that (i) holds.

On the other hand, it follows from (Q_{11}) that

(a, 0) · (u + v) = (a, 0) · (v + u) = (a, 0) · v ⊞ (a, 0) · u ⊞ F (u · (a, 0), v) , and hence

(a(u + v), φ(a, u + v)) = (av, φ(a, v)) ⊞ (au, φ(a, u)) ⊞ [f(θ(a, u), v)]
= (av + au, φ(a, v) + φ(a, u) + g(av, au) + f(θ(a, u), v)) .

So we have shown that a(u + v) = au + av and that (ii) holds. □

Theorem 8.123. X_{0} is a vector space over K, with the scalar multiplication given by the map (t, a) ↦ ta = a · tϵ.

Proof. First of all, we have that 1a = a · ϵ = a for all a ∈ X. By Lemma 8.122, the two distributivity laws hold, since

t(a + b) = (a + b) · tϵ = a · tϵ + b · tϵ = ta + tb

for all t ∈ K and all a, b ∈ X, and

(s + t)a = a · (s + t)ϵ = a · (sϵ + tϵ) = a · sϵ + a · tϵ = sa + ta

for all s, t ∈ K and all a ∈ X. Finally, it follows from Lemma 8.121(i) that

st · a = ts · a = a · (ts)ϵ = a · t(sϵ) = ta · sϵ = s · ta

for all s, t ∈ K and all a ∈ X. □

Lemma 8.124. For all a, b ∈ X, all u, v ∈ V and all t ∈ K, we have that

(i) θ(ta, v) = t^{2}θ(a, v) ; (ii) θ(a, tv) = tθ(a, v) ; (iii) θ(a + b, v) + g(a, b)v = θ(a, v) + θ(b, v) + h(b, av) ; (iv) θ(a, u + v) = θ(a, u) + θ(a, v) .

Proof. Let w := (a, 0) ∈ W. Note that π_{tϵ}(c) = π_{ϵ}(c) = −\overline{c}, for all c ∈ V. It thus follows by substituting tϵ for v and v for c in (Q_{25}) that v · δ(tϵ) · w = v · w(tϵ). Hence

θ(ta, v) = v · (ta, 0) = v · w(tϵ) = v · δ(tϵ) · w
= v · [q(tϵ)] · w = t^{2}vw = t^{2}v · (a, 0) = t^{2}θ(a, v) ,

which proves (i). Since t · vw = tv · w, we have that tθ(a, v) = θ(a, tv), which proves (ii).

It follows from (Q_{12}) that

θ(a + b, v) + g(a, b)v = v · (a + b, g(a, b))
= v · ((a, 0) ⊞ (b, 0))
= v · (a, 0) + v · (b, 0) + H((b, 0), (a, 0) · v)
= θ(a, v) + θ(b, v) + H((b, 0), (av, φ(a, v)))
= θ(a, v) + θ(b, v) + h(b, av) ,

which shows (iii). Finally, it follows from (Q_{4}) that

θ(a, u + v) = (u + v) · (a, 0)
= u · (a, 0) + v · (a, 0)
= θ(a, u) + θ(a, v) ,

which proves (iv). □

Lemma 8.125. For all a, b ∈ X and all t ∈ K, we have that h(ta, b) = h(a, tb) = th(a, b).

Proof. If we substitute tϵ for v in Lemma 8.124(iii), then we get, by Lemma 8.124(ii), that

h(b, a · tϵ) = θ(a + b, tϵ) + g(a, b)tϵ − θ(a, tϵ) − θ(b, tϵ)
= tθ(a + b, ϵ) + tg(a, b)ϵ − tθ(a, ϵ) − tθ(b, ϵ)
= th(b, a) ,

hence h(b, ta) = th(b, a). It follows by (Q_{22}) and Lemma 8.50(ii) that

h(ta, b) = −\overline{h(b, ta)} = −\overline{th(b, a)} = −t\overline{h(b, a)} = th(a, b)

as well, and we are done. □

Lemma 8.126. For all a, b ∈ X, we have that f(h(a, b), ϵ) = g(b, a) − g(a, b).

Proof. If we set v = ϵ, w_{1} = (b, 0) and w_{2} = (a, 0) in (Q_{8}), then we get that

⊟(b, 0) ⊟ (a, 0) ⊞ (b, 0) ⊞ (a, 0) = [f(h(a, b), ϵ)] .

Since

⊟(b, 0) ⊟ (a, 0) ⊞ (b, 0) ⊞ (a, 0) = ⊟((a, 0) ⊞ (b, 0)) ⊞ ((b, 0) ⊞ (a, 0))
= ⊟(a + b, g(a, b)) ⊞ (a + b, g(b, a))
= (0, −g(a, b) + g(b, a)) ,

it follows that f(h(a, b), ϵ) = −g(a, b) + g(b, a). □

Lemma 8.127. For all a, b ∈ X and all v ∈ V, we have that

f(h(a, b), v) = f(h(av, b), ϵ) = f(h(a, b\overline{v}), ϵ) .

Proof. It follows from (Q_{8}) that

F (H(w_{2}, w_{1}), v) = [w_{1}, w_{2}v]_{⊞} = F (H(w_{2}v, w_{1}), ϵ)

for all w_{1}, w_{2} ∈ W. If we choose w_{2} = (a, 0) and w_{1} = (b, 0), then we get that f(h(a, b), v) = f(h(av, b), ϵ). It then follows from Lemma 8.54 that

f(h(a, b), v) = f(\overline{h(a, b)}, \overline{v}) = −f(h(b, a), \overline{v})
= −f(h(b\overline{v}, a), ϵ) = −f(\overline{h(b\overline{v}, a)}, \overline{ϵ})
= f(h(a, b\overline{v}), ϵ)

as well. □

Lemma 8.128. We have that av\overline{v} = q(v)a and au\overline{v} + av\overline{u} = f(u, v)a for all a ∈ X and all u, v ∈ V.

Proof. Let w := (a, 0) ∈ W. It then follows from (Q_{15}) that avv^{−1} = a. Since q(v)v^{−1} = \overline{v}, it follows from Lemma 8.121(i) that q(v)a = q(v)avv^{−1} = av · q(v)v^{−1} = av\overline{v}. It then follows that

f(u, v)a = q(u + v)a − q(u)a − q(v)a
= a(u + v)(\overline{u} + \overline{v}) − au\overline{u} − av\overline{v}
= au\overline{v} + av\overline{u}

as well. □

We have now come to a point which is very similar to Chapter 26 in [20]. For some of the remaining identities, we will thus simply refer to the appropriate place in [20]. Note that [20] uses δ where we use ζ.

Lemma 8.129. For all a, b ∈ X, we have that g(a, b) = f(h(b, a), ζ).

Proof. See [20, (26.20)]. □

Since h is bilinear over K, it follows from Lemma 8.129 that g is bilinear over K.

Lemma 8.130. For all a, b ∈ X and all v ∈ V, we have that

h(a, bv) − h(b, av) = f(h(a, b), ϵ)v .

Proof. See [20, (26.23)]. □

Lemma 8.131. If char(K) ≠ 2, then φ ≡ 0, and for all a ∈ X and all v ∈ V, we have that

(i) g(a, a) = 0 ; (ii) θ(a, v) = \frac{1}{2}h(a, av) .

Proof. See [20, (26.24)]. □

Note that it follows from Lemma 8.131(i) and the fact that g is bilinear over K that g is skew-symmetric if char(K) ≠ 2.

Lemma 8.132. If char(K) = 2, then

(i) h(a, av) = g(a, a)v = f(ϵ, π(a))v; (ii) f(θ(a, v), v) = g(av, av) = g(a, a)q(v) = f(ϵ, π(a))q(v) ; (iii) f(θ(a, u), v) = f(θ(a, v), u) + f(ϵ, π(a))f(u, v) ;

for all a ∈ X and all u, v ∈ V.

Proof. See [20, (26.25)]. □

Lemma 8.133. For all a ∈ X and all u, v ∈ V, we have that

(i) f(θ(a, v), v) = f(ϵ, π(a))q(v) ; (ii) f(θ(a, v), u) + f(θ(a, u), v) = f(ϵ, π(a))f(u, v) .

Proof. See [20, (26.26)]. □

Lemma 8.134. For all a ∈ X, all u ∈ V and all v ∈ V^{∗}, we have that

θ(av^{−1}, u) + φ(a, v^{−1})u = q(v)^{−1}\overline{θ(a, \overline{u})} − f(u, v′)\overline{θ(a, v^{−1})}

− f(\overline{θ(a, \overline{u})}, v)q(v)^{−1}v′ + f(\overline{θ(a, v^{−1})}, v)f(u, v′)v′ ,

where v′ = \overline{v^{−1}} = q(v)^{−1}v.

Proof. Let w := (a, 0) ∈ W, and let c := q(v)^{−1}π_{v}(u) ∈ V. Since δv = [q(v)], it follows by substituting u for c in (Q_{25}) and by Lemma 8.52 that

π_{v}(\overline{u · wv}) = q(v)π_{v}(\overline{u}) · w .

Note that π_{v^{−1}}(\overline{v_{2}}) = \overline{π_{v}(v_{2})} for all v_{2} ∈ V by Lemma 3.18(i). If we replace v by v^{−1}, then it follows that

\overline{π_{v}(u · wv^{−1})} = q(v^{−1})\overline{π_{v}(u)} · w ,

and hence, since q(v^{−1}) = q(v)^{−1}, that

u · wv^{−1} = π_{v}(\overline{\overline{c} · w}) ,

from which it follows that

u · (av^{−1}, φ(a, v^{−1})) = π_{v}(\overline{θ(a, \overline{c})}) ,

and therefore

θ(av^{−1}, u) + φ(a, v^{−1})u = \overline{θ(a, \overline{c})} − f(v, \overline{θ(a, \overline{c})})v′ .

Since \overline{c} = q(v)^{−1}\overline{π_{v}(u)} = q(v)^{−1}\overline{u} − q(v)^{−1}f(v, u)v^{−1} = q(v)^{−1}\overline{u} − f(u, v′)v^{−1}, it follows by Lemma 8.124 that

θ(av^{−1}, u) + φ(a, v^{−1})u = q(v)^{−1}\overline{θ(a, \overline{u})} − f(u, v′)\overline{θ(a, v^{−1})}

− q(v)^{−1}f(v, \overline{θ(a, \overline{u})})v′ + f(v, \overline{θ(a, v^{−1})})f(u, v′)v′ ,

which is what we had to show. □

We now define

v^{∗} := \begin{cases}0 & if char(K) ≠ 2 \\ f(v, ζ)ϵ + f(v, ϵ)ζ + v & if char(K) = 2 \end{cases} ,

for all v ∈ V.

Lemma 8.135. If char(K) = 2, then

(i) φ(a, v) = f(θ(a, v^{∗}), v)

= f(π(a), v)f(ζ, v) + f(θ(a, ζ), v)f(ϵ, v) + f(ϵ, π(a))q(v) ;

(ii) If f(ϵ, v) = f(ζ, v) = 0, then π(av) = π(a)q(v) + f(π(a), v)v; (iii) π(aζ) = \overline{π(a)}q(ζ) + θ(a, ζ) + f(ϵ, π(a))ζ; (iv) θ(av, u) = q(v)\overline{θ(a, \overline{u})} + f(u, \overline{v})\overline{θ(a, v)} + f(θ(a, v), \overline{u})\overline{v} + φ(a, v)u;

for all a ∈ X and all u, v ∈ V.

Proof. By Lemma 8.134, this follows from the proof of [20, (26.30)]. □

Lemma 8.136. For all v ∈ V, all w ∈ W and all a ∈ X, we have that

(i) q(vw) = q(v)q(ϵw) ; (ii) q(θ(a, v)) = q(v)q(π(a)) .

Proof. Since δ · V ⊆ Rad(H), it follows by substituting δ for w and w for z in (Q_{26}) that δ · ϵw · v = δ · vw, hence [q(ϵw)] · v = [q(vw)]. By Lemma 8.51, it follows that [q(v)q(ϵw)] = [q(vw)], which proves (i). Substituting (a, 0) for w in (i) now yields (ii). □

Lemma 8.137. For all a ∈ X, we have that φ(a, π(a)) = 0.

Proof. By Lemma 8.131, we may assume that char(K) = 2. Since f(ϵ, ζ) = 1 = q(ϵ), we have that q(ϵ + ζ) = q(ζ). It then follows, by Lemma 8.124(iv) and Lemma 8.136(ii), that

q(π(a) + θ(a, ζ)) = q(θ(a, ϵ + ζ))
= q(ϵ + ζ)q(π(a))
= q(ζ)q(π(a))
= q(θ(a, ζ)) ,

and hence q(π(a)) = f(π(a), θ(a, ζ)). It thus follows from Lemma 8.135(i) that

φ(a, π(a)) = f(π(a), π(a))f(ζ, π(a)) + f(θ(a, ζ), π(a))f(ϵ, π(a)) + f(ϵ, π(a))q(π(a))
= 0 + q(π(a))f(ϵ, π(a)) + f(ϵ, π(a))q(π(a))
= 0 ,

which is what we had to prove. □

Lemma 8.138. For all v ∈ V and all w ∈ W^{∗}, we have that

w · q(ϵw)^{−1}ϵ · \overline{ϵw} · vw = wv .

Proof. If we substitute λ(w(−ϵ)) for w in (Q_{19}), then we get that

λ(w(−ϵ)) · vw = wv .

If we set v = ϵ in this identity, then we get that λ(w(−ϵ))·ϵw = w, hence λ(w(−ϵ)) = w · (ϵw)^{−1}, and therefore

w · (ϵw)^{−1} · vw = wv .

Note that it follows by substituting [t] for z in (Q_{26}), with t ∈ K, that w·tϵ·v = w·tv. Since (ϵw)^{−1} = q(ϵw)^{−1}\overline{ϵw}, it follows from this identity that

w · q(ϵw)^{−1}ϵ · \overline{ϵw} · vw = wv ,

which is what we had to show. □

Lemma 8.139. For all a ∈ X and all v ∈ V, we have that

(i) q(π(a))av = a\overline{π(a)}θ(a, v) ; (ii) aπ(a)v = aθ(a, v) .

Proof. We may assume that a ≠ 0. First, we substitute (a, 0) for w in Lemma 8.138, and we get that

(a, 0) · q(ϵ(a, 0))^{−1}ϵ · \overline{ϵ(a, 0)} · v(a, 0) = (a, 0)v,

hence

(q(π(a))^{−1}a, 0) · \overline{π(a)} · θ(a, v) = (a, 0)v.

If we calculate the X-component of both sides, then we get that

q(π(a))^{−1}a\overline{π(a)}θ(a, v) = av,

which shows (i).

On the other hand, if we substitute (a, 1) for w in Lemma 8.138, then we get that

(a, 1) · q(ϵ(a, 1))^{−1}ϵ · \overline{ϵ(a, 1)} · v(a, 1) = (a, 1)v,

hence

(q(π(a) + ϵ)^{−1}a, q(π(a) + ϵ)^{−2}) · (\overline{π(a) + ϵ}) · (θ(a, v) + v) = (a, 1)v.

Again, we calculate the X-component of both sides, and we get that

(q(π(a)) + f(π(a), ϵ) + q(ϵ))^{−1}a(\overline{π(a)} + ϵ)(θ(a, v) + v) = av ,

from which it follows that

a\overline{π(a)}θ(a, v) + aθ(a, v) + a\overline{π(a)}v + av = q(π(a))av + f(π(a), ϵ)av + av .

Since f(π(a), ϵ)av = aπ(a)v + a\overline{π(a)}v by Lemma 8.128, it follows by (i) that

aθ(a, v) = aπ(a)v ,

which proves (ii). □

Lemma 8.140. If |K| > 2, then

cθ(a, v) − cπ(a)v = ah(a, c)v − ah(a, cv)

for all a, c ∈ X and all v ∈ V.

Proof. See [20, (26.36)]. □

Lemma 8.141. For all a ∈ X and all v ∈ V, we have that

θ(a, θ(a, v)) = θ(a, v)f(ϵ, π(a)) − q(π(a))v .

Proof. See [20, (26.33)]. □

We can rephrase this identity in terms of W in place of X, which results in a nice identity.

Lemma 8.142. For all v ∈ V and all w ∈ W, we have that vw(⊟w) = −q(ϵw)v.

Proof. Let w = (a, t), with a ∈ X and t ∈ K. Let Q(a) := g(a, −a) = f(π(a), ϵ), and observe that 2Q(a) = 0; see Lemma 8.119 and Lemma 8.131. Note that it follows from Lemma 8.116 that ⊟w = (−a, −t + Q(a)). Hence, by Lemma 8.124 and Lemma 8.141,

vw(⊟w) = v(a, t)(−a, −t + Q(a))
= (θ(a, v) + tv) · (−a, −t + Q(a))
= θ(−a, θ(a, v) + tv) + (−t + Q(a))(θ(a, v) + tv)
= θ(a, θ(a, v)) + tθ(a, v) − tθ(a, v) + Q(a)θ(a, v) − t^{2}v + Q(a)tv
= θ(a, v)Q(a) − q(π(a))v + Q(a)θ(a, v) − t^{2}v + Q(a)tv
= −(q(π(a)) + Q(a)t + t^{2})v
= −q(π(a) + tϵ)v
= −q(ϵw)v ,

and we are done. □

Lemma 8.143. For all v ∈ V and all w ∈ W, we have that

vww = f(ϵ, ϵw)vw − q(ϵw)v .

Proof. By 3.13(i), we have that w(−ϵ) = F (ϵw, ϵ) ⊟ w, and hence, by (Q_{6}), (Q_{12}) and Lemma 8.142,

vww = vw · w(−ϵ)
= vw · (F (ϵw, ϵ) ⊟ w)
= vw · [f(ϵw, ϵ)] + vw(⊟w)
= f(ϵw, ϵ)vw − q(ϵw)v ,

which is what we had to show. □

Definition 8.144. For all v ∈ V^{∗} and all w ∈ W^{∗}, we let [v]_{w} := ⟨v, vw⟩ be the subspace of V (over K) generated by v and vw. Note that [v]_{w} is 2-dimensional if and only if w ∈ W \ Y.

Lemma 8.145. For all v ∈ V^{∗} and all w ∈ W \ Y, we have that [v]_{w} · w = [v]_{w}, i.e. [v]_{w} is a 2-dimensional subspace of V which is irreducible under the action of w.

Proof. It follows from Lemma 8.143 that [v]_{w} ·w = ⟨v, vw⟩·w = ⟨vw, vww⟩ = ⟨vw, f(ϵ, ϵw)vw − q(ϵw)v⟩ = ⟨vw, v⟩ since q(ϵw) ≠ 0. □

Definition 8.146. Let u, v ∈ V^{∗} and w ∈ W \ Y. Then u and v are called w-orthogonal if and only if f([u]_{w}, [v]_{w}) = 0.

Remark 8.147. It is clear that the definition above of [v]_{w} and the notion of w-orthogonality are generalizations of the definition of [v]_{a} and the notion of a-orthogonality as defined in [20]. See [20, (26.37) and (26.38)].

Theorem 8.148. Let a ∈ X^{∗}, and let w := (a, 0) ∈ W^{∗}. Suppose that f(ϵ, π(a)) ≠ 0 if char(K) = 2. Let T be the endomorphism of V given by T (v) := vw for all v ∈ V. Then:

(i) The endomorphism T is a norm splitting map of the quadratic space (V, K, q) ;

(ii) The minimal polynomial of T is

p(x) = x^{2} + f(ϵ, π(a))x + q(π(a)) .

Let E denote the splitting field of p over K, and let γ ∈ E be a root of p. Then E/K is a separable quadratic extension and there is a scalar multiplication from E × V to V extending the scalar multiplication from K × V to V, such that T (v) = γv for all v ∈ V ;

(iii) Let S be a finite set of pairwise w-orthogonal elements of V^{∗}. Then the elements of the set S ∪Sw are linearly independent over K; if this set does not span V, then S can be extended to a larger set of non-zero pairwise w-orthogonal vectors ; (iv) Let ψ : E → [ϵ]_{w} be given by

ψ(r + tγ) := rϵ + tπ(a)

for all r, t ∈ K. Then ψ is an isomorphism of vector spaces and X is a (right) vector space over E with scalar multiplication given by bu := bψ(u) for all b ∈ X and all u ∈ E. If σ denotes the non-trivial element in Gal(E/K), then ψ(u^{σ}) = \overline{ψ(u)} for all u ∈ E. If N denotes the norm of the extension E/K, then N(u) = q(ψ(u)) for all u ∈ E.

Proof. See [20, (26.39)]. □

Lemma 8.149. Let a ∈ X^{∗} be arbitrary, and let w := (a, 0) ∈ W^{∗}. Let D := ⟨ϵ, ϵw, v, vw⟩ for some v ∈ V \ ⟨ϵ, ϵw⟩. Then dim_{K} D = 4, and aDD ⊆ aD (but not necessarily bDD ⊆ bD for other elements b ∈ X).

Proof. See [20, (26.41)]. □

Theorem 8.150. Let dim_{K} V = 4. Then V can be made into a division ring such that X is a right vector space over V with scalar multiplication given by the map (a, v) ↦ av for all a ∈ X and all v ∈ V.

Proof. See [20, (26.42)]. □

It will be convenient now to set v^{σ} := \overline{v} for all v ∈ V.

Theorem 8.151. Suppose that dim_{K} V ∈ {2, 4}. Then there is a multiplication on V which gives V the structure of a division ring with the following properties:

(i) ⟨ϵ⟩ is a subfield lying in the center of V and the map t ↦ tϵ is an isomorphism from K to ⟨ϵ⟩; (ii) σ is an involution of V ; (iii) X is a right vector space over V with scalar multiplication given by the

map (a, v) ↦ av; (iv) q(v) = vv^{σ} = v^{σ}v ∈ ⟨ϵ⟩, and f(u, v) = uv^{σ} + vu^{σ} = u^{σ}v + v^{σ}u ∈ ⟨ϵ⟩ for

all u, v ∈ V ;

(v) h is a skew-hermitian form on X with respect to σ; (vi) (V, ⟨ϵ⟩, σ) is an involutory set ; (vii) θ(a, v) = π(a)v for all a ∈ X and all v ∈ V .

Proof. See [20, (26.43)]. □

Theorem 8.152. Suppose that dim_{K} V ≤ 4. Then dim_{K} V ∈ {2, 4}. Let V be given the structure of a division ring as in Theorem 8.151. Then (V, K, σ, X, π) is an anisotropic pseudo-quadratic space. Moreover, we have that

π(av) = v^{σ}π(a)v − φ(a, v)ϵ

for all a ∈ X and all v ∈ V.

Proof. See [20, (26.44)]. □

Theorem 8.153. Suppose that dim_{K} V ≤ 4. Let (V, K, σ, X, π) be as in Theorem 8.152. Then (V, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{P} (V, K, σ, X, π).

Proof. Let (T, ⊞) be the group defined in section 7.4 applied on the pseudoquadratic space (V, K, σ, X, π). By the definition of T, we have that π(a) − v ∈ ⟨ϵ⟩ for all (a, v) ∈ T. Let χ(a, v) be the unique element t ∈ K such that v − π(a) = tϵ.

Let ϕ be the isomorphism from [V] to V which maps [v] to v for all v ∈ V, and let ψ be the isomorphism from [T] to W which maps [a, v] to (a, χ(a, v)) for all (a, v) ∈ T ⊆ X × V. Then ϕ([1]) = [1] = ϵ and ψ([0, 1]) = (0, χ(0, 1)) = (0, 1) = δ since 1ϵ − π(0) = 1ϵ. (Remember that we have identified K with ⟨ϵ⟩ ⊆ V by Theorem 8.151(i).)

Now, let v ∈ V and (a, x) ∈ T be arbitrary. By Lemma 8.139(ii) and Lemma 8.121(i), we have that

a · xv = aπ(a)v + axv − aπ(a)v
= aθ(a, v) + a(x − π(a))v
= aθ(a, v) + a · χ(a, x)ϵ · v
= aθ(a, v) + χ(a, x)av
= a(θ(a, v) + χ(a, x)v) ,

and hence, by Theorem 8.151(iii), it follows that

xv = θ(a, v) + χ(a, x)v = v · (a, χ(a, x)) .

By Theorem 8.152 and Theorem 8.151(i and iv), we have that

χ(av, v^{σ}xv)ϵ = v^{σ}xv − π(av)
= v^{σ}xv − v^{σ}π(a)v + φ(a, v)ϵ
= v^{σ}(x − π(a))v + φ(a, v)ϵ
= v^{σ} · χ(a, x)ϵ · v + φ(a, v)ϵ
= (χ(a, x)q(v) + φ(a, v))ϵ ,

and hence

χ(av, v^{σ}xv) = χ(a, x)q(v) + φ(a, v) .

It follows that

ϕ([v][a, x]) = ϕ([xv]) = xv = v · (a, χ(a, x)) = ϕ([v])ψ([a, x]) , and
ψ([a, x][v]) = ψ([av, v^{σ}xv]) = (av, χ(av, v^{σ}xv)) = (av, χ(a, x)q(v) + φ(a, v))
= (a, χ(a, x)) · v = ψ([a, x])ϕ([v]) ,

for all v ∈ V and all (a, x) ∈ T.

Hence (ϕ, ψ) is an isomorphism from Ω_{P} (V, K, σ, X, π) to (V, W, τ_{V} , τ_{W} , ϵ, δ).

□
8.7. Quadrangular Systems of Type E_{6}, E_{7} and E_{8}

In this section, we continue to assume that Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is a wide quadrangular system which is the extension of a quadrangular system Λ of quadratic form type, such that Rad(F) = 0. It only remains to consider the case where dim_{K} V > 4.

Lemma 8.154. If char(K) = 2, then there exists an element ξ ∈ X^{∗} such that π(ξ) = αζ for some α ∈ K^{∗}.

Proof. Suppose that g(a, a) = 0 for all a ∈ X. Since g is bilinear, it would follow that g(a, b) = g(b, a) for all a, b ∈ X, and hence, by Lemma 8.116, that W is abelian. It would then follow by (Q_{8}) that Im(H) ⊆ Rad(F). Since Rad(F) = 0 and H ≢ 0, this is a contradiction.

Hence there exists an element a ∈ X^{∗} such that g(a, a) ≠ 0. Let w_{1} := (a, 0) ∈ W^{∗}. By Lemma 8.119, it follows that f(ϵ, ϵw_{1}) = f(ϵ, π(a)) = g(a, a) ≠ 0. Hence (see Definition 8.109)

S_{1} ∩ S_{2} = {v ∈ V | F (ϵ, v) ≠ 0} ∩ {ϵw | w ∈ W } ≠ ∅ ,

since ϵw_{1} ∈ S_{1} ∩ S_{2}. By the definition of ζ, this implies that ζ = f(ϵ, z)^{−1}z for some z ∈ S_{1} ∩ S_{2}. Let z = ϵw_{2} for some w_{2} ∈ W^{∗}. Since f(ϵw_{2}, ζ) = f(z, ζ) = f(z, f(ϵ, z)^{−1}z) = 0, w_{2} is ζ-orthogonal, hence w_{2} = (ξ, 0) for some ξ ∈ X^{∗}. We conclude that π(ξ) = ϵ(ξ, 0) = ϵw_{2} = z = f(ϵ, z)ζ = αζ for α = f(ϵ, z) ∈ K^{∗}. □

Definition 8.155. If char(K) ≠ 2, let ξ be an arbitrary element of X^{∗}. If char(K) = 2, choose ξ ∈ X^{∗} as in Lemma 8.154.

By Lemma 8.154 and Theorem 8.148, the endomorphism T of V which maps v to v(ξ, 0) is a norm splitting map of q.

We have come to a point which is completely similar to the beginning of Chapter 27 in [20], and the rest of the proof could literally be copied from that chapter.

Theorem 8.156. The quadratic space (K, V_{0}, q) is of type E_{6}, E_{7} or E_{8}.

Proof. The proof is exactly as in [20, (27.17)], where we have to use Definition 8.117 and 8.118 and Lemmas 8.124, 8.127, 8.128, 8.129, 8.130, 8.131, 8.132, 8.135, 8.139, 8.140, 8.141 and 8.149. □

Theorem 8.157. (V, W, τ_{V} , τ_{W} , ϵ, δ) ≅ Ω_{E}(K, V_{0}, q).

Proof. It follows from the proof of [20, (27.19)], using Definition 8.109 and 8.118 as well as Lemmas 8.124, 8.128, 8.129, 8.130, 8.131, 8.132, 8.133, 8.135 and 8.139, that the maps h, g, θ and φ are exactly as in section 7.5.

Let ϕ be the map from [V_{0}] to V which maps [v] to v for all v ∈ V, and let ψ be the map from [S] to W which maps [a, t] to (a, t) for all (a, t) ∈ S. Since we have seen in Definitions 8.117 and 8.118 that (a, t) · v = (av, tq(v) + φ(a, v)) and v · (a, t) = θ(a, v) + tv, it is now obvious that (ϕ, ψ) is an isomorphism from Ω_{E}(K, V_{0}, q) to (V, W, τ_{V} , τ_{W} , ϵ, δ). □

This completes the proof of Theorem 8.10, and thereby the proof of the classification of quadrangular systems.

APPENDIX A Abelian Quadrangular Systems

In this appendix, we will describe the quadrangular systems (V, W, τ_{V} , τ_{W} , ϵ, δ) where W is abelian, and we will restate the axiom system for some specific cases.

A quadrangular system Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) will be called abelian if and only if W is abelian. One can check that Ω is abelian if and only if it is of quadratic form type, of involutory type, of indifferent type or of type F_{4}. (Note that, if Ω is of pseudo-quadratic form type with W abelian, then Ω is in fact reduced, and hence of one of these types.) In this case, we simply write + and − in place of ⊞ and ⊟, respectively, and we get the following description.

Consider an abelian group (V, +) and an abelian group (W, +). Suppose that there is a map τ_{V} from V ×W to V and a map τ_{W} from W ×V to W, both of which will be denoted by · or simply by juxtaposition, i.e. τ_{V} (v, w) = vw = v · w and τ_{W} (w, v) = wv = w · v for all v ∈ V and all w ∈ W. Consider a map F from V × V to W and a map H from W ×W to V which are additive in both variables. Suppose furthermore that there exists a fixed element ϵ ∈ V^{∗} and a fixed element δ ∈ W^{∗}, and suppose that, for each v ∈ V^{∗}, there exists an element v^{−1} ∈ V^{∗}, and for each w ∈ W^{∗}, there exists an element w^{−1} ∈ W^{∗}, such that, for all w, w_{1}, w_{2} ∈ W and all v, v_{1}, v_{2} ∈ V, the following axioms are satisfied.

(A_{1}) wϵ = w.

(A_{2}) vδ = v.

(A_{3}) (w_{1} + w_{2})v = w_{1}v + w_{2}v.

(A_{4}) (v_{1} + v_{2})w = v_{1}w + v_{2}w.

(A_{5}) v(−w) = −vw.

(A_{6}) w(−v) = wv.

(A_{7}) Im(F) ⊆ Rad(H).

(A_{8}) Im(H) ⊆ Rad(F).

(A_{9}) δ ∈ Rad(H).

(A_{10}) If Rad(F) ≠ 0, then ϵ ∈ Rad(F).

(A_{11}) w(v_{1} + v_{2}) = wv_{1} + wv_{2} + F (v_{1}w, v_{2}).

(A_{12}) v(w_{1} + w_{2}) = vw_{1} + vw_{2} + H(w_{1}v, w_{2}).

(A_{13}) (v^{−1})^{−1} = v (if v ≠ 0).

(A_{14}) (w^{−1})^{−1} = w (if w ≠ 0).

(A_{15}) wvv^{−1} = w (if v ≠ 0).

(A_{16}) vww^{−1} = v (if w ≠ 0).

(A_{17}) v^{−1}(wv) = \overline{vw} (if v ≠ 0).

(A_{18}) w^{−1}(vw) = wv (if w ≠ 0).

(A_{19}) F (v_{1}^{−1}, \overline{v_{2}})v_{1} = F (v_{1}, v_{2}) (if v_{1} ≠ 0).

(A_{20}) H(w_{1}^{−1}, w_{2})w_{1} = H(w_{1}, w_{2}) (if w_{1} ≠ 0).

Then (V, W, τ_{V} , τ_{W} , ϵ, δ) is an abelian quadrangular system.

A.1. Reduced Quadrangular Systems

If Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is reduced or indifferent, i.e. if H ≡ 0, then Ω is abelian, and we get the following description.

Consider an abelian group (V, +) and an abelian group (W, +). Suppose that there is a map τ_{V} from V × W to V and a map τ_{W} from W × V to W, both of which will be denoted by · or simply by juxtaposition, i.e. τ_{V} (v, w) = vw = v · w and τ_{W} (w, v) = wv = w · v for all v ∈ V and all w ∈ W. Consider a map F from V × V to W which is additive in both variables. Suppose furthermore that there exists a fixed element ϵ ∈ V^{∗} and a fixed element δ ∈ W^{∗}, and suppose that, for each v ∈ V^{∗}, there exists an element v^{−1} ∈ V^{∗}, and for each w ∈ W^{∗}, there exists an element w^{−1} ∈ W^{∗}, such that, for all w, w_{1}, w_{2} ∈ W and all v, v_{1}, v_{2} ∈ V, the following axioms are satisfied.

(R_{1}) wϵ = w.

(R_{2}) vδ = v.

(R_{3}) (w_{1} + w_{2})v = w_{1}v + w_{2}v.

(R_{4}) (v_{1} + v_{2})w = v_{1}w + v_{2}w.

(R_{5}) v(−w) = −vw.

(R_{6}) w(−v) = wv.

(R_{7}) If Rad(F) ≠ 0, then ϵ ∈ Rad(F).

(R_{8}) w(v_{1} + v_{2}) = wv_{1} + wv_{2} + F (v_{1}w, v_{2}).

(R_{9}) v(w_{1} + w_{2}) = vw_{1} + vw_{2}.

(R_{10}) (v^{−1})^{−1} = v (if v ≠ 0).

(R_{11}) (w^{−1})^{−1} = w (if w ≠ 0).

(R_{12}) wvv^{−1} = w (if v ≠ 0).

(R_{13}) vww^{−1} = v (if w ≠ 0).

(R_{14}) v^{−1}(wv) = \overline{vw} (if v ≠ 0).

(R_{15}) w^{−1}(vw) = wv (if w ≠ 0).

(R_{16}) F (v_{1}^{−1}, \overline{v_{2}})v_{1} = F (v_{1}, v_{2}) (if v_{1} ≠ 0).

Then (V, W, τ_{V} , τ_{W} , ϵ, δ) is a reduced or indifferent quadrangular system (and it is reduced if and only if F ≢ 0).

Remark A.1. As we explained in Remark 6.4, axiom (R_{7}) had only been introduced to simplify the classification result of the wide quadrangular systems. In particular, it is not needed for the reduced quadrangular systems, and it is in fact often more convenient – and perfectly allowed – to leave it out.

A.2. Indifferent Quadrangular Systems

If Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) is indifferent, i.e. if F ≡ 0 and H ≡ 0, then Ω is abelian, and we get the following description.

Consider an abelian group (V, +) and an abelian group (W, +). Suppose that there is a map τ_{V} from V × W to V and a map τ_{W} from W × V to W, both of which will be denoted by · or simply by juxtaposition, i.e. τ_{V} (v, w) = vw = v · w and τ_{W} (w, v) = wv = w · v for all v ∈ V and all w ∈ W. Suppose furthermore that there exists a fixed element ϵ ∈ V^{∗} and a fixed element δ ∈ W^{∗}, and suppose that, for each v ∈ V^{∗}, there exists an element v^{−1} ∈ V^{∗}, and for each w ∈ W^{∗}, there exists an element w^{−1} ∈ W^{∗}, such that, for all w, w_{1}, w_{2} ∈ W and all v, v_{1}, v_{2} ∈ V, the following axioms are satisfied.

(D_{1}) wϵ = w.

(D_{2}) vδ = v.

(D_{3}) (w_{1} + w_{2})v = w_{1}v + w_{2}v.

(D_{4}) (v_{1} + v_{2})w = v_{1}w + v_{2}w.

(D_{5}) w(v_{1} + v_{2}) = wv_{1} + wv_{2}.

(D_{6}) v(w_{1} + w_{2}) = vw_{1} + vw_{2}.

(D_{7}) (v^{−1})^{−1} = v (if v ≠ 0).

(D_{8}) (w^{−1})^{−1} = w (if w ≠ 0).

(D_{9}) wvv^{−1} = w (if v ≠ 0).

(D_{10}) vww^{−1} = v (if w ≠ 0).

(D_{11}) v^{−1}(wv) = vw (if v ≠ 0).

(D_{12}) w^{−1}(vw) = wv (if w ≠ 0).

Then (V, W, τ_{V} , τ_{W} , ϵ, δ) is an indifferent quadrangular system. (Note that we do not have to assume a priori that all elements of V and W have order at most 2, but that this follows from these axioms.)

A.3. Radical Quadrangular Systems

An abelian quadrangular system Ω = (V, W, τ_{V} , τ_{W} , ϵ, δ) will be called radical if and only if Rad(F) ≠ 0. One can check that Ω is radical if and only if it is of quadratic form type with ϵ ∈ Rad(f) (and hence char(K) = 2), of indifferent type or of type F_{4}. We will give two different (but equivalent) descriptions. The first one is useful to check whether a certain system is a radical quadrangular system; the second one is more convenient to work with. Note that each of these descriptions is completely symmetrical.

A.3.1. First Description. Consider an abelian group (V, +) and an abelian group (W, +). Suppose that there is a map τ_{V} from V × W to V and a map τ_{W} from W × V to W, both of which will be denoted by · or simply by juxtaposition, i.e. τ_{V} (v, w) = vw = v · w and τ_{W} (w, v) = wv = w · v for all v ∈ V and all w ∈ W. Consider a map F from V × V to W and a map H from W × W to V which are additive in both variables. Suppose furthermore that there exists a fixed element ϵ ∈ V^{∗} and a fixed element δ ∈ W^{∗}, and suppose that, for each v ∈ V^{∗}, there exists an element v^{−1} ∈ V^{∗}, and for each w ∈ W^{∗}, there exists an element w^{−1} ∈ W^{∗}, such that, for all w, w_{1}, w_{2} ∈ W and all v, v_{1}, v_{2} ∈ V, the following axioms are satisfied.

(F_{1}) wϵ = w. (F_{2}) vδ = v. (F_{3}) (w_{1} + w_{2})v = w_{1}v + w_{2}v. (F_{4}) (v_{1} + v_{2})w = v_{1}w + v_{2}w. (F_{5}) Im(F) ⊆ Rad(H). (F_{6}) Im(H) ⊆ Rad(F). (F_{7}) δ ∈ Rad(H). (F_{8}) ϵ ∈ Rad(F). (F_{9}) w(v_{1} + v_{2}) = wv_{1} + wv_{2} + F (v_{1}w, v_{2}). (F_{10}) v(w_{1} + w_{2}) = vw_{1} + vw_{2} + H(w_{1}v, w_{2}). (F_{11}) (v^{−1})^{−1} = v (if v ≠ 0). (F_{12}) (w^{−1})^{−1} = w (if w ≠ 0). (F_{13}) wvv^{−1} = w (if v ≠ 0). (F_{14}) vww^{−1} = v (if w ≠ 0).

(F_{15}) v^{−1}(wv) = vw (if v ≠ 0).

(F_{16}) w^{−1}(vw) = wv (if w ≠ 0).

(F_{17}) F (v_{1}^{−1}, v_{2})v_{1} = F (v_{1}, v_{2}) (if v_{1} ≠ 0).

(F_{18}) H(w_{1}^{−1}, w_{2})w_{1} = H(w_{1}, w_{2}) (if w_{1} ≠ 0).

Then (V, W, τ_{V} , τ_{W} , ϵ, δ) is a radical quadrangular system. It is of type F_{4} if and only if F ≢ 0 and H ≢ 0.

A.3.2. Second Description. Let K and L be two commutative fields with char(K) = char(L) = 2, such that K is a vector space over L and that L is a vector space over K. If t is an element of the field K, then we will denote the corresponding element of the vector space K by [t]; if s is an element of the field L, then we will denote the corresponding element of the vector space L by [s]. Let V be a vector space over K containing [L] as a subspace, and let W be a vector space over L containing [K] as a subspace.

Suppose that q is an anisotropic quadratic form from V to K, with corresponding bilinear form f, and that q̂ is an anisotropic quadratic form from W to L, with corresponding bilinear form f̂, such that [L] ⊆ Rad(f) and [K] ⊆ Rad(f̂). Let ϵ := [1] ∈ [L] ⊆ V and δ := [1] ∈ [K] ⊆ W. Finally, suppose that there is a map τ_{V} from V × W to V which is K-linear on V, and a map τ_{W} from W × V to W which is L-linear on W, both of which will be denoted by · or simply by juxtaposition, i.e. τ_{V} (v, w) = vw = v · w and τ_{W} (w, v) = wv = w · v for all v ∈ V and all w ∈ W. Moreover, suppose that the following axioms hold, for all v ∈ V, w ∈ W, t ∈ K and s ∈ L.

(C_{1}) v[t] = tv.

(C_{2}) w[s] = sw.

(C_{3}) v · sw = vw · sδ.

(C_{4}) w · tv = wv · tϵ.

(C_{5}) [t]v = [tq(v)].

(C_{6}) [s]w = [sq̂(w)].

(C_{7}) vww = v · q̂(w)δ.

(C_{8}) wvv = w · q(v)ϵ.

(C_{9}) v · wv = q(v)vw.

(C_{10}) w · vw = q̂(w)wv.

(C_{11}) v(w_{1} + w_{2}) = vw_{1} + vw_{2} + [f̂(w_{1}v, w_{2})].

(C_{12}) w(v_{1} + v_{2}) = wv_{1} + wv_{2} + [f(v_{1}w, v_{2})].

Then (V, W, τ_{V} , τ_{W} , ϵ, δ) is a radical quadrangular system. It is of type F_{4} if and only if f ≢ 0 and f̂ ≢ 0.

Bibliography

[1] A. A. Albert, On exceptional Jordan division algebras, Pacific J. Math. 15, 1965, pp. 377–404.

[2] R. Baer, Homogeneity of projective planes, Amer. J. Math. 64, 1942, pp. 137–152.

[3] R. H. Bruck and E. Kleinfeld: The structure of alternative division rings, Proc. Amer. Math. Soc. 2, 1951, pp. 878–890.

[4] T. De Medts, Automorphisms of F_{4} quadrangles, to appear in Math. Ann.

[5] M. Hall, Jr.: Projective Planes, Trans. Amer. Math. Soc. 54, 1943, pp. 229–277.

[6] F. D. Jacobson and N. Jacobson: Classification and representation of semisimple Jordan algebras, Trans. Amer. Math. Soc. 65, 1949, pp. 141–169.

[7] N. Jacobson: Structure and Representation of Jordan Algebras, Amer. Math. Soc. Colloquium Ser. 39, Providence, 1968.

[8] : “Lectures on Quadratic Jordan Algebras”, Tata Institute Lecture Notes, 1969.

[9] K. McCrimmon: The Freudenthal-Springer-Tits constructions of exceptional Jordan algebras, Trans. Amer. Math. Soc. 139, 1969, pp. 495–510.

[10] : The Freudenthal-Springer-Tits constructions revisited, Trans. Amer. Math. Soc. 148, 1970, pp. 293–314.

[11] R. Moufang: Alternativkörper und der Satz vom vollständigen Vierseit, Abh. Math. Sem. Univ. Hamburg 9, 1933, pp. 207–222.

[12] B. Mühlherr and H. Van Maldeghem: Exceptional Moufang quadrangles of type F_{4}, Canad. J. Math. 51, 1999, pp. 347–371.

[13] H. P. Petersson and M. L. Racine: Jordan algebras of degree 3 and the Tits process, J. Algebra 98, 1986, pp. 211–243.

[14] : Classification of algebras arising from the Tits process, J. Algebra 98, 1986, pp. 244–279.

[15] J. Tits: Buildings of Spherical Type and Finite BN-Pairs, Lecture Notes in Mathematics 386, Springer-Verlag, New York/Heidelberg/Berlin, 1974.

[16] : Classification of buildings of spherical type and Moufang polygons: a survey, in Coll. Intern. Teorie Combinatorie Acc. Naz. Lincei, Proceedings Roma 1973, Atti dei Convegni Lincei 17, 1976, pp. 229–246.

[17] : Non-existence de certains polygons généralisés, I, Invent. Math. 36, 1976, pp. 275–284.

[18] : Non-existence de certains polygons généralisés, II, Invent. Math. 51, 1979, pp. 267–269.

[19] : Moufang octagons and Ree groups of type ^{2}F_{4}, Amer. J. Math. 105, 1983, pp. 539–594.

[20] J. Tits and R. Weiss: “Moufang Polygons”, Springer Monographs in Mathematics, Springer-Verlag, Berlin/Heidelberg/New York, 2002.

[21] R. Weiss: The nonexistence of certain Moufang polygons, Invent. Math. 51, 1979, pp. 261–266.

[22] M. Zorn: Theorie der alternativen Ringe, Abh. Math. Sem. Hamburg. Univ. 8, 1930, pp. 123–147.