Geometriae Dedicata 69: 67–81, 1998. c 1998 Kluwer Academic Publishers. Printed in the Netherlands.

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Fourgonal Extensions ¨ STEFAN LOWE Technische Universit¨at Braunschweig, Institut fu¨ r Geometrie, Pockelsstr. 14, D-38106 Braunschweig, Germany. e-mail: [email protected] (Received: 22 March 1996; revised version: 7 March 1997) Abstract. A fourgonal extension is an algebraic description of a certain class of generalized quadrangles. It is a special case of the fourgonal families of Kantor. In this article we introduce fourgonal extensions and morphisms of fourgonal extensions. The main result is that the isomorphism classes of fourgonal extensions correspond bijectively to the isomorphism classes of generalized quadrangles obtained from fourgonal extensions. We also characterize the generalized quadrangles arising from fourgonal extensions. For the finite case, we consider the known examples with emphasis on characteristic 2. Mathematics Subject Classification (1991): 51E12. Key words: generalized quadrangles, fourgonal families.

1. Introduction Let Q = (P ; L; I) be a generalized quadrangle with not necessarily finite parameters (s; t). An elation of Q about a point 1 is an automorphism of Q which is either the identity, or fixes any line through 1 but no point in Pn1? . We call Q an elation generalized quadrangle if it has a group G of elations about a point 1 which is regular on the set of points Pn1?. In this case we write (Q(1) ; G) instead of Q. An elation generalized quadrangle (Q(1) ; G) can be recovered from G provided we know the line stabilizers Xi in G of every line `i through an arbitrary but fixed point p not collinear with 1, and all the stabilizers Xi of the points in fp; 1g? . The recovery is done as in [3] for the finite case. Define an incidence structure where the set of points is the union of the following sets: (P1) The elements of G. (P2) The set of cosets of the subgroups Xi . (P3) The set consisting of an additional symbol 1. The set of lines is the union of the following sets: (L1) The cosets of the subgroups Xi . (L2) For every subgroup Xi a symbol [Xi ]. There are three types of incidence:

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(I1) 1 is collinear with all lines of type (L2). (I2) A line [Xi ] of type (L2) is incident precisely with the points Xi g of type (P2). (I3) All other incidences are induced by the containment relation. This incidence structure is a generalized quadrangle isomorphic to the original generalized quadrangle. Which properties of the group G and the two families of subgroups make this construction work? Suppose that we have an arbitrary group G with two families X and X of subgroups. The elements of X should correspond to the subgroups Xi , the elements of X should correspond to the subgroups Xi . So let us impose the conditions that jXj = jX j = t + 1, all elements of X have cardinality st, every element X of X has cardinality s and is contained in an element X  of X. If (K1) for two different X , Y 2 X, X \ Y  = f1g and XY  = G, (K2) for three pairwise different XS , Y , Z 2 X, (X
Y ) \ Z = f1g, (K3) for every X 2 X, G = X  [ Y 2X X
Y ,

then the above construction gives an elation generalized quadrangle Q = (Q(1) ; G) with parameters (s; t). The elements of X are the stabilizers of the lines through 1, the elements of X are the stabilizers of the points X  of type (P2). We call such a successful pair (G; X) a fourgonal family. We drop X in our notation , since
every S  X is uniquely determined as the complement of Y 2XnfX g X
Y nf1g .

2. Fourgonal Extensions We start with some preliminary definitions. Let F and V be two Abelian groups of cardinality t and s2 . A central extension of V by F is an exact sequence 0 ! F ! G ! V ! 0 such that the image of F is contained in the center of the middle group G. Every possible central extension is given by a map f : V  V ! F such that the 2-cocycle condition holds For every x;

y; z 2 V; f (x; y) + f (x + y; z) = f (y; z ) + f (x; y + z ):

Trivial examples for 2-cocycles are the 2-boundaries. These are defined as follows: for an arbitrary function h: V ! F , let @h: V  V ! F , (x; y ) 7! h(x) + h(y) , h(x + y). Then @h is a 2-boundary. Less trivial examples are provided by biadditive functions f . On the set F  V we define a multiplication

[a; x]
[b; y] = [a + b + f (x; y); x + y]; which defines a group structure G on F  V which is a central extension of V by F . Here the injection of F is a 7! [a; 0], and the projection onto V is [a; x] 7! x. DEFINITION 1. A fourgonal extension with parameters (s; t) is a quintuple E = (F; V; U; f; h) where

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(1) F is an Abelian group of cardinality t, (2) V is an Abelian group of cardinality s2 , (3) U is a system of t + 1 subgroups of V of cardinality s, (4) f : V  V ! F is a 2-cocycle, normalized such that f (0; (5) h: V ! F is a map,

: ) = f ( : ; 0) = 0,

such that (FE1) Two different members of U have trivial intersection and generate V . (FE2) For all X 2 U and all x, y 2 X , f (x; y ) + @h(x; y ) = 0. (FE3) For any three pairwise different X , Y , Z 2 U, if x 2 X and y 2 Y such that x + y 2 Z , then f (x; y ) + @h(x; y ) = 0 implies x = y = 0. (FE4) For all X 2 U and all z 2 V nX let

(

VX;z := (x; y) 2 X  Then the map VX;z

[

Z 2U

)

Z j x+y =z :

! F , (x; y) 7! h(x) + h(y) + f (x; y) is surjective.

With E there is associated a central extension of V by F . We denote its middle group by G(E ) or G when there is no danger of confusion. Remark 1. The datum h is superfluous as we may replace f by the cohomologous cocycle f , @h. We keep it since it is sometimes desirable to witch to special cocycles f . Even then, only its values on the members of U matter. Our first aim is to show that E gives rise to a generalized quadrangle. The geometric reason is that from E we can construct a triangle-free cover of the net (V; U) which can be ‘projectively’ completed to a generalized quadrangle, see [5]. We will now see how fourgonal extensions are related to fourgonal families. To this extent we define two operators A and A on the set U, which give subgroups of the middle group G of E .

X 2 U let A(X ) := f[h(x); x] j x 2 X g and A (X ) := f[a; x] j x 2 X; a 2 F g

DEFINITION 2. For

Note that the definition uses only (1)–(5) of Definition 1, so we can employ this construction in very general circumstances. The following theorem shows that (FE1–4) are needed to create generalized quadrangles. THEOREM 1. Let the data E := (F; V; U; f; h) have the properties (1)–(5) of Definition 1. Let X := fA(X ) j X 2 Ug and define the map  on X by A(X ) ! A (X ). Then X is a fourgonal family of G if and only if E is a fourgonal extension. Proof. Suppose that E is a fourgonal extension. The sets A(X ) and A (X ) are by (FE2) subgroups of G which have the correct cardinality. Let us show that the properties (K1), (K2), and (K3) are also fulfilled. First we have to show that for different X , Y 2 U, A(X ) \ A (Y ) = f0g and A(X )
A (X ) = G. From (FE2)

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and (4) of Definition 1 we obtain 0 = f (0; 0) = h(0). If [a; x] 2 A(X ) \ A (Y ) then x 2 X \ Y = f0g by (FE1), and a = h(x) = h(0) = 0 so the intersection is trivial. Let now [a; z ] 2 G. There exist unique x 2 X , y 2 Y with z = x + y . Then [h(x); x][b; y ] = [h(x) + b + f (x; y ); z ]. Choose b := a , h(x) , f (x; y ) to obtain [a; z ] 2 A(X )A (Y ), which finishes the proof of (K1). For (K2) let X , Y , Z 2 U be pairwise different and let [h(x); x]
[h(y ); y ] = [h(x + y); x + y] where x 2 X , y 2 Y such that x + y 2 Z . Then h(x + y) = h(x) + h(y) + f (x; y), which implies x = y = 0 by (FE3). To prove (K3), let [a; z ] 2 GnA (X ). Then z 62 X , so we may apply (FE4) to find x 2 X , y 2 [Z 2U Z such that x + y = z and h(x) + h(y ) + f (x; y ) = a. Let Y 2 U such that y 2 Y . Then [h(x); x] 2 A(X ), [h(y ); y ] 2 A(Y ), and [a; z ] = [h(x); x][h(y); y] 2 A(X )
A(Y ). Suppose now that X is a fourgonal family. From (K1) it follows in particular that X \ Y = f0g for different X , Y 2 U, which implies (FE1). As every A(X ) is a subgroup of G, we have

[h(x); x][h(y); y] = [h(x) + h(y) + f (x; y); x + y]; for x, y 2 X , which must be equal to [h(x + y ); x + y ]. Thus (FE2) holds. Let X , Y , Z 2 U and x 2 X , y 2 Y such that x + y 2 Z . Suppose that f (x; y) = ,@h(x; y). Then [h(x); x][h(y ); y ] = [h(x + y ); x + y ], thus

[h(x + y); x + y] 2 A(X )A(Y ) \ A(Z ) = f0g; by (K2). We obtain x = y = 0, so (FE3) holds. Let X 2 U, z 2 V nX , and a 2 F . Then [h(x); x] 2 A(X ), [h(y); y] 2 A(Y ) such that

[a; z ] 62 A (X ), so there exist

[a; z ] = [h(x); x][h(y); y] = [h(x) + h(y) + f (x; y); x + y]: Since x + y

= z we have (x; y) 2 VX;z , which proves (FE4).

This shows that with every fourgonal extension there is associated a generalized quadrangle which we will denote by q(E ) or sometimes Q if there is no danger of confusion. DEFINITION 3. Let E = (F; V; U; f; h) and E 0 = (F 0 ; V 0 ; U0 ; f 0 ; h0 ) be two fourgonal extensions with parameter (s; t). A morphism m: E ! E 0 is a triple (; ;  ) where (1) : F ! F 0 is a group homomorphism, (2) : V ! V 0 is a group homomorphism, (3)  : V ! F 0 is a map,

such that

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(MOR1) U  U0 , (MOR2) (f (x; y )) = f 0 ( x; y ) , @ (x; y ) for all X , Y 2 U, x 2 X , y (MOR3) For all X 2 U and all x 2 X , (h(x)) ,  (x) = h0 ( x).

2Y,

In particular, m induces a group homomorphism of the middle group of E to the middle group of E 0 , mapping [a; x] to [(a) +  (x); (x)]. We shall abuse notation and call this homomorphism m as well. In general, it is not true that every homomorphism between middle groups of central extensions arises in this way. Composition of two morphisms m = (; ;  ) and m0 = (0 ; 0 ;  0 ) is the morphism m0  m = (0 ; 0 ; 0  +  0 ). The identity on E is given by (idF , idV ; 0). If m is invertible, its inverse sis given by

m,1 = (,1 ; ,1 ; ,,1  ,1 ): In particular, m is invertible if and only if  and are group isomorphisms. Naturally a morphism m: E ! E 0 induces a geometric morphism q(m): q(E ) ! q(E 0 ), which maps 1 to 10 , X 2 U to X 2 U0 , and maps the remainder of Q as m maps the corresponding subsets of G(E ). Clearly, q(id) = id. The main problem is to study the morphisms between Q and Q0 in terms of the morphisms between the underlying fourgonal extensions. 3. Isomorphisms of Generalized Quadrangles Let E = (F; V; U; f; h) be a fourgonal extension. Then q(E ) has by definition a special point 1 which is a regular point, see [9, 1.3]. Let E 0 be another fourgonal extension. In general we will distinguish the objects associated with E from the corresponding objects of E 0 with a prime 0 . Suppose that Q and Q0 are isomorphic generalized quadrangles. In this section we want to prove that then there exists an isomorphism : Q ! Q0 such that (1) = 10 . We will need this type of isomorphisms in the next paragraph. We further give geometric characterizations of some subgroups of G, using ideas of [3]. Every element g 2 G induces an isomorphism  (g ): Q ! Q0 , which fixes 1 and all the lines through 1. The action on the rest of the quadrangle is induced by right multiplication with g ,1 on G. The group  (G) is regular on the set of points not collinear with 1, so (q(E )(1) ;  (G(E ))) is an elation generalized quadrangle. The following lemma is well-known, at least in the finite case, see [3, 3.1(i)]. LEMMA 1. Let H be the full automorphism group of (Q(1) ; G). Then the orbit H (1) is (a) either f1g, and 1 is the only fixpoint of H , or (b) a line through 1, and H has two orbits on the points, or (c) the full set of points of Q.

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Proof. We shall exploit that G 6 H is transitive on := fq j q  1g and on every line through 1. This implies in (a) that 1 is the only fixpoint. For (b), let L = H (1) and let p := (1) 6= 1 be a point on L. Then for a line M 6= L through 1, (M ) has a point in , so M nf1g  H (! ) for an arbitrary ! 2 . Assume finally that neither (a) nor (b) hold. It is then sufficient to show that H (1) contains a point of , and contains from every line through 1 a point 6= 1. By our assumption there exist , 2 H such that (1) 6= 1, (1) 6= 1, and (1)  (1). If 1ILI(1) and 1IM I (1) we have L, M  H (1). Further, at least one of the lines (L), (M ) through (1) is not L, hence it must contain points of . We may therefore assume right from start that (1)  1, which implies  H (1). Let Li be a line through 1. Then every line (Li ) contains a point 6= (1), so Li contains a point from H (1). THEOREM 2. Let (Q(1) ; G) and (Q0 (1 ) ; G0 ) be two elation generalized quadrangles. Suppose that Q and Q0 are isomorphic. Then there exists an isomorphism : Q ! Q0 with (1) = 10 . ,
Proof. Let : Q ! Q0 be an isomorphism. Then by Lemma 1, H (1) = H 0(10). 0

COROLLARY 1. Let E and E 0 be two fourgonal extensions. Suppose that q(E )  = q(E 0 ). Then there exists an isomorphism : q(E ) ! q(E 0 ) with (1) = 10 and (1) = 10 . We shall need a more detailed knowledge of the action of automorphisms and the substructure of their fixpoints and fixlines. We start with a fundamental lemma for an arbitrary generalized quadrangle Q. LEMMA 2. Let  be an automorphism of Q.

(a) Let L be a fixpoint line and M be a fixline which does not intersect L. Then M is a fixpoint line. (b) Let every line through p be a fixpoint line. If there exists a fixpoint q  p, then  = id. (c) Suppose that L is a fixpoint line and p is a fixline point on L. If there exists a fixpoint q  p, then  = id. Proof.

(a) For every xIM there exists an unique  (x)IL such that x   (x). Then M I(x)  (x), hence (x) = x. (b) Every line through q is a fixline. From (a) it follows that it even is a fixpoint line. Let L be a line not through p or through q . Then there exists unique points x and y with LIx  p and LIy  q . Call L of class I if x 6= y . We see immediately that lines of class I are fixpoint lines. Suppose now that x = y . We want to ensure the existence of z 6= x and M 6= L with LIz IM and M

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missing fp; q g? . Suppose that by bad luck we have taken a point z 0 IL such that all lines through z 0 meet fp; q g? . Then we may take any z 6= x; z 0 on L and choose any line M 6= L through z . Now M must be of class I, hence z is a fixpoint. Since x also is a fixpoint, the hypothetical z 0 must be a fixpoint as well. So L is a fixpoint line, showing that  = id. (c) Let M be a line through q missing L, and let p I L0  M . Then M is a fixline, hence, by (a), a fixpoint line. So apart from the line M0 through q intersecting L, all lines through q are fixpoint lines. It follows sequentially that L0 and M0 are fixpoint lines, and now (b) proves the assertion. We obtain the following characterization of the subgroups A(X ) and F of the middle group G of a fourgonal extension. This will be very important, as these subgroups generate G. LEMMA 3. Let E be a fourgonal extension. For every X 2 U, let S (X ) be the set of automorphisms ! 2 H of q(E ) which fix all lines through 1, all lines through A (X ), and which fix the line [X ] of type (L2) pointwise. Then S (X ) =  (A(X )). Proof. Clearly  (A(X )) is a subgroup of S := S (X ). Let L be a line 6= [X ] through A (X ) and A (X ) 6= x I L. For every ! 2 S there exists a g 2 A(X ) such that ! (x) =  (g )(x), hence ! ,1  (g ) 2 S fixes x. Since x cannot be collinear with 1 and A (X ) simultaneously, we obtain from Lemma 2(c) that ! =  (g ). LEMMA 4. Let E be a fourgonal extension. Let T be the set of all automorphisms ! 2 H which fix all lines through 1 pointwise. Then T =  (F ). Proof. Clearly  (F ) is a subgroup of T . Let M 6= [X ] be a line trough A (X ). For every ! 2 T there exists a g 2 F such that ! (M ) =  (g )(M ), hence M is a fixline for ! ,1  (g ). From Lemma 2(a) it follows that M is a fixpoint line, hence 2(b) shows that ! =  (g ). 4. The Equivalence We set up a bijection between the isomorphism classes of fourgonal extensions and the isomorphism classes of generalized quadrangles constructed from a fourgonal extension. In the next section we characterize these quadrangles geometrically. Let E be the category of fourgonal extensions with parameter (s; t). Let Q be the full subcategory of generalized quadrangles arising from fourgonal extensions. Our map q is a covariant functor from E to Q, and the main task is to study the morphisms in Q in terms of the morphisms q(m), where m runs through the morphisms of E. LEMMA 5. Let E , E 0 2 E. Let : q(E ) ! q(E 0 ) be an isomorphism such that (1) = 10 and (1) = 10 . Then there exists an isomorphism m: E ! E 0 such that q(m) = .

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Proof. Let Q := q(E ) and Q0 := q(E 0 ). Since (1) = 10 and G is the set of points of Q not collinear with 1, it follows that (G) = G0 . Let  be the restriction of  onto G. Furthermore,  induces an isomorphism  from H := Aut(Q) to H 0 := Aut(Q0 ) which maps ! to   !  ,1. We first show that

 ( (G)) =  0 (G0 ):

(1)

We investigate the image of the generating set A(U) [ F of G. Let X 2 U. The geometric characterization of  (A(X )) in Lemma 3 together with (1) = 10 and (1) = 10 proves the existence of X 0 2 U0 such that  ( (A(X ))) =  0 (A(X 0 )). The geometric characterization of  (F ) in Lemma 4 shows that  ( (F )) =  0(F 0 ). Hence Equation (1) is proved. Let g 2 G. Then by (1) there exists an unique  (g ) 2 G0 such that  ( (g )) = 0  ((g)). Since  0 is an injective homomorphism it follows that  is a homomorphism as well. The image of 10 under  ( (g )) is (g ,1 ); under  0 ( (g )) it is (g),1 = (g,1 ), hence we have  =  and  is a group isomorphism. The proof of Equation (1) implies that (F ) = F 0 and (A(U)) = A(U0 ). Denote the restriction of  to F by . Further  induces an isomorphism : V = G=F ! G0 =F 0 = V 0 . There exists a map  : V ! F 0 such that ([a; x]) = [a + x; x]. We claim that m := (; ;  ) is an isomorphism E ! E 0 of fourgonal extensions. Since  is a group isomorphism,  and are group isomorphisms as well. It remains to show that  fulfills (MOR1–3). But (A(U)) = A(U0 ), which implies (MOR1) and (MOR3). Since  is a homomorphism, a direct calculation shows (MOR2). Clearly, q(m) = . The isomorphism problem in Q can be studied entirely inside E.

THEOREM 3. Let E , E 0 2 E. Then E and E 0 are isomorphic if and only if q(E ) and q(E 0 ) are isomorphic. Proof. Since q is a functor, the image q(m) of an isomorphism m: E ! E 0 is an isomorphism. Suppose that q(E ) and q(E 0 ) are isomorphic. By Theorem 2 there exists an isomorphism : q(E ) ! q(E 0 ) such that (1) = 10 and (1) = 10 . The previous lemma finishes the proof. Remark 2. A certain class of isomorphisms between generalized quadrangles can be written as q(m) where m is an isomorphism of the corresponding fourgonal extensions. In fact, the automorphisms of a generalized quadrangle fixing 1 and 1 have this property. From this one deduces easily that the stabilizer of 1 in H is the semidirect product of Aut(E ) and  (G), where  (G) is normal in H . 5. A Geometric Characterization We want to characterize the elation generalized quadrangles which arise from a fourgonal extension. For that purpose we need to study elation generalized quadrangles (Q(1) ; G) with the following properties

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(a) There exists a central subgroup F of G which, for every p  1, acts regularly on the set of lines L, where p I L 6 I 1 (b) For every p 2 Pn1? , and every line L through p, the stabilizer of the point pL := L \ 1? in G is a normal subgroup of G.

We shall briefly write (Q(1) ; G; F ) for such an elation generalized quadrangle. Some further notation is introduced in the statements of the following three lemmata, compare [3]. LEMMA 6. Let (Q(1) ; G; F ) be an elation generalized quadrangle. Let p  1. Then the stabilizer Bp 6 G of p is regular on p?n1? . Proof. Let q; r be two different points in p? n1? . There exists an unique g 2 G such that g (q ) = r . Suppose g (p) 6= p. Since p  q we have g (p)  g (q ), hence p  g(p)  r  p, which is a triangle. This contradiction shows that g 2 Bp . LEMMA 7. Let (Q(1) ; G; F ) be an elation generalized quadrangle. Let L be a line with p I L 6 I 1 and AL 6 G be the stabilizer of L. Then AL  Bp , AL \ F = fidg, and Bp = AL
F . Further, AL is regular on Lnfpg. Proof. Clearly, AL is a subgroup of Bp . Since F operates regularly on the lines through p missing 1, AL \ F = f idg. For two points q , r 2 Lnfpg there exists an unique g 2 G with g (q ) = r . It follows that g (L) = L, hence g 2 AL , implying that AL is regular on the points 6= p on L. For every g 2 Bp there exists an unique h 2 F such that g (L) = h(L), so L is a fixline of g ,1 h. Since F is central, we obtain Bp = AL
F . LEMMA 8. Let (Q(1) ; G; F ) be an elation generalized quadrangle. Let p  q be two points collinear with 1; let L be a line through p and M be a line through q which intersect in r 6= 1. Then AL \ Bq = fidg and AL
Bq = G. Proof. Let g 2 AL \ Bq . Since Bq is normal in G, q 1 is a fixpoint line of g . From (a) in Lemma 2 it follows that L is a fixpoint line of g . We obtain consecutively, that all lines through 1 except possibly p1 are fixpoint lines, that M is a fixpoint line, and that p1 is a fixpoint line. Now (b) in Lemma 2 shows that g = id. Now take g 2 G arbitrary. Let q 1 I y  g (r ) and L I y 0  y . By Lemma 7, there exists an unique h 2 AL such that h(y 0 ) = r . Then h(yy 0 ) = M , so h(g (r ))  q . By Lemma 6, there exists an unique k 2 Bq such that k (r ) = h(g (r )), hence g = h,1k. COROLLARY 2. Bp \ Bq = F . Proof. From Lemma 7, Bp = AL F . Every g 2 Bp \ Bq can be written as g = az , where a 2 AL, z 2 F . Then az 2 Bq , whence a 2 Bq \ AL = f id g. Let us now fix notation for the remainder of this section. Pick a point 1 not collinear with 1, and let L ,  2 I , be the lines through 1. Let p be the point on L collinear with 1. Now put A := AL and B := Bp . The natural projection

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: G ! G=F maps the normal subgroups B onto normal subgroups of G=F . We are ready to prove the following key lemma.

LEMMA 9. The group G=F is Abelian. Proof. The corollary shows that, for  6=  ,  (B ) \  (B ) = f id g. Since the (B ) are normal subgroups, [(B ); (B )] 6 (B ) \ (B ), whence (B ) and  (B ) centralize each other. Applying the same sort of argument for  6= ;  , we see that every B iscontained in the center of G=F . From Lemma 8 we obtain that  (B ) and  (B ) generate G=F , which proves the lemma. Up to now we have shown that G is the middle group of the central extension of the Abelian group G=F by the Abelian group F . The following theorem gives the missing parts which turn this extension into a fourgonal extension. We continue the notation of Lemma 9. THEOREM 4. Let V := G=F and U := f (B ) j  2 Ig. There exist a normalized cross-section s: V ! G such that for every  the restriction of s to  (A ) is an injective group homomorphism. Let

f (x; y) := s(x)s(y)s(x + y),1 be the 2-cocycle associated with s, and let h: V ! F be the zero map. Then (V; F; U; f; h) is a fourgonal extension, and the fourgonal family associated with it is the fourgonal family associated with (Q(1) ; G). Proof. Recall that  (B ) =  (A ). First we construct the cross-section s. Since  embeds A injectively into V , for x 2  (A ) there exist a unique yx 2 A such that  (yx ) = x. If x 62 [2I  (A ), we choose a fixed yx 2  , (x). Then s(x) := yx is a well-defined cross-section, and its restriction to any (A ) is a homomorphism. We claim that E := (V; F; U; f; h) is a fourgonal extension. It is clear from the previous lemmata that all objects of E have the correct cardinality, so (1)–(5) of Definition 1 are fulfilled. We now have to check the properties (FE1– 4). The first, (FE1), has already been shown in the proof of Lemma 9. Since the restriction of s to  (A ) is a homomorphism, we have immediately that any  (A ) is a totally isotropic subspace for f , ie. the restriction of f to  (A )   (A ) is zero. This proves (FE2). For (FE3) take ,  , and  pairwise different, let x 2  (A ), y 2  (A ) such that x + y 2  (A ). From f (x; y ) = 0 it follows that s(x + y ) = s(x)s(y ). We claim that f1;  (s(x))(1);  (s(x + y ))(1)g is a triangle. For s(x) 2 A , whence  (s(x)) stabilizes L , thus 1 and  (s(x))(1) are collinear. By the same argument, 1 and  (s(y ))(1) are collinear, thus  (s(x))(1) and  (s(x))   (s(y ))(1) are collinear as well. From s(x + y ) = s(x)s(y ) we obtain  (s(x + y )) =  (s(x))   (s(y )), and since s(x + y ) 2 A , we conclude as before that 1 and  (s(x + y ))(1) are collinear. Now since triangles are degenerate in a generalized quadrangle we conclude that 1,  (s(x))(1), and  (s(x + y ))(1) are collinear. But a point 6= 1 on L is never

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collinear with a point 6= 1 on L , so together with the regularity of G we obtain that s(x) = s(y ) = s(x + y ) = id. This implies x = y = 0. Finally, let z 2 V nX where X =  (A ). Let g 2 F be arbitrary. We must find x 2 X and a suitable  with y 2 Y := (A ) such that s(x)s(y)s(x + y),1 = g. Applying  shows that then automatically x + y = z . By assumption, s(z )(L ) \ L = ;, so there exists an unique line M through gs(z)(1) intersecting L . Let M intersect the line through 1 which contains p . There exists an unique x 2 X such that s(x)(1) = M \ L . Since s(x)(L ) = M , there exists a unique y 2 Y with s(x)s(y )(1) = gs(z )(1), which proves (FE4). If we apply the isomorphism between G and the central extension of G=F by F , we see immediately that the fourgonal family associated with (Q(1) ; G; F ) corresponds to the fourgonal family associated with E . This means, in particular, that the quadrangles arising from G and from E are isomorphic. We may summarize the results of this section in the following THEOREM 5. The generalized quadrangles arising from a fourgonal extension are the elation generalized quadrangles (Q(1) ; G; F ).

6. The Known Examples In this section we list the known examples of finite fourgonal extensions E = (F; V; U; f; h) from [8]. Let V be a vector space of dimension 4 over the field F = GF(q). Let U0 and U1 be two arbitrary (but different) elements in U. After introducing bases in both subspaces, we write every other element of U as the graph of a linear map U0 ! U1 with matrix Ma , a 2 F nf0g. Note that M0 := 0. To facilitate the comparison with [8], we use the same enumeration of the examples until Example 8. In contrast to [8] we consider the constructions which also work in characteristic 2 separately in 6.2. We give a quick review of the setup in [8]. There the fourgonal extension is described using matrices fAa g such that Ma = Aa + Atra . The cocycle, except in , 0 0
Example 8, is given by the matrix I 0 , where I is the 2  2 unit matrix. Finally, h(x; xMa ) = xAa xtr , and h(0; x) = 0. In odd characteristic, but not for Example 8, let

f 0 = 12

0

,I !

I

0

and h0 = 0. Then the fourgonal extensions (F; V; U; f; h) and (F; V; U; f 0 ; h0 ) are isomorphic via the isomorphism (idF ; idA ;  ), where  : V = F 2  F 2 ! F; (x; y) 7! 12
xytr . We see immediately that

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– All U 2 U are maximal totally isotropic subspaces of V with respect to the symplectic form f 0 , – if W is a totally isotropic subspace meeting two elements of U in non-zero vectors, it meets all other elements of U only in 0. We distinguish two classes of examples, according to the characteristic of F .

6.1. CLASS I: q ODD

We first consider the examples where f is F -bilinear.

EXAMPLE 1. (Tits, see [1, p. 304] and [9, 10.6.1]): Let X 2 + uX + v be irreducible over F . For a 2 F ,

ua ! Ma := : ua 2va EXAMPLE 2. [3]: Let jF j  2 (mod 3). ! 2 a 3 a2 Ma := : 3a2 6a3 EXAMPLE 3. [4]: Let  2 Aut(F ) and m be a non-square in F . a 0 ! Ma := : 0 ,ma EXAMPLE 4 (4). : Let q  2 (mod 5). ! 2 a 5 a3 Ma := : 5a3 10a5 2a

EXAMPLE 5. [10]: This example is space-consuming; we refer to the reference or [8]. EXAMPLE 6. [2, Sect. 6]: Let q be a power of 5 and k be a non-square.

! a2 Ma := 2 ,1 : a 2k a + 4a3 + 2ka5 EXAMPLE 7. (see [2, Sect. 8] and [8]): Let q be a power of 3, let n be a non-square. ! a a3 Ma := 3 : a ,(na + n,1a9) 2a

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In the following example we have the unique case that f is not F -bilinear. EXAMPLE 8. Here we consider the famous ‘Bad Roman Eggs’ of Payne [7], see also example 8 in [8], as a fourgonal extension. We have q > 9 a power of 3. The maps Ma are particularly simple: x 7! ax. The complexity of this example is contained in the 2-cocycle of the extension. Let n be a non-square in F . Then

0 0 BB 0 f (a; b) := a B BB @ ,1

0 1 0 0 0 0

n

0

2 00 66 BB 0 6 B + 66a B 64 BB@ 0

1 2 00 CC tr 666 BBB 0 ,n C C b + 66a BB 0C A 4 @0

0

0

0

0

0

0

0

0

0

0 n,1 0

0 0 1

1 0

,1 1 31=3 CC tr 777 ,1 0 C C b 77 0 0C A 5 0

0

0

1 31=9 CC 777 ,n,1 C C btr 77 : 0 C CA 75 0

0

We note that this cocycle is linear with respect to the fixed field of x 7! x1=3 . Remark 3. Before we study examples in characteristic 2, let us briefly mention geometric interpretations of the above fourgonal extensions. In 6.1, the common interpretation is based on a happy confusion between the 2-cocycle f and the commutator c: V ^ V ! F of the extension. Since f = 12 c, we may as well study the elements of U with respect to c and arrive thus at the notion of BLT sets. We might consider Example 8 as an instance of the following abstract example. Over the field F := GF(q ) take n symmetric 2n  2n matrices S1 ; : : : ; Sn and q n 2n  2n matrices fMa j a 2 GF(q n )g such that

M0 = 0; For different a, b 2 GF(q n ), the matrix Ma , Mb is invertible,

Sj Matr = Ma Sj

for all j

= 1; : : : ; n

and

a 2 GF(qn);

For pairwise different a, b, c 2 GF(q n ), each of the following two sets of quadratic forms has only the null vector as common zero

f(Ma , Mb)Sj j 1 6 j 6 ng; f(Ma , Mb)(Mb , Mc),1 (Mc , Ma )Sj j 1 6 j 6 ng:

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Given these data, a fourgonal extension F n

 0 ,1  S := 1 0 :

! G ! F 4n is constructed. Let

The 2-cocycle f is given by the n-tuple of bilinear maps (S1 S; : : : ; Sn S ), and U consists of 0  F 2n together with the graphs of the linear maps fMa j a 2 GF(q n )g. 6.2. CLASS II: q EVEN The original examples 1, 2, and 4 in [8] also give examples of fourgonal extensions in characteristic 2. EXAMPLE 9.

a ua ! Aa := : 0 va

EXAMPLE 10. Let q be an odd power of 2.

a a2 ! Aa := : 0 a3

EXAMPLE 11. [6] Here q is an odd power of 2.

a a3 ! Aa := : 0 a5

Remark 4. A geometric interpretation as in odd characteristic is not possible here, since cocycle and commutator are substantially different and not related in any canonical way. If we replace the cocycle by

00 BB 0 f~ := B BB @0

1 C 0C CC ; 0C A

0 1 0 0 0 0 0

0 1 0 0 we have that the elements of U are totally isotropic subspaces with respect to f~, which gives us half of the geometric interpretation in odd characteristic. However, we must then replace h by a new map ~h which, contrary to the situation in odd characteristics, is not zero.

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Acknowledgements This article is based on parts of my Habilitationsschrift. I am very grateful to Professor Ott for his generous and encouraging support of my work. I would also like to thank the patient referee; with his help I was able to improve the exposition, eliminate many errors, and generalize everything to the infinite case. References 1. Dembowski, P.: Finite Geometries, Springer, Berlin, Heidelberg, New York, 1968. 2. Gevaert, H., and Johnson, N.: Flocks of quadratic cones, generalized quadrangles, and translation planes, Geom. Dedicata 27 (1988), 301–317. 3. Kantor, W.: Generalized quadrangles associated with G2 (q ), J. Comb. Th. (A) 29 (1980), 212–219. 4. Kantor, W.: Some generalized quadrangles with parameters (q 2 ; q ), Math. Z. 192 (1986), 45–50. 5. L¨owe, S.: On the structure of generalized quadrangles with a regular point, submitted, 1996. 6. Payne, S.: A new infinite series of generalized quadrangles, Congr. Numer. 49 (1985), 115–128. 7. Payne, S.: An essay on skew translation generalized quadrangles, Geom. Dedicata 32 (1989), 93–118. 8. Payne, S.: A census of generalized quadrangles, in: W. Kantor et al. (eds), Finite Geometries, Buildings and Related Topics, Conference on Buildings and Related Geometries (Pingree Park, 1988), Clarendon Press, Oxford, 1990. 9. Payne, S., and Thas, J.: Finite Generalized Quadrangles, Pitman Res. Notes Math. Ser. 110, Pitman, Boston, 1984. 10. Thas, J.: Generalized quadrangles and flocks of cones, Europ. J. Combin. 8 (1987), 441–452.

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