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l. two possibilities: = PG(3n — 1, g) contains -K, a contradiction. So O is a Kantor-Knuth egg. Next, assume that the subspace PG(3n - 1, q) of PG(4n - 1, q) contains at least five elements of 0\{ir}, say ni, 7T2,..., 7r5. We rely again on Theorem 5.2.1. Assume, by way of contradiction, that O is not classical. Then, by the first part of the proof, O is a nonclassical Kantor-Knuth egg. Let W n 7TJ = {r,}, with i = 1,2, . . . , 5 . Clearly (ri,r 2 ,... ,r 5 ) ^ PG(4,q n ), as otherwise O is contained in PG(3n — 1,q). Now suppose that (r\,rz,...,r$) = PG(3, qn). Then, with the notation of Theorem 5.2.1, M n PG(3, gn) ^ 0, and as in the previous section we find that IT c PG(3n - l,q), a contradiction. Consequently ri, r2,..., rs are contained in a common plane PG(2,qn). Clearly PG(2,g n ) n M = 0, as otherwise TT C PG(3n - l,q). The set W U jVl is the projection of a quadric Veronesean V| from a point p in a conic plane 7 of V| onto a hyperplane PG(4,g n ); the point p is an exterior point of the conic C of V| in 7. Let n be the projection of the point r[ e V|, with i = 1,2,..., 5. If (rlr'z, ...,r'5)= PG'(2, g"), then r'1; r'2,..., r'5 are contained in a common conic of V|, hence C n PG'(2, g") ^ 0, and so PG(2,g") n M ^ 0, a contradiction. Consequently, ( r i , ^ , . . • ,7-5) = PG'(3,g n ), withp G PG'(3,g n ). Clearly PG'(3,g n ) n 7 = {p}, as otherwise PG(2,g n ) C\M ^ 0. As ir ^ PG(3n - l,g), no three points of {r'i,r'2,.. .,r'5} are contained in a conic plane of V|. Let £ be the bijection of a plane PG* (2, gn) onto V| which maps the lines of PG* (2, qn) onto the conies of V|. Further, let C'c = L, {r1^1 = pi} with i = 1,2,... ,5. Then no three points of {pi,p2, • • • ,Ps} are collinear. So pi,p 2) • • • ,Ps belong to just one conic C" of PG*(2, qn). It follows that r[,r'2,..., r'5 generate a hyperplane of PG(5, qn) D V|; see page 117. So we have a contradiction, and conclude that O is classical. • Remark 5.2.4 (a) If O is a nonclassical Kantor-Knuth egg of PG(4n - 1 , q), with q odd, then from Thas [180] it easily follows that there exist (3ra - 1)-dimensional subspaces in PG(4n - l,q) which contain exactly four elements of O. (b) Independently and in a completely different way, Lavrauw [94] also proved the last part of Theorem 5.2.3.
Then we have
(a) the set of standard elations about pis a group; (b) s = t2, s is a power of 2, and there is a subGQ isomorphic to W(i) containing p which is fixed pointwise by an involution ofS^p\ Proof. See the proof of Theorem 7.2 of [130].
•
Remark 4.14.4 Examples of (b) yield counter examples to Question (3). Such STGQs will play a central role in §4.15. From the next theorem of [214], it will follow that "most of the time", the answer to Question (2) is that the set of elations about an elation point is not a group, and it also explains precisely why. SETTING. In this section, S = (V, B, I) is a GQ of order (s, t), s, t > 1, and x is an elation point for the elation group G. We let W be the group of all whorls about x.
Now let H < W be any subgroup of W that acts transitively on V \ x1-. We apply Burnside's Lemma on the permutation group (H, V \ xx) to obtain s2t-l 2
\H\ =s t+ Y^
i6(
S)i
where S(j) denotes the number of elements of H that fix precisely j points
olV\xL. Let E be the number of elations in H (not including 1) about x; then, as s2t-l
s2t-l 2
l + E+Y,
5(i) = s t+ J2
i=l
i<5
W'
i=l
it follows that the number of elations (now including 1 = id!) is given by the following: s2t-l 2
1 + E = s t+ ^ which is clearly at least s2t. The following theorem now easily follows.
(i-l)S(i),
Chapter 4. Generalized Quadrangles and Flocks
98
Theorem 4.14.5 Let («S(p\G) be an EGQ, and let W be the group of all whorls about p. Then the set of elations about p is a group if and only if there is no nontrivial element in W fixing more than one point not collinear with p if and only if W is a Frobenius group. • The Known Examples It is convenient to mention the known (classes) of examples of EGQs with elation point p for which the set of elations about p is not a group. These classes are treated in detail in [214]. • The classical and dual classical examples. H(4,q2);H(4,g2)^.
W(q) with q odd; H(3, q2);
• Flock GQs. The flock GQs with an even number of points on a line. • Dual flock GQs which are EGQs. As the order is (q, q2) for some q, it can be shown that not more than one point not collinear with the elation point can be fixed by a nonidentity collineation. So there are no examples possible. • TGQs. There are no examples possible. • Dual TGQs which are EGQs. GQs «SD, where S is a TGQ of order (
4.15 Elation Generalized Quadrangles with Nonisomorphic Elation Groups The following fundamental question (especially in construction theory for GQs) was posed by Payne [123] during the 2004 Pingree Park Conference "Finite Geometries, Groups and Computation": QUESTION (4). Let S^ morphic elation groups?
be an EGQ. Can x be an elation point for noniso-
The following theorem considers a class of GQs which do admit nonisomorphic elation groups, thus answering Payne's question affirmatively. The only known examples of this class are the GQs H(3, q2) with q even.
4.15. Elation Generalized Quadrangles with Nonisomorphic Elation Groups
99
Theorem 4.15.1 Let S = {S^P\H) be an EGQ of order (q2,q), where q is even, which contains a subGQ S' of order (s, q), s > 1, fixed pointwise by a nontrivial automorphism 9 of S. If H2 < Z{H), and if for all Lip and xILlp, with p y£ x, we have that \H(x,L,p)\ = q2, where H(x,L,p) is the group of whorls about x, L and p, then there is an automorphism group H' ofS such that H' ^ H and (S^ ,H')is an EGQ. Proof.
SeeK. Thas [207].
•
Corollary 4.15.2 Let S = (S^,!!) be an EGQ of order (q2,q), where q is even, which contains a subGQ S' of order (s, q), s > 1, which is fixed pointwise by a nontrivial automorphism 8 of S. Let z ^ p and suppose z ~ zi ~ pfor i = 0,l,...,q. If all groups H(p,pzi,Zi)nH are elementary abelian and have size q2, then there is an automorphism group H' of S such that H' ^ H and (S
100
Chapter 4. Generalized Quadrangles and Flocks
He then constructs q2 -1 distinct elation groups Kt = 1,2,..., q2 — 1 of size q5 with base-point (CXD), and shows that all Ki are mutually isomorphic. The Kis have nilpotency class 3, while K has nilpotency class 2, so that K p Ki for all i. The proofs are long and technical. For details and several other results, see Rostermundt [134].
Chapter 5
Good Eggs In this chapter we will define good eggs, explain the relationship with Property (G) and flock GQs, and give some interesting characterization theorems. There is also a section on the coordinatization of good eggs and their translation duals, and coordinates of all known examples are given.
5.1 Good Eggs and Good Translation Generalized Quadrangles Definition 5.1.1 The egg 0(n,2n,q) = O is good at its element TT if any PG(3n - 1, q) containing ir and at least two other elements of 0(n, 2n, q) contains exactly qn + 1 elements of 0(n, 2n, q); the space n is called a good element of 0(n, 2n, q). In such a case we also say that 0(n, 2n, q) is good. If O is good at ir, then we say that the TGQ T(O) is good at its line IT, and ir is a good line of T ( 0 ) ; shortly we say that the TGQ T(O) is good. We emphasize that each known egg 0(n,2n,q) good.
or its translation dual is
Now we give the relation between good and Property (G); see Thas [176]. Theorem 5.1.2 The egg 0(n,2n,q) — O satisfies Property (G) at the flag ((oo),7r) if and only if the translation dual T(G*) of the TGQ T(0) is good at its element r with T the tangent space of O at n. Proof. Suppose that T((9) satisfies Property (G) at the flag ((oo), ir), with •n e O. The tangent space of O at it is denoted by r. Let nj,irk be distinct 101
102
Chapter 5. Good Eggs
elements of 0\{n} and put P G ( n - 1, q) = r n Tj H rk, with T; the tangent space of Oatiri, I e {j,k}. By Theorem 3.10.1(i) each point of P G ( n - l , g ) is contained in exactly qn + 1 tangent spaces of O. Let PG(4n, q) be the space in which T(O) is defined, and let P G ( 4 n - l,q) be the space of O. Further, let u £ P G ( n - 1, q), let N be a line of PG(4n, q) through u not contained in PG(4n - 1, q), and let u\,v,2 be distinct points of N not contained in PG(4n - 1, q). Then each of the points (N, T), {N, TJ), (N, rk) of Type (ii) of T(0) is collinear with each of the points (oo), ui,u2 of T ( 0 ) . Clearly {(oo),u1,u2}J- is the set {(N,T), (N,Tj), {N,Tk}, • • • } with T,T0,Tk,. • • the qn + 1 tangent spaces of O through u. As T(O) satisfies Property (G) at the flag ((oo),7r), we have \{(oo),ui,U2}1"L\ = qn + 1. Hence (N, r) n (AT, T,-} n (N, rfe) n • • • = r) contains qn points of PG(4n, q) \ P G ( 4 n - 1, q). Consequently rj is an n-dimensional space, that is, TDTJ f) Tfc n • • • is (n - l)-dimensional. So the intersection of these qn + 1 tangent spaces is P G ( n - l,q). We conclude that P G ( n - l,q) is contained in exactly qn + 1 tangent spaces of O. Hence the translation dual O* of O is good at its element T. Conversely, let O* be good at its element T. SO for any two distinct elements Tj,Tk € 0*\{T} the space P G ( n - l,q) = TDTJ C\Tk is contained inexactly qn + 1 tangent spaces of C Let T, TJ, r^ be points of Type (ii) of T(O) collinear with (oo), where r c r, r,- c Tj, T^ C r^. Then T fl Tj fl Tfc is an n-dimensional space P G ( n , g ) . Clearly P G ( n — l,q) = P G ( n , q) n PG(4n - 1, g). If r, Tj, Tfc,... are the g™ + 1 tangent spaces of O containing P G ( n - l , q ) , then the ij™+ 1 spaces (PG(n,q),r) = T, (PG(n,g),Tj) = Tj, (PG(n,q),Tk) = Tfc,... are points of Type (ii) of T(O), each of which is collinear in T(O) with (oo) and the qn points of Type (i) in PG(n,q). Hence the triad {f,Tj,Tk} is 3-regular. So we conclude that T(O) satisfies Property (G) at the flag ((oo), ir). • Remark 5.1.3 As the translation group of T(O) is transitive on the points of Type (ii) of T(O) incident with n, in the statement of Theorem 5.1.2 the flag ((oo),7r) may be replaced by a point pair ((oo),7r), with n a point of Type (ii) incident with -IT. Corollary 5.1.4 If the egg 0(n, In, q) = O, with q odd, is good at its element IT, then T(C*) is the point-line dual of a flock GQ with elation point T, where T is the tangent space ofO at n. Conversely, if the flock GQ S has a translation line, then the point-line dual ofS is isomorphic to a T(O) for which O* is good at its element which corresponds to the elation point of S.
5.1. Good Eggs and Good Translation Generalized Quadrangles
Proof.
From Theorems 4.10.1, 4.10.6 and 5.1.2.
103
•
Theorem 5.1.5 Ifqis even and if the TGQ T ( 0 ) satisfies Property (G) at the flag ((oo),7r), with n G O, then the translation dual T(C*) satisfies Property (G) at the flag ((oo), T), with r the tangent space of O at TT. Proof. Suppose that T(O) satisfies Property (G) at the flag ((oo), TT), with q even. Let r, rj, TJ be points of Type (ii) of T ( 0 ) collinear with (oo), where T C T . T J C Tiy Tj c Tj, with r, r,, T,- the tangent spaces of O at the respective (distinct) elements TT, iri,itj of O. Then, by Theorem 2.6.2, {¥, Ti,Tj}± and {T, T^T,}- 1 - 1 - are contained in a subquadrangle <S' of order qn of T(0). The space 7 n f , n T., is an n-dimensional space PG(n,q), and the space P G ( n - 1, q) = P G ( n , q) n PG(4n - 1, q) = r n Tj n TJ is contained in exactly qn + 1 tangent spaces T,n,Tj,... of O. Then the points of P G (4n, g) \ P G ( 4 n - l , g ) in (PG(n,q),iv), (PG(n,g),7r i ), ( P G ( n , g ) , f , ) , " - are the points of Type (i) in S'. Now we consider a line of Type (a) concurrent with 7r and incident with a point of Type (i) in the space (PG(n, q), Wi). As this line is also a line of S', it is incident with a point of Type (i) in each of the qn - 1 spaces (PG(n, q), TVJ), • • •. Hence the spaces (PG(n, q), TTJ), • • • are contained in the 3n-dimensional space (PG(n, g),7r,7r,) = PG(3n, q). Consequently the spaces P G ( n — 1, q), IT, m, TTJ, ... are contained in a common P G ( 3 n - 1, q) = PG(3n, q) n PG(4n - 1, q). So, by Theorem 3.8.1 (i), the (3n - 1)-dimensional space (TT, m, TTJ) contains exactly qn + 1 elements of O. Now, by Theorem 5.1.2, the GQ T(C*) has Property (G) at the flag ((oo),r). • Corollary 5.1.6 If q is even and if the egg 0(n, 2n, q) is good at its element IT, then the translation dual 0*(n,2n,q) is good at the tangent space r of 0(n, 2n, q) at IT. Proof.
Directly from Theorems 5.1.2 and 5.1.5.
•
Remark 5.1.7 An alternative proof of Corollary 5.1.6, without using GQs, goes as follows. Assume that q is even and that the egg 0(n,2n,q) = O is good at its element IT. Let 7^,71-., be distinct elements of 0\{TT}, and let T,Ti,Tj be the respective tangent spaces of O at ir,Tri,iTj. Further, let TT, 7Tj, TTJ, ... be the qn + 1 elements of O in (n, TTi,TTj) = PG(3n - l,q) and let O' be the pseudo-oval of P G ( 3 n — 1,) consisting of these elements. By Theorem 3.9.1 the tangent spaces of O' contain a common (n - 1)dimensional space 77, which was there called the kernel of O'. Clearly the
104
Chapter 5. Good Eggs
tangent spaces of O at the elements of O' all contain 77. It follows that the translation dual O* of O is good at its element r. Assume that the egg 0(n, 2n, q) = O, with O = {IT, m,..., 7rg2n}, is good at its element n, let r be the tangent space of O at -K, and let n be the tangent space of O at 7TJ, with i = 1,2,..., q2n. Further, let C be a (3n - 1)dimensional subspace of the P G ( 4 n - 1, q) containing O, with 77 n C = 0Now we project the elements of 0\{TT} from n onto (. The projection of 7TJ from 7r onto C is denoted by 7^, that is, n[ = {71-, 7r;)n C, with i = 1,2,..., g 2 n . Then 7ri U 71-3 U • • • U 7i^2n is C\T', where r ' is the (2n - 1)-dimensional space T f l ( . The set of all intersections (7r,7rj,7rj) n ( , i ^ j , is denoted by B. Then, with respect to inclusion, the elements of V = {TT[, TT'2, . . . , ir'q2ri} and B, are the points and lines, respectively, of a 2 - (q2n, qn,l) design A, that is, of an affine plane of order qn = s. The affine plane A will be called the derived or internal plane of the egg O at the good element n. If p' and p" are distinct parallel lines of A, then p ' n p " is necessarily an (n-1)-dimensional subspace of r'. Hence all lines of A parallel to p' contain a common (n — 1)dimensional subspace of r'. These qn + 1 (distinct) subspaces of r ' are denoted by v0, v\,..., vqn. If for some j ^ j we have J/J n i/j ^ 0, say re £ ViC\Vj, then the lines p' and p" of ^l containing respectively ^ and i/j, do not intersect in a point of A, so are parallel, that is, vt = z/,-, a contradiction. Hence {^o, I/J, . . . , vq™} is a partition (an (n - 1)-spread) of T' . The elements va,vi,...,vqn and r ' extend A to a projective plane II of order qn. Now by a theorem of Segre [140] such a plane II is Desarguesian, so in C there are n planes £i> £2, • • •, £n over GF(
5.1. Good Eggs and Good Translation Generalized Quadrangles
105
Assume again that the egg O = 0(n,2n,q) is good at its element w. Let T be the tangent space of O at IT, and let r, be the tangent space of O at 7Tj, with 7Ti e 0\{TT} and i = 1,2,..., <j2n. Further, let O be contained in P G ( 4 r a - l , q), let P G ( 4 n - l , g) be ahyperplane of PG(4n, g), and let T ( 0 ) be the corresponding TGQ of order (qn, q2n). If irk, 7Tj are distinct elements of 0\{ir} then PG(3n - 1, q) = {n, iTj,nk) contains exactly qn + 1 elements of O. Let PG(3n - l,q) C PG(3n,g) c PG(4n,g), with PG(3n,g)
106
Chapter 5. Good Eggs
Theorem 5.1.10 Let 0(n,2n,q) = O be an egg in PG(4n - l,q), q even, which is good at IT e O. If the q2n + qn pseudo-ovals on O containing IT are regular, then O is regular, so the corresponding TGQ T(O) is isomorphic to a T3(0')ofTits.
Proof. Let O = {n, TTI, TT2, • • •, irq^} be an egg in P G ( 4 n - 1, q), q even, which is good at IT. Suppose that the q2n + qn pseudo-ovals on O containing IT are regular. Clearly we may assume that n > 1. Next, assume that qn = 4, so q = n = 2. As qn = 4, the q2n + qn regular pseudo-ovals on O containing IT are pseudo-conies. Now, by Brown and Lavrauw [26], cf. Corollary 7.8.4 below, the egg is classical, hence regular. From now on we assume that qn > 4. Let PG(3n - 1, q) be a subspace of P G ( 4 n — 1, q) which is skew to IT and let {7r,7Ti)n PG(3n - l,q) = 7T-, with i = 1,2,... ,q2n. Let r be the tangent space of O at IT, and let n be the tangent space of (!) at 7r„ with i = 1,2,..., q2n. If T' = r n P G ( 3 n - 1, q), then the intersections of r ' with the tangent spaces at IT of the pseudo-ovals on O containing IT, form an (n —1)-spread T* of r'. Then T* U {TT[, ir'2,... ,7r' 2 „} is a regular (n — 1)spread T of PG(3n — 1, q). This spread T is the point set of the projective completion II of the derived plane A of O at IT. If we extend the elements of T to G¥(qn), then these extensions have a point in common with each of n planes £i, & , . . . , £„ over GF(qn). Let 7f • n £j = {pij}, with 7f • the extension of TT[ to GF(g"), i = 1,2,..., q2n and j = 1,2,..., n. Then £.,- = {pij,p2j, • • • ,Pq2™j} U I/j, with L,- the intersection of £j with the extension of r' to GF(q n ), with j = 1,2,..., n. Let Mj be a line of £,-, with Lj ^ Mj. Then (MJ,TT), with 7f the extension of IT to GF(q"), intersects the extensions to GF(qn) of all elements of a pseudooval O' on O containing IT. AS 0 ' is a regular pseudo-oval, there are planes 7^ over GF(g") intersecting the extensions of all elements of O', with k = 1,2,...,n. Then {<¥,7fc) D PG(3n - 1,qn) \\ k = 1,2,...,n} = {(O1) n ^ II j = 1,2,..., n}; here we rely on the fact that a regular (n - 1)-spread in a P G ( 2 n - 1, q) uniquely defines n transversals over GF(<7n). So we may assume that (7f,7 J )nPG(3n-1,<7™) = (0')n£j = M,-,withj = 1,2, . . . , n . Clearly the extensions of the elements of O' intersect ^3 in the points of an oval G'j over GF(qn), with O) n n = {rj},j = 1,2,..., n. This oval O) is contained in (w, £j). Now let iV, be the tangent of Oj at rj. Then (A^-, W} intersects ^ in a point of Lj. If CK and O" are two such ovals, with O'j ^ O", then we have the following cases.
5.1. Good Eggs and Good Translation Generalized Quadrangles
107
• If the corresponding lines of £, intersect in a point not on Lj, then {O'j n Q'!)\K is a point. • If the corresponding lines of £,- intersect in a point of Lj, then (O' n 0")\W is empty. Let the distinct points Vij-.Vi'j-.Vv'j define a triangle of £j. The ovals defined by the lines VijVvj->VviVvjiVviVij a s above, are denoted by uf', Oj , Oj ', respectively. Then one easily sees that all q2n + qn ovals O'j are contained in the space (Oj ,Oj\ Oj). So all these ovals generate a P G ( j ) (m, qn), with 3 < m < 5.
(a) The Case m = 3 Then (TF, PG (i) (3,g™)) = {W,Q is (n + 2)-dimensional, so TfPiPGU)(3,qn) is a point u. This point u belongs to the q2n+qn ovals Oj. The union of these ovals is a set W of size g2™ + 1 . As no three points of W are collinear, this set W is an ovoid O of P G ^ ' (3, qn). Hence the extensions of the elements of O meet the n spaces P G ( 1 ) ( 3 , ? n ) , P G ( 2 ) ( 3 , g " ) , . . . , P G ( n ) ( 3 , q n ) and the egg O is regular (the TGQ T(O) is isomorphic to the TGQ T 3 ( 0 ) of Tits).
(b) The Case m = 5 Then (TF, P G ( i ) ( 5 , g n ) } = (¥,£,-) is (n +2)-dimensional, so TF n P G ( j ) ( 5 , g n ) is a plane 5j. Let Wi be the extension of 7TJ to GF(g n ) and let Wi n PG^(5,<7 n ) = {si}, with i = l , 2 , . . . , g 2 r a . Now we project from sis 2 onto a subspace P G ( 3 , g n ) of P G ^ (5, qn) which is skew to sis 2 . Let the qn ovals containing si but not s 2 be O n , O i 2 , . . . , Oig™, and let the qn ovals containing s 2 but not s\ be 021,022, • • •, 02 g »- Further, let Lu be the tangent line of On at si, with Z = 1, 2 and i = 1,2,..., qn. By projection of Ou\{si} from s\S2 onto PG(3,g"), there arise qn points of a line Tu. Let (si,s2,Lu) n P G ( 3 , g") = {in}; then fe e 7^. Assume, by way of contradiction, that Tu C\Tik ^ 0 for i ^ k. Then all ovals in PG ( j ' ) (5,g") are contained in (Tij,T( fc ,si,s 2 ) which is at most 4-dimensional, a contradiction. So Tn,Ti2,...,Tiqn are mutually disjoint, with I = 1,2. If O is the oval containing si and s 2 , then 0 \ { s i , s 2 } is projected onto a point i of P G ( 3 , g n ) . The notation is chosen in such a way that 0 2 i is the oval on s 2 which contains no point of OH\7F, with i = 1,2,..., qn.
108
Chapter 5. Good Eggs
So Tu n T2fc ^ 0 for k =£ i. As qn > 4, it follows that there is a hyperbolic quadric H in PG(3,g n ) such that {Tn,Ti2,. • . , 2 V } is a subset of a regulus ft; of H, with Z = 1,2. Let O H n W = On n 6j = {n*}, and <si,s 2 ,ni> n P G ( 3 , g " ) = {SH}. If n is the extension of n to GF(g n ), then Tt n P G ( i ) (5, qn) is a plane Pji, with j = 1,2,..., n and i = 1,2,..., q2n. So the lines Ln,Li2,..., L(g« are contained in the plane /a,-;. It follows that Ui,ti2, •••, Uq™ are points of some line T/, with I = 1,2. Clearly {t} = T^nT^ and SH = s2i = s'j, with i = 1,2,..., g n . Hence ovals which define a common point of Lj intersect 8j in a common point. The points s^s^,..., s'qn belong to (Sj,si,s2) n PG(3,g"), so are contained in a plane ip. Hence s[,s'2, ...,s'qn belong to the nonsingular conic C = ip n H. Also, we have
tec. Now we consider an oval O'j not containing si nor s 2 . By projection of O'j from S1S2 we obtain the intersection of H with a plane of PG(3,g") through t, but not containing T[ nor T 2 . Hence this intersection is a nonsingular conic, and so O'j is a conic. It follows that the corresponding pseudo-oval O is a pseudo-conic. Now, by Brown and Lavrauw [26], see also Corollary 7.8.4, the egg is classical. But in such a case m is necessarily three, a contradiction.
(c) The Case m = 4 Then (7f,PG (j) (4,g ri )) = ( ¥ , ^ ) is (n + 2)-dimensional, SOTF n P G ( i ) ( 4 , g " ) is a line Wj. Let Wi be the extension of 7TJ to GF(g") and let 7fj n P G ( i ) (4, qn) = {Si}, with i = 1,2,..., g 2 n . Let w e Lj. With the g n lines of ^ through w, but distinct from Lj, correspond qn ovals. The tangent lines of these ovals at their intersection point with Wj are contained in the plane P G ^ (4, g n ) n (TT, W). Assume, by way of contradiction, that the point u G U = W3 is on at least g™ + 1 ovals O'j. So at least two of these ovals, say Oj and Oj, have a point s^ <£ U in common. Let Oj be a third oval containing u. As all ovals generate P G ( i ) ( 4 , qn), the oval of^ cannot contain a point of both 0^\{u, st} and
of\{u,Si}. First, assume that Oj contains s,. If r* is the extension of n to GF(g n ), then T j n P G ( i ) (4, g n ) is a plane pjt, with j = 1,2 . . . , n and i = 1,2,..., g 2 n . So the tangent lines at st of the ovals Oj , Oj , of' are coplanar. It follows that these three ovals are contained in the threedimensional space (pji, u),
5.1. Good Eggs and Good Translation Generalized Quadrangles
109
and so all q2n+qn ovals O'j are contained in (pji, u), clearly a contradiction. So any oval of containing u, with of ^ of ^ of\ does not contain s^ Let of and of] be distinct ovals containing u, with \of n of\ = \of] n of\ = 2 and I e {1,2}. Then o}fc) n OJ.fc'} = {«}, as otherwise (0J 0 , Of \ of]) is 3-dimensional. That means that all ovals containing u are elements of two classes W\ and W2, where of G Wi and any two distinct ovals of Wj just have u in common, with i = 1,2. Now we project from u onto a PG(3, qn) in PG ( i ) (4, qn), with w £ PG(3, g n ). The ovals of W, are denoted by o g = of\o$,..., with Z = 1,2. Let Tfe(0 be the tangent line of ofl at u, and let T^ n PG(3,? n ) = {4°}- Then the points 4 > 4 > • • • a r e distinct, as otherwise all ovals are contained in a 3-dimensional space, with i = 1,2; as ovals of different classes W\ and W2 always have a point of PG^(4,g")\[7 in common, we clearly have 4 1} ^ 4^- L e t t h e projection from u onto PG(3,g") of ofk\{u} belong to the line Vfew. Then 4^ e Vfe(i). Now we shall prove that the lines v}l),V2(l\ ..., with I = 1,2, all belong to a common hyperbolic quadric H of PG(3, qn). Suppose, by way of contradiction, that these lines do not belong to a common hyperbolic quadric. As T{',T2 ,... define a common point on Lj, the points t\', 4 , • • • are collinear, with I = 1,2. Assume that \Wi\ < \W2\. Then the only possibility is that |Wi| = 1 and \W2\ = qn, as otherwise all lines Vk ' belong to a common hyperbolic quadric. The points tf, 4 2 ) , • • • are on a line R; let u' be the (g™ + l)-th point of that line. Then {u1} = UC\ PG(3, qn). With the q2n - 1 ovals not containing u correspond, by projection from u, q2n - 1 ovals through u'. Any such projection is a subset of V1(2) U V2(2) U • • • U V^2) U {u'}. The planes of these projections, together with the qn + 1 planes on R in PG(3, qn) and the plane («', V^ ) are the g2n + qn + 1 planes of PG(3, gn) through w'. Now consider a line V, with Vf> ^V ^R, which intersects v}2) ,V^2\v£2). Then (V, u') intersects v}2) U V2(2) U • • • U Vg(2) U {u'} in g" + 1 points which form an oval, a contradiction as this intersection contains three distinct collinear points. Consequently the lines V^ all belong to a common hyperbolic quadric H ofPG(3, g n ). Assume \WX\ < \W2\ and let U n PG(3,g n ) = {u'}; from the foregoing it follows that | W\ | > 2. Then the points u', t^p, 4 ' \ • • • a r e o n a common line Ru with I = 1,2. Assume, by way of contradiction, that we are not in the case \WX\ = \W2\ = qn. Then there exists an oval C , with £>'• n of =
Chapter 5. Good Eggs
110
0. The projection O'j of O'j from u onto PG(3,q") contains u', and the projection of the tangent line of £K at O'j nU is the line Rx. It follows that (O'j) contains Rx. As i?i c H, the oval £^ contains \W2\ > qn - 1 points of a line of H, clearly a contradiction. Hence \W\\ = \W2\ = qn. Any oval O'j not in W\ U ^ 2 is then projected onto a conic of H, and so C^ is a conic. It follows that the corresponding pseudo-oval O is a pseudo-conic. Now, by Brown and Lavrauw [26], see Corollary 7.8.4, the egg is classical. But in such a case m is necessarily three, a contradiction. So we have shown that each point u of U is contained in exactly qn of the q2n + qn ovals. Consider again distinct ovals of^of*1 assume that Oj
n Oj
containing the point u e U and
= {u, s j , with s» ^ [/. Then as before all ovals
containing u are elements of two classes W\ and W2, where Oj e Wj and any two distinct ovals of W; just have u in common, with I = 1,2. Assume that |Wi| < \W2\. We proceed as before, using similar notation. Then, if the lines Vfe ' do not belong to a common hyperbolic quadric H, we have \Wi\ = 1 and |VF2j = qn — 1. In such a case the planes of the q2n ovals not on u together with the qn + 1 planes through R in P G ( 3 , q") and the plane (u', V}1'), are the q2n + qn + l distinct planes in P G ( 3 , qn) containing u'; the unique oval O* having no point in common with the union of the elements of W2 defines a plane of PG(3,g") which contains R (the projection O* onto PG(3,g n ) of this oval is tangent to R at u'). Now consider a line V, with Vi(1) ^ V ^ R, which intersects v/ 2 ) , V2{2), V3(2). Then (V,u') intersects v / 2 ' U V2(2) U • • • U T/(„2^ U (5* U {u'} in qn + 1 points which form an oval, a contradiction as this intersection contains three distinct collinear points. So the lines Vk' belong to a common hyperbolic quadric H of P G ( 3 , qn). Also, from the foregoing reasoning it follows that \W\\ > 2. Consider an oval O'j with of] n O'j = 0. The projection O'j of O'j from u onto P G ( 3 , g") contains u', and the projection of the tangent line of O'j at O'j C\U is the line Rx containing u', tf ) , i^ 1} ,.... It follows that (O'j) contains Ri. As Ri contains at least three points of H, it belongs to H. It follows that O'j contains \W2\ > qn — 2 points of a line of H, clearly a contradiction. So if Oj ,Oj ]
C>W n of PG
(i)
are any two distinct ovals containing a point u of U, then = {u}. Let u e U and project from u onto a P G ( 3 , f ) C
B
( 4 , g ) , with u £ P G ( 3 , g n ) . Let of] ,of],...
containing u, and let Tt be the tangent line of Oj
,0{f] at u.
be the ovals Further, let
5.1. Good Eggs and Good Translation Generalized Quadrangles
111
Ti n P G ( 3 , f ) = {U} and let the projection of of\{u} belong to the line Vi, with i = l,2,...,qn. Then ij e Vi. The points ti, t2, •.., V are not necessarily distinct. Also, as Ti, T 2 , . . . , Tq™ define a common point on Lj, the points ti,t2, • • • , V are collinear. If U n PG(3,g") = {V}, then the points u',ti,t2, • •. ,tqn are on a common line 17' of P G ( 3 , qn). Let it; e U\{u}, with / = 1,2,..., qn. The tangent lines at m of the
= t2 = ••• = tq-n. Hence Ti = T2 = • • • = Tqn.
A similar conclu-
sion holds for each point m e U, with I = 1,2,..., qn. So in PG^ j ) (4, q") the nuclei of the q2n + qn ovals are on qn + 1 lines Bi,B2l..., Bqn.+1. As Bi,B2,..., Bqn+1 define all qn + 1 points of L,, these lines are mutually skew. In the tangent space r of O at 7r, the space 7r together with the nuclei of the pseudo-ovals on O containing IT form a regular (n — 1)-spread T* in T; see Remark 5.1.7 and the text preceding Lemma 5.1.8. As the nuclei of the q2n + qn pseudo-ovals are distinct, also the nuclei of the q2n + qn ovals are distinct. The extensions to G¥(qn) of the elements of T* meet n planes ££, ££> •••>££ o v e r GF("). Hence Bi,B2,..., -Bg«+i are contained in qn +1 distinct planes £*, a contradiction as qn +1 > n. Now the theorem is completely proved. • As a corollary of the previous theorem we will prove in a geometric way a result which is equivalent to Johnson's Theorem 4.5.1. Theorem 5.1.11 If the TGQ T(0) of order (s, s2), with s even, is the dual of a flock GQ, then T ( 0 ) is classical.
112
Chapter 5. Good Eggs
Proof. Let T(O) be a TGQ of order (s, s2), s even, which is the dual of a flock GQ. By Theorems 4.10.1, 5.1.2 and 5.1.5 the egg O is good at some element IT = PG(n - l,q), where s = qn. Let O' be any of the pseudoovals on O through IT. By Payne [117] the point-line dual of the TGQ T ( C ) of order qn is isomorphic to a T 2 (0') of Tits, with O' some oval of PG(2, qn); the translation point of the point-line dual of T2(<5') is the line 7T of T (£>'). As the elements of O' are regular points of the point-line dual of T 2 (£>'), the GQ T(O') is a T 2 (0") of Tits, where O" is a translation oval of PG(2, s); see Payne [109] and Section 12.5 of Payne and Thas [128]. By Corollary 3.5.3 and Section. 3.6 the pseudo-oval O' is regular. Now by Theorem 5.1.10 the egg O is regular, that is, T(0) is isomorphic to a T 3 (0) of Tits, with O some ovoid of PG(3, qn). With the notation of Section 4.10, the ovoids Oz are isomorphic to the ovoid O. As T(0) is the dual of a flock GQ, all ovoids Oz are elliptic quadrics; see also the paragraph preceding Theorem 4.10.7. Hence O is an elliptic quadric, that is, T 3 (0) = T(O) is classical. • If T is a semifield flock of the quadratic cone K in PG(3, q) with q even, then, by Section 4.7, the corresponding flock GQ <S(.F) has a translation line. So by Theorem 5.1.11 S(T) is classical, so T is linear (see e.g. Payne [120, 121]). This is Johnson's Theorem 4.5.1. Theorem 5.1.11 also follows from the theorem of Johnson. Consider a flock GQ S{T) having a translation line L, with T a flock of the quadratic cone K of PG(3, q), with q even. If S(F) is not classical, then, by Payne and Thas [129], the elation point (oo) of SiJ7) is fixed by the full automorphism group of S(F). So we may assume that L contains the point (oo). By Payne [121] we have S(T) = S{T'), where in S(.T) the line L is the line [oo] (we "recoordinatize" the flock GQ). Then, by Section 4.7, T' is a semifield flock. Hence, by Theorem 4.5.1, T' is linear, and consequently, by Payne and Thas [128], the GQ S{T) = <S(.F') is classical. • We now consider TGQs T(0) of order (s, s2) for which the egg O has more than one good element. Theorem 5.1.12 If q is odd and the egg 0(n,2n,q) = O has at least two good elements, then O is classical and so the TGQ T(0) is classical. If q is even and the egg 0(n,2n,q) = O has at least four good elements, not contained in a common pseudo-oval on O, then O is regular and so T(0) is aT3{0')ofTits.
5.1. Good Eggs and Good Translation Generalized Quadrangles
113
Proof. Assume that q is odd and that the egg 0(n, 2n, q) = O has at least two good elements. Let n be a good element of O. By Corollary 5.1.4 the translation dual T(C*) of T(O) is the point-line dual of a flock GQ with elation point r, where r is the tangent space of O at TT. Let TT' e 0\{TT} be good, and let TT" e 0\{TT, TT'}. There is an elation of T(C*) with center T, mapping the tangent space r ' of O at 7r' onto the tangent space T" of O at 7r". By Theorem 3.7.2 there is an automorphism 9 of PG(4n - 1, q) D 0 , fixing 0*, fixing T and mapping r ' onto T". Clearly 0 fixes 0 , fixes 7r and maps TT' onto 7r". Hence all elements of 0 are good. By Theorem 3.10.7(ii) the egg is regular, and so by Section 3.6 the egg 0 is classical. Next, assume that q is even and that the egg 0 ( n , 2n, q) = O has at least four good elements, not contained in a common pseudo-oval on 0 . Let TT and TT' be distinct good elements of 0 . Now we consider a pseudo-oval 0 i on 0 containing TT, but not containing TT'. Let P G ( 3 n - l,q) be the subspace of P G ( 4 n - 1, q) D 0 generated by Oy. By projection of 0\{TT'} from TT' onto PG(3n - 1, q) there arises the derived plane A of 0 at TT'; see Section 5.1. Clearly 0\ is an oval of A, and so 0 i is regular. As there are four good elements, not contained in a common pseudo-oval on O, it easily follows that all pseudo-ovals on O containing TT are regular. Now by Theorem 5.1.10 the egg O is regular. • From Theorem 4.7.3 and Theorem 5.1.12, one can now deduce the following result, which is essentially contained in Bader, Lunardon and Pinneri [4]. Theorem 5.1.13 Suppose T is a Kantor-Knuth semifield flock of the quadratic cone in PG(3,g), q odd, and let S(J-)D = T(O) he the associated dual flock GQ. Let TT be a good element of O, and let TT* be a good element of O*. Then there exists an isomorphism between T(0) and T(C*) that maps TT onto TT*. If T is nonlinear, then any isomorphism between T(O) and T(C*) has this property. Moreover, in the latter situation, the good element of O* is the tangent space of O at TT. Proof. If F is linear, then T(O) ^ T(£>*) ^ Q(5,), and the theorem easily follows. Let T be nonlinear. Then O and O* have precisely one good element by Theorem 5.1.12, say, respectively, TT and TT*. SO any isomorphism a between T(O) and T(C*) maps TT onto TT*. Let S be an anti-automorphism of the space P G ( 4 n - l,q) containing O, for which Os = O*. Suppose
Chapter 5. Good Eggs
114
TTS = 7r* ^ T, with r the tangent space of O at n; if IT* is tangent to 0 at ir', then, by Corollary 5.1.4, n' corresponds to the point (oo) of S(T). Let Ll(oo) correspond to the translation point of T(O). Then since S(T) is an EGQ, the stabilizer of the flag ((oo), L) in the automorphism group oiS(F) acts transitively on the points incident with L and different from (oo). It readily follows that O* contains more than one good element, contradicting the hypothesis that T is nonlinear. So -K* = T, and the theorem is proved. • The next theorem provides an interesting condition for a good TGQ to be a Kantor-Knuth GQ. Theorem 5.1.14 (Blokhuis, Lavrauw and Ball [16]) Assume that T(0) is a good TGQ of order (qn,q2n), q odd, where GF(g) is the kernel of the TGQ, with the additional condition that q > An2 - 8n + 2. Then T(O) is isomorphic to a Kantor-Knuth GQ. Sketch of Proof. Since T(£>) is good, by Corollary 5.1.4 T(0*)D is a flock GQ. By Theorem 4.7.2, T(£>*)D ^ S(T), where T is a semifield flock of the quadratic cone K, with equation X\ — X§X\ = 0 in PG(3,g"), and where the kernel of the corresponding TGQ is GF(q). In the dual space of PG(3, q), the generators of /C become a set of qn + 1 lines in the plane irv which is the dual of the vertex v of the cone. These lines are the tangents to a conic C. If the dual is taken w.r.t. the standard inner product, then C' is given by the equations X3 = X\ - 4X 0 X! = 0. In the dual space, let T be the set of qn points which corresponds to T. Since every intersection line of distinct flock planes is skew to every generator of K,, it follows that in the dual space the line joining two points of T meets nv in an internal point of C (the external points and the points of C are incident with a tangent); see also Section 4.4. Let I be the set of these internal points. With the notation of Section 4.6, and taking into account that f(t) = —zt and g(t) = yt,t e GF(qn), are additive, we have X = { ( * , - / ( * ) , 5(4), 0) || t € G F ( g n ) } .
5.1. Good Eggs and Good Translation Generalized Quadrangles
115
A nondegenerate quadratic form is either always a square on the external points of the conic it defines, and a nonsquare on the internal points, or vice versa. The quadratic form Q(X0,X1,X2)
=
X2-4X0X1
is a square on all external points of the conic C; (0,0,1) is incident with a tangent line and Q(0,0,1) = 1. So it is a nonsquare on the set of internal points I , implying that g(t)2 + 4tf(t) is a nonsquare for all t £
GF(qn).
The argument just given can be reversed, so that any pair of functions / and g which are GF(q) -linear with the property above, gives rise to a semifield flock and TGQ with kernel GF(q). There are two cases to consider now. CASE (1): J is CONTAINED IN A LINE. Then the flock planes share a common point. Theorems 4.4.5 and 4.4.6 imply that the flock is a Kantor-Knuth flock.
CASE (2): W E ARE NOT IN CASE 1. Then n > 3 and X generates the plane TTV (over GF(qn)). The set J is a P G ( n - 1, q) when interpreted over GF(g), so X contains a subplane of order q of nv which generates nv and which is contained in the set of internal points of the conic C.
The following lemma of Weil can be found in [135]. Lemma 5.1.15 (Weil) The number N of solutions in GF(q) of the hyperelliptic equation Y2 = g(X), where g e GF(q)[X] is not a square and has degree 2m > 2, satisfies \N-q+l\<
(2m-2)y/q.
BlokhuiSi Lavrauw and Ball [16] use Lemma 5.1.15 to obtain
Chapter 5. Good Eggs
116
Lemma 5.1.16 (Blokhuis, Lavrauw and Ball [16]) Let f(X) = X2 + uX+v e GF(qn)[X] be a nonzero square in GF(qn)for all X = x e GF(q), where q is odd and q > An2 — 8n + 2. At least one of the following holds: (i) f(X) is the square of a linear polynomial; (ii) n is even and f has two distinct roots in
GF(qn/2);
(iii) the roots of f are a and aa for some a £ Gal(GF(g")/GF(g)), where a e GF(qn). • Using the latter lemma, it is then shown that when q > An2 — 8n + 2, no subplanes of order q can be contained in the internal points of a conic in PG(2,g"). •
5.2
Good Eggs and Veronese Surfaces
There is a strong connection between good eggs and the Veronese surface V| of all conies of PG(2,g). So we include a short section on Veronese varieties; a good reference is Chapter 25 of Hirschfeld and Thas [74]. The Veronese variety of all quadrics of PG(n, K), n > 1 and K any commutative field, is the variety V = {(XQ, XJ,...,X1,
(x0,xi,...
X0Xi,X0X2,
. . . , X0Xn,
XxX2, • • • , XxXn, . . . , Xn-iXn)
7xn) is a point of PG(n,K)}
of PG(iV,K) with N = n(n + 3)/2. The variety V has dimension n and order 2 n [74]; for V we also write Vn or V2". It is also called the Veronesean ofquadrics of PG(n, K), or simply the quadric Veronesean of PG(n, K). It can be shown that the quadric Veronesean is absolutely irreducible and nonsingular [74]. Let PG(7V, K), with TV = n(n + 3)/2, consist of all points (yoo ) 2/11) • • • ; 2/nn) 2/01) 2/02) • • • ) 2/Ori) 2/12) • • • j 2/ln) • • • )
Vn—l,nji
for yij we also write y^. Let C, : PG(n,K) ^ PG(iV,K), with n > 1, be defined by (xQ,xi,...,xn) H^ {yoo,yn,...,yn-i,n), with y^ = XiXj. Then £ is a bijection of PG(n, K) onto the quadric Veronesean V of PG(n, K). It then follows that the variety V is rational.
\\
5.2. Good Eggs and Veronese Surfaces
117
The quadrics of PG(n, K) are mapped by £ onto all hyperplane sections of V. It follows that no hyperplane of PG(iV, K) contains the quadric Veronesean V. The lines of PG(n, K) are mapped onto irreducible conies of V; for K ^ GF(2) each irreducible conic of V is the image of a line of PG(n, K). Hence for |K| > 2 any two distinct points of V are contained in a unique irreducible conic of V. If K = GF(q), then it is clear that Vn contains 9(n) = qn + qn-1-\ \-q + l points. As no three points of Vn are collinear, the quadric Veronesean Vn is a 0(n)-cap of PG(iV, q), with N = n(n + 3)/2. Let n = 2 and assume from now on that K = GF(q). Then V is a surface of order 4 in PG(5, K). Apart from Vf, which is a conic, the variety V| is the quadric Veronesean which is most studied and characterized. With the conies (irreducible or not) of PG(2,q) correspond all hyperplanes sections of V|. The hyperplane is uniquely determined by the conic if and only if the latter is not a single point. If the conic of PG(2, q) is one line, then the corresponding hyperplane of PG(5, q) meets V| in an irreducible conic; the surface V| contains no other irreducible conies. It follows that V| contains exactly q2 + q + 1 irreducible conies, that any two distinct points of V| are contained in a unique irreducible conic of V|, and that any two distinct irreducible conies on V| meet in a unique point. If the conic C of PG(2, q) is two distinct lines, then the corresponding hyperplane PG(4, q) meets V| in two irreducible conies with exactly one point in common; if C is irreducible, then PG(4, q) meets V| in a rational quartic curve. The planes of PG(5, q) which meet V| in an irreducible conic are called the conic planes of V|. Any two distinct conic planes of V| have exactly one point in common, and this common point belongs to V|. The tangent lines of the irreducible conies of V| are called the tangents or tangent lines of V|. Since no point of the surface V| is singular, all tangent lines of V| at the point p of V| are contained in a plane ir(p). This plane ir(p) is called the tangent plane of V| at p. Since p is contained in exactly q + 1 irreducible conies of V| and since no two conic planes through p have a line in common, the tangent plane 7r(p) is the union of the q+1 tangent lines of V| through p. Also V|ri7r(p) = {p}. For any two distinct points pi and p 2 of V|, the tangent planes ir(pi) and TT(J>2) have exactly one point in common. If C is a nonsingular conic on \>2, if 7T is the plane of C and if p' e V|\C, then ir(p') n n = 0. Suppose that q is odd. Then PG(5, q) admits a polarity that maps the set of all conic planes of V| onto the set of all tangent planes of V| • Suppose that q is even. Then the nuclei of the q2 + q + 1 irreducible conies
118
Chapter 5. Good Eggs
on Vf are the points of a plane PG(2, q), called the nucleus or kernel of V|. In Thas [180, 183], the following classification of good eggs is obtained. Theorem 5.2.1 Consider a TGQ T(n,2n,q) = T(0), q odd, with O = 0(n, 2n, q) = {TT, TTI , . . . , irq2n } . If O is good at IT, then we have one of the following. (a) There exists a P G ( 3 , qn) in the extension PG(4n - 1, qn) of the space PG(4n - 1, q) of 0(n, 2n, q) which has exactly one point in common with the extension 7Fj of 7r, to GF(qn), with i = 1,2,..., q2n. The set of these q2n + 1 points is an elliptic quadric of P G ( 3 , qn) and T(O) is isomorphic to the classical GQ Q(5, qn). (b) We are not in Case (a) and there exists a PG(4,g") in PG(4n 1, qn) which intersects the extension W of IT to GF(qn) in a line M and which has exactly one point ri in common with each space TTI, with i = 1,2,..., q2n. Let W = {n \\ i = 1,2,..., q2n} and let M be the set of all common points of M and the conies which contain exactly qn points of W. Then the set W U M is the projection of a quadric Veronesean V| from a point pin a conic plane of V| onto a hyperplane PG(4, qn); the point p is an exterior point of the conic o / V | in the conic plane. Abo, the point-line dual ofT(0) is isomorphic to the flock GQ S(T) with T a Kantor-Knuth flock, that is, O is a Kantor-Knuth egg. (c) We are not in Cases (a) and (b) and there exists a P G ( 5 , g n ) in PG(4n - 1, qn) which intersects if in a plane /x and which has exactly one point n in common with each space Wi, with i — 1,2,..., q2n. Let W = {ri || i = 1,2,..., q2n} and let C be the set of all common points of \x and the conies which contain exactly qn points of W. Then the set W U C is a quadric Veronesean in P G ( 5 , qn). • Thas [180] contains a long technical proof of Theorem 5.2.1, buth with a weaker version of Case (b); Case (b) could be improved relying on Thas [183], see Section 9 of this paper. An alternative proof is contained in Lavrauw and Lunardon [96]. Remark 5.2.2 If we project the Veronesean V | in PG(5,g r a ) onto some PG(3,g") c P G ( 5 , g n ) from a line TV that intersects V| in two points of PG(5, q2n) \ P G ( 5 , qn) which are conjugate with respect to the extension GF(g 2 n ) of GF(g"), then we obtain an elliptic quadric of P G ( 3 , qn).
5.2. Good Eggs and Veronese Surfaces
119
If the egg 0(n, 2n, q) = O, with q odd, is good at its element n, then, by Corollary 5.1.4, the translation dual T(C*) of T(O) is the point-line dual of a flock GQ S {J7) with elation point T, where r is the tangent space of O at 7r. The point (oo) of T(£>*) is a translation line of S{T). We may assume that (oo) is the line [oo] of <S(JF) and then T is a semifield flock; see Section 4.7 and Payne and Rogers [125]. Conversely, if T is a semifield flock, then, by Section 4.7 and Corollary 5.1.4, S{T) = T(0) where O* is good at its element r which corresponds to the elation point of T. If J 7 is a Kantor-Knuth flock which is not linear, then S{T) = T(O) ^ T(C*) with O of Type (b) in Theorem 5.2.1; see Theorem 4.7.3. If T is a Ganley flock or the Penttila-Williams-Bader-Lunardon-Pinneri flock, then S{F) = T(0*) with O of Type (c) in Theorem 5.2.1. If the TGQ T(O) of order (s 2 , s), with s odd, is good and the kernel has either size s or size
120
Chapter 5. Good Eggs
exactly qn + 1 elements of O. If m, with i e {3,4}, does not belong to
5.3.
Coordinatization and Applications
121
5.3 Coordinatization and Applications We now introduce a coordinatization method of eggs and their duals which was first used by Payne in [118], and later on generalized by Lavrauw and Penttila [97].
5.3.1 Coordinatization of Eggs and Dual Eggs The next theorem states the connection between additive g™-clans and eggs, and is taken from [97]. We use the notation of §4.7.2.
Theorem 5.3.1 The set C = {At = ( *
Vt
J \\ t £ GF(qn)}
is an additive
qn-clan if and only if the set O = {TT(A) || A G G F ( g n ) 2 U {oo}}, with TT(A)
= {(*, -gt(X), -A 5 ') || t G GF(qn)},
A G GF(
and TT(OO) =
{(0,*,(0,0)) I)
teGF(qn)},
where the triples defining w(X) and 7r(oo) are considered as vectors of the in-dimensional vectorspace over GF(q), is an egg o / P G ( 4 n — l,q), where gt(A) = A^4tAT and XSt = X(At + AT). The tangent space at the element n(X), X E GF(qn)2, is r(A) = {(*, PXT + A5'AT - gt(X), p)\\te
GF(qn), (3 e G F ( ^ ) 2 } ,
and the tangent space at IT{oo) is r(oo) = {(0,i,/3) || t G GF(qn),(3 G GF(g") 2 }.
• For the rest of this section, it is convenient to write F instead of
GF(qn).
If At is additive, so defines an egg in P G ( 4 n — l,q), we can write At as
A
>=t (-:)"'
Chapter 5. Good Eggs
122
with (a0, a i , . . . , a„_i) = (1,0,..., 0), for certain bt, Ci £ F. If an egg O is written in this form, we denote it by 0{b,c) where b = {bo,bi,... , 6„-i) and c = (c 0 ,ci,... ,c„_i). If O is as such, it is easy to explicitly write down the elements of the dual egg, and the tangent spaces to the latter. This was first done by Lavrauw and Penttila [97] in its full generality, based on observations of Payne in [118], who carried out the calculations for the eggs corresponding to the Roman TGQs. We start with a lemma. Lemma 5.3.2 Let tr be the trace map from F to GF(q). Then /n-l
\
tr Y; "**** for allt <E F if and only if n-l
E°r1-i=°i=0
Proof. Since the trace function is additive and tr(a;) = tr(a;'?) for x e F, we have that
5>r-> B
fn-l
tr J2 ^
}=
tr
\i=Q
vi=0
The fact that tr(ax) = 0, Vx e F implies that a = 0, ends the proof.
•
Theorem 5.3.3 The elements of the translation dual 0*(b,c), and the corresponding tangent spaces, of the egg 0(b, c) are given by TT*(A)
= {{-9*tW,t, -Ai) || t G F}, A E F 2 , 7T* (C») = {(«, 0 , ( 0 , 0 ) ) | | t 6 F } ,
r*(A) = {(/*(/?, A) +g*t (A), t,/3) \\teW,0£
F 2 }, A e F 2 ,
r * M = {(t,0,/3)||ieF,/3GF 2 }, with
n-l
2 2 1 1 9t (a, 6) = £ ( a i a + bia& + Cift ) /'** /'* i=0
5.3. Coordinatization and Applications
123
with (a 0 , ai, • • •, a„_i) = ( 1 , 0 , . . . , 0) and n-l
f*((a, b), (c, d)) = ^ ( 2 a i a c + &i(ad + 6c) + 2ci&d)1/\ i=0
Proof. We calculate the vector space dual of 7r(A) and r(A) in V(4n, q) with respect to the standard inproduct ((x, y, 2;, w), (a;', ?/', 2:', to')) i-» Xx{xx' + yy' + zz' + ww'), with tr the trace map from F onto GF(g). If (x, y, z, w) is an element of the vector space dual of r( A), thentr[xt+y((3\T+\St \T -gt(\)) + (z,w)/3T] = 0, 2 for all i e F and all (3 e F . With A = (a, 6) and j3 = (c, d), we obtain tr[xi + y(ac + bd + XStXT - gt(X)) +zc + wd] = 0,
(5.1)
for all c,d,t e F. When t = 0, this equation is satisfied for all c, d e F if and only if w = -by and z = -ay. Substituting this back into Equation (5.1), we obtain that tr[a;i + y(XStXT - gt{X))} = 0 for all t e F. Using the formulas for gt and St this is equivalent to n-l
tr[(x + y{a0a2 + b0ab + c0b2))t + ^
y(aia2 + btab + Cib2)tq%] = 0,
i=l
for all t e F. By Lemma 5.3.2, it follows that (x, y, 2;, to) is of the form n-l
( - ^ ( a , a 2 + 6,06 +
Ci 6
2 1/< 1/qi
)
i
,t, - a t , -bt),
i=0
for some i e F . The tangent spaces are handled in the same way.
5.3.2
•
The Known Examples Revisited
Using Theorem 5.3.3, we now write down, for q odd, the known examples of good eggs, and their translation duals; for q even each known egg arises by field reduction either from an elliptic quadric of P G ( 3 , q) or from a Tits ovoid. We only provide elements of type 7r(A) and 7r*(A), A e F = GF(q"), as the elements 7r(oo) and 7r*(oo) are all of the form {(0, t, 0,0) II t e F} and {(£, 0,0,0) || t e F} in this representation.
Chapter 5. Good Eggs
124
•
In this case, the egg is isomorphic to its translation dual; see Payne [118]. Let q be odd, a an automorphism of F, and m a given nonsquare in F. Then KANTOR-KNUTH EGGS.
TT(A) = {(t, -a2t + mb2ta, -2at, 2bmta) \\ t G F} and TT*(A) = { ( - a 2 i + ( m 6 2 ) C T _ V ~ \ t , -at,-bt)
•
\\
te¥}.
GANLEY AND ROMAN EGGS. For q > 9, the Ganley eggs are not isomorphic to their translation dual; see Payne [118]. They exist for q = 3h, h > 1. Let m be a nonsquare in F. Then TI-(A)
= {(*, - ( a 2 - b2m)t - abt3 + b2m~H^, at - bt3, -at3 + m~H9))
-b(mt
\\teW}
and 7r*(A) = { ( - ( a 2 - b2m)t -
{abf'3tll3
+ (^m" 1 ) 1 / 9 * 1 / 9 , t, -at, -bt) || * G F}. • THE PENTTILA-WILLIAMS-BADER-LUNARDON-PINNERI TRANSLATION DUAL.
EGG AND ITS
The Penttila-Williams-Bader-Lunardon-Pinneri
egg is not isomorphic to its translation dual. Here q = 3 5 . Then 7r(A) = {(t, ~a2t - abt3 + b2t27, at - bt3, -at3 - bt27)
\\te¥}
and ,r*(A) = {(-a2t
- {abflh1'3
+ {b2)l'27t1'27,
t, -at, -bt) \\ t G F}.
5.3.3 A Geometric Connection between Semifield Flocks and Good Eggs Let J 7 be a semifield flock of the quadratic cone K with vertex v in P G ( 3 , qn), q odd. If one dualizes in P G ( 3 , qn), the qn flock planes become the affine points of an n-dimensional affine space AG(n,q) over GF(q). The (n — 1)-dimensional space at infinity P G ( n - 1, q) of AG(n, q) is contained in the set of internal points of the conic C, that corresponds to the
5.3. Coordinatization and Applications
125
set of generators of /C, in the plane TTV corresponding to v. Over GF(q) the conic C becomes a pseudo-conic C in the space P G ( 3 n — l,q) which corresponds to 7r„. Dualizing in P G ( 3 n - 1, q), the space P G ( n - 1, q) becomes a (2n — l)-space PG(2n — 1, q) skew to the elements of the pseudo-conic C obtained by dualizing C. Now suppose O is a good egg in P G ( 4 n — l,q) = (O), q odd. Suppose O' is one of the pseudo-conies containing the good element TT (see Theorem 5.1.9), and let TT' e O \ O'. Suppose T' is the tangent space of O atir'; then T' n (£>') = /3 is a (2n - l)-space in P G ( 3 n - 1, q) = (£>') skew to £>'. Now reverse the argument of the first paragraph to obtain a semineld flock. In [95] Lavrauw proves that such a semifield flock is isomorphic to the semifield flock defining the good egg O.
5.3.4
Further Characterizations of Good Eggs in Odd Characteristic
The following interesting theorems are due to Lavrauw [94]. Theorem 5.3.4 Let O he an egg o / P G ( 4 n - 1, q), q odd, which is good at TT. Then the following properties are equivalent. (i) O is classical. (ii) There exists a triple (ir',w",TT'"), with TT',TT",TT'" distinct elements of 0\{n} and where n' is not contained in the pseudo-conic O' of O in (IT, TT", TT'"), such that the tangent space of O at n' contains two distinct elements of the regular (n - l)-spread of (w, TT", IT'"} which contains O'. (iii) Each triple (TT', n", TT'"), with n', IT", TT'" distinct elements ofO\{ir} and where TT' is not contained in the pseudo-conic O' of O in (n, IT", TT'"), is such that the tangent space of O at TT' contains two distinct elements of the regular (n - l)-spread of (TT, TT", TT'"} which contains O'. Proof. Clearly (i) implies (iii), and (iii) implies (ii). It remains to show that (ii) implies (i). So assume that (ii) is satisfied. By Corollary 5.1.4 and Section 4.7 the translation dual O* of O defines a semifield flock T. In the (2n - 1)-dimensional space (TT, TT", TT'") n r ' = £, with T' the tangent space of O at TT', a (n - 1)spread is induced by the regular (n - 1)-spread defined by O' (the space £
126
Chapter 5. Good Eggs
is a line of the PG(2, qn) defined by the regular (n - l)-spread). It follows that the semifield flock T is linear, and so O is classical. • Theorem 5.3.5 Let O be an egg o / P G ( 4 n — \,q),q odd, which is good at TT. Then the following properties are equivalent. (i) O is a Kantor-Knuth egg. (ii) There exists a triple (TT',TT",TT'"), with TT'' ,TT"\TT'" distinct elements of 0\{TT} and where TT' is not contained in the pseudo-conic O' of O in (TT, TT", TT'"), such that the tangent space of O at TT' contains an element C of the regular (n - l)-spread of (TT, TT", TT'") which contains O'. (hi) Each triple (TT', TT", TT"'), with TT', TT", TT'" distinct elements ofO\{Tr} and where TT' is not contained in the pseudo-conic O' of O in {TT, TT",TT'"), is such that the tangent space of O at TT' contains an element ( of the regular (n - l)-spread of (TT, TT", TT'"} which contains O'. Proof. Assume that O is a Kantor-Knuth egg and let T be the corresponding flock. Let TT',TT",TT'" be distinct elements of 0\{7r} where TT' is not contained in the pseudo-conic O' of O in {TT,TT",TT'"). By Theorem 4.4.6 all planes of the elements of T contain a common point. It follows that (n, TT", TT'") n T' = £, with r ' the tangent space of O at TT', contains an element C of the regular (n - l)-spread denned by O'. Hence (i) implies (iii). Since (iii) obviously implies (ii), it remains to show that (ii) implies (i). So assume that (ii) is satisfied. The egg O* defines a semifield flock T. Also, all planes of the elements of T contain a common point. By Theorem 4.4.6 the flock T is a Kantor-Knuth flock, and so O = O* is a Kantor-Knuth flock. • Remark 5.3.6 If O is not classical, then the element ( in Theorem 5.3.5 defines an exterior point of the conic C of P G ( 2 , g n ) defined by the pseudoconic O'; see Theorem 4.4.6. Theorem 5.3.7 Let O he an egg in PG(4n - 1, q), q odd, and let r\ he the subspace generated by three distinct egg elements. (i) If O is a Kantor-Knuth egg, then any element TT' of O is either contained in r\ or intersects r\ in at most one point. (ii) If O is a Ganley-Payne egg, then any element TT' of O is either contained in rj or intersects r\ in at most a line.
5.3. Coordinatization and Applications
127
(iii) If O is the translation dual of the Penttila-Williams-Bader-LunardonPinneri egg, then any element IT' of O is either contained in n or intersects n in at most a line. Proof. Using coordinates, see Lavrauw [94].
•
Chapter 6
Generalized Quadrangles, Nets and the Axiom of Veblen Each regular point of a thick GQ <S of order (s, t) defines a dual net N® • If TVjP is not a dual affine plane and satisfies the Axiom of Veblen, then M® is isomorphic to the "classical" dual net H™, for some q and n. In such a case strong properties for the GQ S can be deduced. In this chapter it is also shown that the existence of particular automorphisms of A/jP, where the regular line L is incident with some coregular point x of S, forces S to be a TGQ.
6.1 Generalized Quadrangles and the Axiom of Veblen Consider a GQ T 3 (C) of Tits, with O an ovoid of PG(3,g). Here s = q and t = q2. Then the point (oo) is coregular, that is, each line incident with (oo) is regular. It is an easy exercise to check that for each line incident with (oo) the corresponding dual net is isomorphic to H^; see Section 2.2 for the definition of H | . Hence for each line incident with the point (oo) the corresponding dual net satisfies the Axiom of Veblen. We now prove the converse. Theorem 6.1.1 and its corollary are due to Thas and Van Maldeghem [193]. Theorem 6.1.1 Let S = {V, B, I) be a GQ of order (s, t), with s ± t, s > 1 and t > 1. If S has a coregular point x and if for each line L incident with x the corresponding dual net N® satisfies the Axiom of Veblen, then S is isomorphic to a T 3 (O) of Tits. 129
130
Chapter 6. Generalized Quadrangles, Nets and the Axiom ofVeblen
Proof. Let Li,L2, L3 be three lines no two of which are concurrent, let Mi, M 2 , M 3 be three lines no two of which are concurrent, let Li / M, if and only if {i,j} = {1,2} and assume that xlL\. By 5.3.8 of Payne and Thas [128] it is sufficient to prove that for any line L4 e {Mi,M2}± with Li >Li,i = 1,2,3, there exists a line M 4 concurrent with L1,L2,L4. So let L4 e {Mi,M 2 }- L with L4 / Li} i = 1,2,3. Consider the line R containing L2 n M 2 and concurrent with L\. Further, consider the line R' containing M2 n L4 and concurrent with L\. By the regularity of Li there is a line R" e {Mi, Ms}- 1 - 1 through the point L 3 n M 2 . Clearly the lines Lx and .R" are concurrent. So the line L\ is concurrent with the lines R, R', R"; also the line M 2 is concurrent with the lines R, R', R". By the regularity of Li the line R" belongs to the line {R, R'}1-1 of the dual net A ^ denned by L1. Hence the lines {R, R'}-1-1 and {Mi,M3}xlof A ^ have the element R" in common. By the Axiom ofVeblen, also the lines {Mi,i?'} ± J - and {M3,R}±Jof A/j^ have an element M 4 in common. Consequently M 4 is concurrent with Li,L2,L4. Now fro m 5.3.8 of Payne and Thas [128] it follows that S is isomorphic to a T 3 ( 0 ) of Tits. • As a corollary the following characterization of the classical GQ Q(5, s) is obtained. Corollary 6.1.2 Let S be a GQ of order (s, t) with s ^ t, s > 1 and t > 1. (i) J/s is odd, then <S is isomorphic to the classical GQ Q(5, s) if and only if it has a coregular point x and if for each line L incident with x the corresponding dual net A/jP satisfies the Axiom of Veblen. (ii) If s is even, then S is isomorphic to the classical GQ Q(5, s) if and only if all its lines are regular and for at least one point x and all lines L incident with x the dual nets A/JP satisfy the Axiom of Veblen. Proof. Let (x, L) be a flag of the GQ Q(5, s). By 3.2.4 of Payne and Thas [128] there is an isomorphism of Q(5,s) onto T 3 (C), with O an elliptic quadric of P G ( 3 , s), which maps x onto the point (oo). It follows that A/JP satisfies the Axiom of Veblen. Conversely, assume that the GQ S of order (s, t), with s odd, s / (, s > 1 and t > 1, has a coregular point x such that for each line L incident with x the dual net A/p1 satisfies the Axiom ofVeblen. Then by Theorem 6.1.1 the GQ 5 is isomorphic to T 3 (C). By Barlotti [8] and Panella [106] each ovoid
6.2. Translation Generalized Quadrangles and the Axiom ofVeblen
131
O of P G ( 3 , s), with s odd, is an elliptic quadric. Now by Theorem 1.5.2(ii) we have S a T 3 (C) ^ Q(5, s). Finally, assume that for the GQ S of order (s, t), with s even, s / t, t > 1, all lines are regular and that for at least one point x and all lines L incident with x the dual nets MP satisfy the Axiom of Veblen. Then by Theorem 6.1.1 the GQ S is isomorphic to T 3 (C). Since all lines of S are regular, by §2.1 we finally have S =* T 3 (C) =* Q(5, s). •
6.2
Translation Generalized Quadrangles and the Axiom of Veblen
This section is also taken from Thas and Van Maldeghem [193]. It explains the relationship between the Axiom ofVeblen and good eggs. Theorem 6.2.1 Let S^ he a TGQ of order (s, s2), s / 1, with base-point x. Then the dual net MP defined by the regular line L, with xlL, satisfies the Axiom ofVeblen if and only if the egg 0(n, 2n, q) which corresponds to S^ is good at its element -Ki which corresponds to L. Proof. Assume that the dual net MP satisfies the Axiom of Veblen. Let the egg 0(n, 2n, q) correspond to S^ and let •K1 correspond to L. We have s = qn. The dual net has qn + 1 points on a line and q2n lines through a point. By Theorem 2.2.3 the dual net MP is isomorphic to H;j„. Consider the TGQ T(n, 2n, q) = 5^) and let PG(3n, q) be a subspace skew to •Ki in the projective space PG(4n,q) in which T(n,2n,q) is defined. Let 0(n,2n,q) = {7r0,7Ti,... ,Tcq2n}, let (7r,,7rj) n PG(3n, q) = TT'J for all j / i (TT'J is (n-l)-dimensional), let P G ( 4 n - l , 9 ) n P G ( 3 n , g ) = P G ( 3 n ~ l , g ) , with PG(4n - 1, q) the space of 0(n, 2n, q), and let rt n PG(3n, q) = T[, with Ti the tangent space of 0(n, 2n, q) at Hi {r[ is (2n - 1)-dimensional). Then the dual net MP is isomorphic to the following dual net MD: • POINTS of MD are the q2n spaces -K'J, j ^ i, and the q3n points of PG(3n,q) \PG(3n-l,q); • LINES of MD are the q4n subspaces of PG(3n, q) of dimension n which are not contained in PG(3n - 1, q) and contain an element it'p j ^ i, and •
INCIDENCE
is the natural one.
132
Chapter 6. Generalized Quadrangles, Nets and the Axiom of Veblen
Clearly the points TT'J, j ^ i, of ND form a parallel class of points. Let M be a line of ND incident with 7^ and let it'k ^ ir'j, with k ^ i ^ j . As J\fD = H 3 „ the elements -n'k and M of MD generate a dual affine plane AD in AfD', and the plane AD contains qn points ix[,l ^i. Clearly the points of AD not of type ir[ are the q2n points of the subspace (ir'k, M) of PG(3n, q) which are not contained in P G ( 3 n -l,q). Hence the qn points of AD of type -K\ are contained in (ir'k,M) n P G ( 3 n - 1, q). It follows that these qn elements 7r,' are contained in a (2n - 1)-dimensional space PG'(2n - 1, q); also, they form a partition of P G ' ( 2 n - 1, q)\r!-. Consequently for any two elements -K[^'V, I ^ i ^ I', the space (7^,71-;',) contains exactly q n elements Tv'r, r ^ i. Hence for any two spaces 717 and -KV of 0(n, 2n,q , )\{7r I }, the (3ra - 1)-dimensional space (71-4,717,717/) contains exactly <jn + 1 elements of 0(n, 2n, 5). We conclude that 0(n, 2n, q) is good at 717. Conversely, assume that 0(n, 2n, q) is good at the element 717 which corresponds to L. As in the first part of the proof we project onto a PG(3n, q) and we use the same notation. Since 0(n, 2n, q) is good at 717, for any two elements 7Tj,7r[,, with I ^ % ±V', the space (71-;, 7^,) contains exactly qn elements n'r, r ^ i; these qn elements form a partition of the points of (717', ix'v) which are not contained in T[. If M, M' are distinct concurrent lines ofAfD, then it is easily checked that M and M' generate a dual affine plane AD of order qn in AfD. As AD satisfies the Axiom of Veblen, also MD does. • Corollary 6.2.2 Let S^ be a TGQ of order (s,s2), s odd and s ^ 1, with base-point x. If the dual net N® defined by some regular line L, with xlL, satisfies the Axiom of Veblen, then S^ contains s 3 + s2 classical subquadrangles Q(4, s) containing L. Proof.
This follows immediately from Theorem 6.2.1 and Theorem 5.1.9.
6.3 Property (G) and the Axiom of Veblen Property (G) for GQs of order (s, s 2 ) was introduced in Section 4.10. If the GQ <S of order (s, s 2 ), s / 1 , satisfies Property (G) at the line L, then we shall prove that L is regular and shall consider the dual net Af® in some detail. Theorem 6.3.1 Let S = (P, B, I) be a GQ of order (s, s2), s ^ 1, satisfying Property (G) at the line L. Then L is regular.
6.3. Property (G) and the Axiom ofVeblen
133
Proof. Let S = (V, B, I) be a GQ of order (s, s2), s even, satisfying Property (G) at the line L. Consider distinct points x and y incident with L, and let x be a center of the triad {y, z, u}. By Theorem 2.6.2, S has a subquadrangle S' of order s which contains {y, z, u}1- U {y, z, u}±J-. Let w be a point of <S', with w / i and w •/ y. Further, let r, r', r" be distinct points of {x, w}^' (here " 1 " ' denotes "perp" in <S'), with rlL. Then {r, r', r"}- 1 and {r, r', r"}^1- are contained in a subquadrangle of order s of 5. Clearly the lines xr,xr',xr",wr,wr',wr" are common lines of S' and 5". By 2.3.1 of Payne and Thas [128], see also Section 1.1, the common points and lines of S' and S" form a subquadrangle S' n <S" of order (s,t'), with £' > 1, of S". Now Lemma 1.1.1 readily implies that S' n S" = S' = 5". It follows that {r,^}- 1 '- 1 ' = {r,r',r"}±±, {r,^}1' = {r,r',r"}x, and so {r,r'} and {x, w} are regular in S'. Hence x is regular in S'. With the notation of the previous paragraph and interchanging the roles of x and y, we see that also y is regular in S'. Since in the previous paragraph y and r play the same role in S', also r is regular in S'. It follows immediately that all points of L are regular in <S'. By Theorem 2.1.5 also the line L is regular in <S'. Consider now a line M of S, with M ^ L. Let y ~ x'lM, x ~ y'lM, and x" e {y,y'}-L\{a;,a;'}. The subquadrangle 5 of order s defined by x, x', x" contains L and M. Since L is regular in <S, the span {L, M}-11- has size s + 1. We conclude that L is regular in S. Next assume that s is odd, s ^ 1, and that the GQ <S = (V, B, I) of order (s2, s) satisfies Property (G) at the point x. By Theorem 4.10.6 the GQ S is a flock GQ. Now we consider the construction of Knarr, see Section 4.9. If the point y is not collinear with x, that is, if y is a point of PG(5, s) not in x^, with £ the symplectic polarity used to construct S in PG(5, s), then {x, y}-11- consists of the s + 1 points of the line (x, y) of PG(5, s). As |{x, y}"1"^! — s + 1 the point x is regular. • Remark 6.3.2 In the first part of the proof, where in the even case we show that x is regular in S', it is sufficient to assume that S satisfies Property (G) at the flag (x,L). By Theorem 4.10.7 it is sufficient in the odd case to assume that <S satisfies Property (G) at some point pair {x, y}, with xlLly. In fact, for all s ^ 1, the GQ S of order (s, s2) is the point-line dual of a flock GQ. Hence L is an axis of symmetry (cf. Section 4.6), and consequently regular. The following theorems of this section are also taken from Thas and Van Maldeghem [193].
134
Chapter 6. Generalized Quadrangles, Nets and the Axiom ofVeblen
Theorem 6.3.3 Let S = (V, B, I) be a GQ of order (s2,s), s even, satisfying Property (G) at the point x. Then the dual net Mx satisfies the Axiom of Veblen, and consequently Mx = H^. Proof. Let S = (V, B, I) be a GQ of order (s 2 , s), s even, satisfying Property (G) at the point x. By Theorem 6.3.1 the point x is regular. Let y be a point of the dual net Mx, let A\ and A2 be distinct lines of Mx containing y, let B\ and B2 be distinct lines of Mx not containing y, and let Ai n Bj ^ 0 for all i,j e {1,2}. Let {z} = Ax n Bx and let zlM, with x IM. Further, let xlL, with z IL, let u be the point of A\ on L, and let u' be the point of B\ on L. The line of S incident with u, respectively u', and concurrent with M is denoted by C, respectively C; the line incident with z and a; is denoted by N. Since <S satisfies Property (G) at x, the triple {C, C, N} is 3-regular. By Theorem 2.6.2 the lines of S concurrent with at least one line of {C, C", N}1- and one line of {C, C", A^}-1-1-, are the lines of a subquadrangle S' of order s of 5. As x is regular for S it is also regular for S'. By Theorem 2.2.1 the point x in S' defines a projective plane wx of order s. Clearly A\,A2, B\,B2 are lines of the projective plane TXX. Hence B\ and B2 intersect in irx. Consequently Nx satisfies the Axiom ofVeblen, and so N° = H^ by Theorem 2.2.3. • Theorem 6.3.4 Let S^ = T(O) be a TGQ of order (s, s2), s / 1, with base-point x. Then the dual net N® defined by the regular line L, with xlL, satisfies the Axiom ofVeblen if and only if the translation dual <S'(x -1 = T(C*) of S^ satisfies Property (G) at the flag {x' ,L'\ where L' corresponds to L, that is, where L' is the tangent space of O at L. In the even case, A/jP satisfies the Axiom ofVeblen if and only ifS^ satisfies Property (G) at the flag (x, L). Proof. By Theorem 6.2.1 the dual net Af® satisfies the Axiom ofVeblen if and only if O = 0(n, 2n, q) is good at the element L. By Theorem 5.1.2 the egg O is good at L if and only if T(0*) satisfies Property (G) at the flag (x', V). By Theorem 5.1.5, for s even, T(0*) satisfies Property (G) at the flag (x', V) if and only if T(0) satisfies Property (G) at the flag (x, L). • Theorem 6.3.5 Let S^ be a TGQ of order {s,s2), s odd and s ± I, with base-point x. If xlL and if the dual net N® satisfies the Axiom of Veblen, then all lines concurrent with L are regular. Proof. Let N be concurrent with L, x IN, and let the line M of S^ be nonconcurrent with N. From Theorem 5.1.9 easily follows that the lines
6.4. Flock Generalized Quadrangles and the Axiom ofVeblen
135
M,N are lines of a subquadrangle of S^ isomorphic to Q(4, s). Hence {M, N} is a regular pair of lines. We conclude that the line N is regular in
S(*\
*
6.4 Flock Generalized Quadrangles and the Axiom of Veblen Let T be a flock of the quadratic cone K, of P G ( 3 , q). Then, by Theorem 4.6.3, with T corresponds a GQ S{T) of order (q2,q). In Payne [118] it was shown that <S(.F) satisfies Property (G) at its point (oo); see Theorem 4.10.1. Hence by Theorem 6.3.1 we have the following result. Theorem 6.4.1 For any GQ 5 (J") of order (q2, q) arising from a flock T, the point (oo) is regular. • (Of course, we also know that (oo) is regular because it is a center of symmetry.) The following theorem is taken from Thas and Van Maldeghem [193]. Theorem 6.4.2 Consider the GQ <S(JF) of order (q2,q) arising from the flock T. If q is even, then the dual net A / ^ always satisfies the Axiom of Veblen and so AfB^ = H^. If q is odd, then the dual net M^ satisfies the Axiom of Veblen if and only if T is a Kantor-Knuth flock. Proof. Consider the GQ S(T) of order (q2,q) arising from the flock T. Then S(T) satisfies Property (G) at the point (oo). Hence for q even, by Theorem 6.3.3, the dual net NP , satisfies the Axiom ofVeblen, and consequentlyA/^^H^. Next, let q be odd. Suppose that J 7 is a Kantor-Knuth flock. Then by Theorem 4.7.2 the point-line dual of S(T) is a TGQ, so is isomorphic to some T ( 0 ) with O = 0(n, 2n, q') and q = q'n and by Theorem 4.7.3, T(O) = T ( 0 * ) . In Payne and Thas [129] it is shown that if S(T) is not classical, then the elation point (oo) of S(T) is fixed by the full automorphism group of <S(JF). Hence if T is not linear, then the point (oo) of S{T) corresponds to some line n of Type (b) of T(O), that is, to some element n of O; if J77" is linear, so S{T) is classical, then we clearly may assume that (oo) corresponds to some element n of O. Consequently T(O) satisfies Property (G) at n. By Theorem 5.1.2 the translation dual O* of O is good
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136
at T, with T the tangent space of O at n. Hence, by Theorem 6.2.1, the dual net AfjP which corresponds with the regular line r of T(0*) satisfies the Axiom of Veblen. By Theorem 5.1.13 also the dual net M£ which corresponds with the regular line -K of T(0) satisfies the Axiom of Veblen. It follows that the dual net MP-, satisfies the Axiom of Veblen. Hence A/"/^ — H^. Conversely, assume that q is odd and that the dual net A / ^ satisfies the Axiom of Veblen. Hence A/"/^-, = H^. In the representation of Knarr, see Section 4.9, this dual net looks as follows: •
POINTS of A/"^\ are the lines of P G ( 5 , Q ) not containing p = (oo) but contained in one of the planes pLt = 7rt, with Lt &V and V the BLT-set which corresponds to T,
•
of NR^ can be identified with the threedimensional subspaces of p£ not containing p (£ is a symplectic polarity of P G ( 5 , q)), and
•
INCIDENCE
LINES
is inclusion.
By point-hyperplane duality in p^, the net A / ^ ) , which is the point-line dual ofAfPy is isomorphic to the following incidence structure: •
POINTS of A/(oo) are the points of p^ \ <5, with S some threedimensional subspace of pt,
•
of A/"(OO) are the planes of p^ not contained in 8, but containing one of the lines of a BLT-set V in 5, where V ~ V', and
•
INCIDENCE
LINES
is the natural one.
As the net A / ^ ) is isomorphic to the dual of Hjj, it is easily seen to be derivable; see e.g. De Clerck and Johnson [43]. Let V = {L'0, L[,..., L'q} be contained in the GQ W()'. Then the line set T = {L^L'J- 1 '- 1 ' u {L'0, L'z)1-'1-' U • • • U {L'0, L'q}±,A-', where "±"' denotes "peip" in W(g)', is a spread of W(q)'; see e.g. Thas [172]. As Af(oo) is derivable, by Biliotti and Lunardon [12] there are two distinct lines in 6 intersecting the q + 1 lines of V". Hence the image of V' on the Klein quadric is contained in a threedimensional space, and so the q planes of the elements of the flock T all contain a common point. Now by Theorem 4.4.6, T is a Kantor-Knuth flock. •
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137
Corollary 6.4.3 Suppose that the TGQ T(0), with O = 0(n,2n,q) and q odd, is the point-line dual of a flock GQ T, where the point (oo) of S^) corresponds to the element IT of O. Then O is good at the element TT if and only if T is a Kantor-Knuth flock. Proof. Assume that J 7 is a Kantor-Knuth flock. Then by Theorem 5.1.2 and Theorem 4.7.3 the egg O is good at TT. Conversely, assume that the egg O is good at TT. Then by Theorem 6.2.1 the dual net J\f^ = NR^s satisfies the Axiom of Veblen. By Theorem 6.4.2 the flock T is a Kantor-Knuth flock. • In Bader, Lunardon and Pinneri [4] a stronger version of Corollary 6.4.3 is obtained; the authors rely on Corollary 6.4.3, however. Theorem 6.4.4 Suppose that the TGQ T(0) with O = 0{n, 2n, q) and q odd, is the point-line dual of a flock GQ <S(.F), or, equivalently, assume that O* is good at some element. Then T(C*) is the point-line dual of a flock GQ, or, equivalently, O is good at some element if and only if T is a Kantor-Knuth flock. Proof. Assume that T is a Kantor-Knuth flock. Then by Theorem 5.1.2 and Theorem 4.7.3 the eggs O and O* are good. Conversely, assume that the egg O is good at some element IT. By Corollary 5.1.4 the GQ T(0*) is the point-line dual of a flock GQ with elation point T, where r is the tangent space of O at IT. Hence T(C*) has an elation group with base-line T. SO G acts transitively on the lines, distinct from r, incident with the point (oo) of T(C*). By Theorem 3.7.2, G is induced by a collineation group G' of PG(4n,q) (with PG(4n,g) the space in which T(0*) is constructed), which fixes O*, T, and acts transitively on 0*\{T}. Hence there is a collineation group G of (O) = PG(4n - l,q) fixing O, IT, and acting transitively on 0\{-7r}. AS T(O) is the point-line dual of a flock GQ, the GQ T(C) has an elation group with base-line IT' e O. Hence there is a collineation group G of PG(4n - 1, q) fixing O, IT', and acting transitively on 0\{IT'}. If IT ^ IT', then the collineation group (G, G) acts 2-transitively on O. Consequently, O is good at each of its elements. Now by Theorem 5.1.12, O is classical, so the flock T is linear. If IT = IT', then by Corollary 6.4.3, T is a Kantor-Knuth flock. •
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138
6.5
Subquadrangles and the Axiom of Veblen
In this section, which is also taken from Thas and Van Maldeghem [193], we describe the impact of the Axiom of Veblen on the existence of subquadrangles of order s of GQs of order (s 2 , s). Theorem 6.5.1 Let S = (V, B, I) be a GQ of order (s, t), s / 1 / t, having a regular point x.Ifx and any two points y, z, with y / x and x ~ z / y, are contained in a proper subquadrangle S' ofS of order (s',t), with s' / 1, then s' = t = y/s and the dual netMx satisfies the Axiom of Veblen. It follows that s and t are prime powers, and that for each subquadrangle S' the projective plane irx of order t defined by the regular point x of S' is Desarguesian. Conversely, if the dual net Mx satisfies the Axiom of Veblen, then either (a) s = t, or (b) s = t2, s and t are prime powers, x and any two points y, z, with y / x and x ~ z / y are contained in a subquadrangle Sf of S of order (t, t), and the projective plane irx of order t defined by the regular point x of S' is Desarguesian. Proof. Let S = (V,B,I) regular point x.
be a GQ of order (s,t), s ^ 1 ^ t, having a
First, assume that x and any two points y, z with y ^ x and x ~ z •/• y, are contained in a proper subquadrangle S' of <S of order (s', t), with s' ^ 1. As x is also regular for S', the GQ S' contains subquadrangles of order (1, t). Then Lemma 1.1.1 implies s' = t = y/s. By Theorem 2.2.1 the dual net N'x arising from the regular point x of S', is a dual affine plane of order t. Hence J\f'x satisfies the Axiom of Veblen. Now consider distinct lines Ai,A2,B1,B2 of the dual net Afx , where Ai n A2 = {z}, z <£ Bu z g B2, and AitlBj / 0 f o r a l H , j e {1,2}. Let AtnBi = {u}, A2C\B2 = {w}, and let y e {u,«;} J -\{a;}. Let S' be a subquadrangle of order t containing the points x, y, z of <S. Then Ai,A2, Bi,B2 are lines of the dual net Af'x . As Af'x satisfies the Axiom of Veblen, we have Bi n B2 / 0. It follows that the dual net J\f® satisfies the Axiom of Veblen. Consequently Mx = H^, and so s and t are prime powers. For any subquadrangle S' the dual net Af'x is a dual affine plane of order t, which is isomorphic to a dual affine plane of order t in Hf. Hence Af'%, and also the corresponding projective plane TTX, is Desarguesian.
6.6. Nets and Characterizations of Translation Generalized Quadrangles
139
Conversely, assume that the dual net N® satisfies the Axiom of Veblen. Also, suppose that s^t, and so s > t by Theorem 2.1.3. Then, by Theorem 2.2.3 and Corollary 2.2.4, we have N® = Hf and s = t2, with s and t prime powers. Now consider any two points y, z, with y / x, x ~ z ^ y. As .A/jf = Hf it is easily seen that z and {x, y}1- generate a dual affine plane A of order t inAfJ^. LetAi,A2,..., Ati be the lines of A. Further, let V' be the point set of <S consisting of the points of Aj- U A£ U • • • U A^-2 and the points of A. Clearly V' contains z and y, and \V\ = t3 + t2 + t + 1. Further, any line of <S incident with at least one point of V' either contains a; or a point of A; the set of all these lines is denoted by B'. Also, any point incident with two distinct lines of £>' belongs to V'. Then by 2.3.1 of Payne and Thas [128], see also Section 1.1, S' = (V',B', I'), with I' the restriction of I to (V x B') U (B1 x V') is a subquadrangle of S of order t. As in the first part of the proof one now shows that for any such subquadrangle S' the projective plane itx denned by x is Desarguesian. • Remark 6.5.2 Note that in the first part of the statement of Theorem 6.5.1, x need not be a regular point a priori; one easily deduces the regularity of x from the assumption about the subquadrangles, and Lemma 1.1.1.
6.6 Nets and Characterizations of Translation Generalized Quadrangles First we give the definition of a T-net, then we prove a characterization theorem for TGQs in terms of T-nets, and finally several interesting corollaries will be obtained. This section is taken from Thas [185]. Let J\f = (V, B, I) be a net of order k and degree r. Further, let R be a line of M and let T be the parallel class of B containing R, that is, T consists of R and the k - 1 lines not concurrent with R. An automorphism 9 of N is called a transvection with axis R if either 9 = 1 or if T is the set of all fixed lines of 9 and R is the set of all fixed points of 9. The net J\f is a T-net if for any two nonparallel lines M,N e B\T there is some transvection with axis belonging to T and mapping M onto N. In particular, let M = (P, B, I) be an affine plane of order k and let V be the corresponding projective plane. Then M is a T-net if and only if the point z of V defined by T is a translation point of V; see Hughes and Piper [75].
140
Chapter 6. Generalized Quadrangles, Nets and the Axiom of Veblen
If M is the dual of H™, then it is easy to check that TV is a T-net for any parallel class T. Theorem 6.6.1 Let (S^X\G) be a TGQ of order (s,t), s ^ 1 ^ t. Then for any line L incident with x, the dual net M£ defined by L is the dual of a T-net ML, with T the parallel class of ML defined by the point x. Proof. Let M, N be two distinct nonconcurrent lines of <S, concurrent with L but not incident with x. So M, N are nonparallel lines of the net ML, not belonging to the parallel class T defined by the point x. Let R e {M, iV} ± J -, with xiR, and let LimlM, LlnlN. In S there is a symmetry 9 about R with me = n. Then Me = N. The symmetry 8 induces an automorphism 9 of the net ML which fixes the parallel class T of ML elementwise, and all points of ML incident with the line R of T; it is easily checked that no other elements of ML are fixed. Also, Me = N. Hence 8 is a transvection with axis R of ML mapping M onto N. It follows that ML is a T-net. • Now we state the main theorem of this section. Theorem 6.6.2 Let S coregular point x. Iffor the dual of a T-net ML a TGQ with base-point
= (V,B,I) be a GQ of order (s,t), s ^ 1 ^ t, with at least one line L incident with x the dual net M® is with T the parallel class of ML defined by x, then S is x.
Proof. Let 9 ^ 1 be a transvection with axis R of the net ML, with ReT. Then 9 fixes all lines of S, distinct from L, incident with x and all sets {U, R}-1-1, with x IU, U ~ L. Now we prove that 9 induces a symmetry 7 about R of the GQ S. First, let r be a point of S not in x1. Put rlUlylL and rlMlzlR. The line Ue is concurrent with M; let UeIr'lM. Now we put r' = r 7 ; clearly r' ^ r and distinct points not in x1- are mapped by 7 onto distinct points not in xL. We show that r ~ u, r ^ u, r g x±, u £ xL, implies that r 7 ~ u1, r 7 ^ u~«. UrlA - L, then r^lA6 - L; iiulD ~ L, then u
6.6. Nets and Characterizations of Translation Generalized Quadrangles
141
regular and ru ~ uu1 ~ i?, we have uu1 ~ W. So a triangle with sides uu 7 , W,De arises, clearly a contradiction. It follows that u 7 l W , and so r 7 ~ u 7 . Clearly r 7 7^ u 7 . Now we define Z 7 for any line Z in 23. lfxIZ, then we define Z 7 = Z. Next, assume x IZ. Let n l Z l r 2 , n ^ r2, n 76 x •/ r 2 . Then Z 7 — 7-7r-7. From the foregoing section follows that in this way Z1 is well-defined. Clearly Z*i = Z for all lines Z ~ R. From the foregoing it also follows that for x IZ the lines Z and Z 7 are concurrent with a common line incident with x. Let Zi, Z2 be two distinct lines incident with a common point z G ar 1 . We will prove that Z 7 , Z 7 are incident with a common point z' G a;-1 with z ~ z'; also, Z7 7^ Z 7 . If at least one of the lines Z\,Z2 is incident with x, this is obvious. So assume x I Zit with i = 1,2. If zli?, then clearly Z 7 I z l Z 7 and Z 7 7^ Z 7 . Now let z IX. Then Zj and Z 2 are parallel lines of the net ML, SO also Zf and Zf are parallel in ML- Hence Z 7 = Z{, ZJJ = Zf, L are concurrent in <S; also Zj ^ Z^. So let z I i? and H I , say zlVlx. From the foregoing we have Zj ~ V — Z^. Assume, by way of contradiction, that Z7 7^ Z 7 . Let ailZj, a 2 i z ; j , ai I V I a 2 , ai ~ a 2 . Since 7 defines a permutation of the set of all pairs consisting of distinct collinear points of S not in x-1, we have a 7 ~ aT, , with aj I Z I and aT, IZ 2 , clearly a contradiction. Hence Z7, Zg, V are concurrent in S; also Z7 ^ Z7. Consider any point z e x1- and let ZiIzlZ2, with Z\ ^ Z 2 . Then, by 1 definition, z is the point incident with both Zl and Z.J. Now it is clear that 7 is an automorphism of S which fixes all lines concurrent with R. Consequently, 7 is a symmetry about R. Let li and l2 be incident with L, l\ 7^ x 7^ l2 and l\ 7^ l2. Now we choose two lines A1,A2, with hlAi, l2lA2 and R e {Ai,A2}±J-. Then there is some transvection 6 with axis R' of ML, for some R! e T, mapping Ai onto A 2 . As ({R', Ai}-11)6 = {R^A^-11= {R1, A2}-L±, it follows that R = R'. For the corresponding symmetry 7 about R we clearly have lj = l2. Consequently, R is an axis of symmetry of S. Hence any line R incident with x, but distinct from L, is an axis of symmetry of S. Now by Theorem 6.7.10 of K. Thas [206] <S is a TGQ with base-point x. • This theorem has several interesting and strong corollaries.
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Chapter6. Generalized Quadrangles, Nets and the Axiom of"Veblen
Corollary 6.6.3 Let S = (V, B, I) be a GQ of order s, s ^ 1, with coregular point x. Iffor at least one line L incident with x the corresponding projective plane TTL is a translation plane with as translation line the set of all lines of S incident with x, then S is a TGQ with base-point x. Proof. Follows immediately from Theorem 2.2.1, the first part of Section 6.6, and Theorem 6.6.2. • Corollary 6.6.4 Let S = {V, B, I) be a GQ of order s, s ^ I, with coregular point x. If for at least one line L incident with x the projective plane -KL is Desarguesian, then S is a TGQ with base-point x. If in particular s is odd, then S is isomorphic to the classical GQ Q(4, s). Proof. As any line of a Desarguesian plane is a translation line, by Corollary 6.6.3 the GQ S is a TGQ with base-point x. Now let s be odd. Then, by Theorem 3.9.6, we have <S^Q(4,s). • Corollary 6.6.5 Let S = {V,B,I) be a GQ of order s, with s e {5,7}. If S contains a coregular point, then S = Q(4, s). Proof. As there is a unique projective plane of order s, s e {5, 7}, by Corollary 6.6.4 we have S = Q(4, s). • A GQ of order 3 is isomorphic either to Q(4,3) or to its dual W(3); see Section 1.8. Corollary 6.6.6 Let S = (V,B,I) be a GQ of order (s,s2), s ^ I, with a regular line L for which the dual net A/jP defined by L satisfies the Axiom of Veblen. If L is incident with a coregular point x, then S is a TGQ with base-point x. Proof. Immediate from Theorem 2.2.3, the first part of Section 6.6 and Theorem 6.6.2. • Corollary 6.6.7 Let S = (V, B, I) be a GQ of order (s, s2), s even, satisfying Property (G) at the line L. If Lis incident with a coregular point x, then S is a TGQ with base-point x. Proof.
Immediately from Theorem 6.3.3 and Corollary 6.6.6.
•
6.6. Nets and Characterizations of Translation Generalized Quadrangles
143
Corollary 6.6.8 Let <S(J") be the GQ of order (q, q2), q even, arising from a flock T. If the point (oo) of <S(JF) is collinear with a regular point x, with (oo) 7^ x, then S(F) is isomorphic to the classical GQ H(3, q2) arising from a nonsingular Hermitian variety of P G ( 3 , q2).
Proof. The GQ S(T) is an EGQ with elation point (oo); see Section 4.6. It follows that the corresponding elation group acts transitively on the points distinct from (oo) which are incident with the line L = (oo)a;. Hence L is coregular. By Theorem 6.4.2 and Corollary 6.6.6, S(T) is a TGQ with baseline L. Now by Johnson's Theorem 5.1.11, S^) is isomorphic to H(3, q2).
The last application of Theorem 6.6.2 is on subquadrangles. Corollary 6.6.9 Let S = (V, B, I) be a GQ of order (s, t), s + 1 ^ t, having a coregular point x. If for a fixed line L, with xlL, and any two lines M, N, with M / L and L ~ N ^ M, there is a proper subquadrangle S' of S of order (s, t'), with t' ^ 1, containing L, M, N, then t' = s = VI, S is a TGQ with base-point x, and S' is a TGQ with base-point x. For s even S satisfies Property (G) at the flag (x, L) and for s odd the translation dual S* of S satisfies Property (G) at the flag (x', L'), with x' the base-point of S* and L' the line of S* corresponding to the line L of S. For s odd the subquadrangles S' are isomorphic to Q(4, s) and S* is isomorphic to the dual of a flock GQ 5(JF). If sis even and ifS* is isomorphic to the dual of a flock GQ S(T), then 5^Q(5,s). Proof. By Theorem 6.5.1 we have that t' = s = y/i and the dual net A/jP satisfies the Axiom of Veblen. Hence by Corollary 2.2.4 A/"f = H;?. By Corollary 6.6.6 the GQ S is a TGQ with base-point x. As S' contains x, S' is also a TGQ with base-point x. Let «S = T ( 0 ) , with O = 0(n, 2n, q), and where -K e O corresponds to L. Now we show that the condition on the existence of subGQs implies that for any two spaces 7TJ,7T.,- G 0\{X}, with i ^ j , the (3n - 1)-dimensional space {-K^^-KJ) contains exactly qn + 1 elements of O. Let PG(3n, q) be any 3n-dimensional space in the PG(4n,g) containing T ( 0 ) , which contains (IT, ni,TTj) but is not contained in P G ( 4 n — l,q) = (O). Next, let /x be an n-dimensonal space in PG(3n,q) containing n, with \i % (ir,iTi,-Kf), let n be an n-dimensional space in PG(3n,q) containing wi, with -n % (TT,^,^) and fj, n r\ = 0, and let p be the n-dimensional space containing TTJ and intersecting p, and 77.
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Chapter 6. Generalized Quadrangles, Nets and the Axiom ofVeblen
By the condition on the existence of subquadrangles, there is a subGQ of order qn = s of T(C) containing the lines it, 77, fi, p. It easily follows that this subGQ contains all points of (7r,7Tj,p)\PG(4n — l,q) = AG(3n,q). Consequently the qn + 1 lines of the subGQ incident with (00) belong to a common PG(3n - l,q), which clearly is (it,iti,itj). So O is good at it, and consequently, by Theorem 5.1.2, the translation dual S* of <S satisfies Property (G) at the flag (x', V), with x' the base-point of S* and V the line of S* corresponding to the line L of S. So, by Theorem 5.1.5, for s even the GQ S satisfies Property (G) at the flag (x,L). Now let s be odd. Then, by Theorem 5.1.9, all subquadrangles S' are isomorphic to Q(4, s). By Corollary 5.1.4 the GQ <S* is isomorphic to the point-line dual of a flock GQ <S(T). Finally, let s be even and assume that S* is isomorphic to the point-line dual of a flock GQ <S(JF). Then, by Theorem 5.1.11 the GQ S*, and consequently also S, is isomorphic to Q(5, s). •
Chapter 7
Ovoids and Subquadrangles The first part of this chapter elaborates on ovoids of the classical GQ Q(4, q). In particular subtended ovoids and translation ovoids will be considered. The second part of Chapter 7 deals with subGQs of order q of TGQs of order (q,q2). Also, there are sections on subquadrangles of flock GQs, on TGQs of order (q, q2) having at least one subGQ of order q and on EGQs admitting certain subGQs.
7.1 Ovoids of Q(4, q) An ovoid O of the GQ Q(4, q) is a set of q2 + 1 points, no two of which are collinear; equivalently, O has a point in common with each line of Q(4, q). If the GQ Q(4, q) is contained in PG(4, q) as a nonsingular quadric Q, and if the hyperplane PG(3, g) intersects the quadric Q in an elliptic quadric O, then O is an ovoid of Q(4, q); such an ovoid is called classical. For q an even power of 2 no other ovoids of Q(4, q) are known. For q = 2 2/l+1 , with h > 1, one other type of ovoid is known; its projection from the nucleus n of Q onto a hyperplane not containing n, is the Tits ovoid; see Thas [158] (see also Subsection 12.3.2 below). In fact, in Thas [158] it is shown that with any ovoid of Q(4,g) with q even, corresponds, by projection, an ovoid of PG(3, q), and, conversely, with any ovoid of PG(3, q) with q even, corresponds an ovoid of Q(4, q). For q odd, Kantor [82] constructed two types of nonclassical ovoids. Starting from a Kantor-Knuth flock T, the construction in Section 4.2 defines an ovoid O of Q(4, q). By Theorem 4.4.6 these ovoids of Q(4, q), q odd, are characterized by the property that they are the union of q conies which are mutually tangent at a common point. 145
Chapter 7. Ovoids and Subquadrangles
146
Such an ovoid is classical if and only if the corresponding automorphism a of GF(q) is the identity. An ovoid of Q(4, ) arising from a Kantor-Knuth flock will be called a Kantor-Knuth ovoid and will be denoted by K\. If q = 32h+1, with h > 1, then Kantor [82] proves that Q(4, q) has an ovoid which arises from the Ree-Tits ovoid on the nonsingular quadric in PG(6, q); such an ovoid is always nonclassical. An ovoid of Q(4, q) of this type will be called a Kantor-Ree-Tits ovoid and will be denoted by /C2. Further, Penttila and Williams [131], with a slight assist from a computer, discovered a new ovoid of Q(4,35). This ovoid will be called the Penttila-Williams ovoid. Finally we mention that for q odd one other class of ovoids of Q(4, q) is known; see Section 7.2. With each ovoid O of Q(4, q) corresponds a spread of the dual GQ W(g). And with a spread ofW(g) corresponds, by the standard argument, a translation plane n of order q2; see e.g. Section 3.1 of Dembowski [45]. The plane -n is Desarguesian if and only if O is classical. If O is of Tits type, then 7r is the Liineburg plane; see Thas [158]. To an ovoid /Ci corresponds a Knuth semifield plane; see Kantor [82].
7.2
Subquadrangles and Ovoids
Let S' be a subGQ of order (s, t') of the GQ S of order (s, t), with t' < t. If z is a point of S which is not contained in S', then the 1 + st' points of S' collinear with z form an ovoid Oz of S'; see 2.2.1 of Payne and Thas [128] (or §1.1). Such an ovoid Oz of S' is called a subtended ovoid. For example, if we consider the subGQ Q(4, q) of the GQ Q(5, q), then Oz is an elliptic quadric on the quadric Q denning Q(4, q). Consider an egg O = 0(n, 2n, q) and assume that 0(n, 2n, q) is good at its element TT. By Theorem 5.1.9 the TGQ T(0) contains exactly q3n + q2n subGQs of order qn containing the line 7r of T(C), and for q odd each of these subGQs is isomorphic to the classical GQ Q(4, qn). Fix such a subGQ S' ^ Q(4, qn). Then by Lavrauw [93, 95] and Theorem 10.6.3 of K. Thas [206], up to isomorphism the subtended ovoid Oz of S' is independent of the choice of the point z. Further, Lavrauw [93, 95] proves that up to isomorphism the ovoid Oz is independent of the choice of the subGQ S'. Let O be an ovoid of Q(4, q), let x G O and let L be a line of Q(4, q) incident with x. We call O a translation ovoid with respect to the flag (x, L) if there is an automorphism group GXIL of Q(4, q) which fixes O, which fixes x linewise and L pointwise, and which acts transitively on the points
7.3. Translation Ovoids and Semifield Flocks
147
of (O \ {x}) n y1- for each t / / i incident with L. We call O a translation ovoid with respect to the point x if O is a translation ovoid with respect to the flag (x, L), for each line L incident with x. In Bloemen, Thas and Van Maldeghem [14] it is shown that the ovoid O of Q(4, q) is a translation ovoid with respect to the point x e O if and only if Q(4, q) admits an automorphism group of size q2 which fixes O and all lines incident with x. It is easy to see that the ovoids Oz of S1 in the previous paragraph are translation ovoids of S1 with respect to the point (oo). For more on translation ovoids we refer to Bloemen, Thas and Van Maldeghem [14] and Lunardon [100]. Let 0(n, 2n, q) = O be a good egg in PG(4n— 1, q), with q odd. By Corollary 5.1.4 and Theorem 4.7.2 the TGQ T(C*), with O* the translation dual of 0, is the point-line dual of a flock GQ Stf), with T a semifield flock. Conversely, if T is a semifield flock, then S{F) is the point-line dual of a TGQ T(£>*), where O is good (see Theorem 4.7.2). So with any semifield flock corresponds an ovoid of Q(4, s), which is a translation ovoid with respect to some point. Also, with any translation ovoid with respect to some point of Q(4, s) corresponds a semifield flock. For details we refer to Lunardon [100]. In the next section we will show how to construct, in a purely geometrically way, the semifield flock from the translation ovoid, and conversely. In Thas [176] it was conjectured that the translation ovoids of Q(4,3 h ), h > 2, arising from the Ganley flocks (See Section 4.5) are new. This conjecture was proved by Thas and Payne [190], and the new ovoids are called Thas-Payne ovoids. Together with the ovoids of Q(4, q) mentioned in Section 7.1, these are the only known ovoids of the classical GQ Q(4, q).
7.3 Translation Ovoids and Semifield Flocks The following construction is taken from Thas [181], see also Lunardon [100]. The plane PG(2,qn), with q = ph and p prime, can be represented in PG(3n - l,q) in such a way that points of the plane become (n - 1)dimensional subspaces belonging to an (n - l)-spread T of PG(3n - 1, q) and lines of the plane become (2n - 1)-dimensional subspaces of PG(3n 1, q); see also Sections 3.6 and 4.5. In each of these (2n - 1)-dimensional subspaces corresponding to lines an (n - 1)-spread is induced by T. Let C be a nonsingular conic of PG(2, qn). With C corresponds a set C of qn + 1
148
Chapter 7. Ovoids and Subquadrangles
(n-l)-dimensional subspaces of PG(3n—1, q). This s e t C is a pseudo-conic ofPG(3rc-l,g). Let 7r be a (2n — 1)-dimensional subspace of P G ( 3 n - 1, q) having no point in common with the elements of C. Now we consider the GQ T 2 (C) = Q(4,9 n ) of Tits defined by C. Then T 2 (C) ^ T(C'), where C corresponds to C and the points of Type (i) of T 2 (C), which are the points not collinear with the point (oo) of T 2 (C), correspond to the points of PG(3n, q) \ PG(3n -l,q) where PG(3n, q) is the space containing T(C'); see Sections 3.6 and 4.5. Let n be a 2n-dimensional subspace of PG(3n, q) containing re, but not contained in PG(3n -l,q). Then (n \ PG(3n - 1, q)) U {(oo)} is an ovoid of the GQ T(C'), so defines an ovoid of T 2 (C), and hence defines an ovoid of Q(4, qn). Also, this ovoid of Q(4, qn) is a translation ovoid with respect to the point which corresponds to the point (oo) of T(C) (respectively, T 2 (C)). By Lunardon [100] any ovoid of Q(4, qn) which is translation with respect to some point can be obtained in this way. Now we dualize in PG(3n - 1, q); choose e.g. a polarity in P G ( 3 n - 1, q) and consider the polar spaces of the subspaces of PG(3n—1, q). With C corresponds a set C' of qn +1 (2n-1)-dimensional subspaces of PG(3n -l,q); C' can be considered as a dual conic C in a PG(2, qn). With n corresponds an (n — 1)-dimensional subspace n of PG(3n - l,q) which is skew to all elements of C'. By Section 4.5 this subspace TT defines a semifield flock, and, conversely, any semifield flock can be obtained in this way. Hence any ovoid O of Q(4, s) which is a translation ovoid with respect to some point, defines a semifield flock To, and, conversely, any semifield flock T of the quadratic cone K in P G ( 3 , s) defines an ovoid 0? of Q(4, s) which is a translation ovoid with respect to some point. Let J 7 be a semifield flock. Then the point-line dual of the GQ S(T) is a TGQ T(£>*) of order (s,s2), where the egg O is good. By Section 7.2, O defines a subtended ovoid Oz of the subquadrangle S' = Q(4, s). By Lavrauw [93, 95] this ovoid Oz is isomorphic to the translation ovoid 0? of Q(4, s), which corresponds to the flock T.
7A
Coordinates of the Known Nonclassical Ovoids of Q(4,<7)
Suppose that Q(4, q) has equation X\ = X0Xi the known ovoids of Q(4, q) in coordinates.
+ X3X4. We now describe
7.4. Coordinates of the Known Nonclassical Ovoids o/Q(4, q)
149
First suppose that q is even. [219]. Define 7 : i i-> tr+1, with t G GF(q) and g = 2 2 e + 1 , e > 1. Then the Tits ovoids Tl are described as follows (see also §12.3.2):
TITS OVOIDS
r i = {(0,0,0,1,0)} U {(x, y,x^2+l
+ y^2, xy + x^+ 2 + y\ 1) || x, y G GF(g)}.
Next suppose that q is odd. KANTOR OVOIDS /CI [82].
Let n be a nonsquare of GF(q)
and let a ^ 1
be an arbitrary automorphism of GF(g). Then the Kantor ovoids K\ are given by / d = {(0, 0,0,1,0)} U {nx°,x, y, y2 - nx°+1,1)
|| x, y G GF(g)}.
/C2 [82]. Define 4> : x ^ x3h, with x G GF(g) and q = 3 2 / l _ 1 , h > 2. Then /C2 is described as follows: KANTOR OVOIDS
/C2 = {(0, 0 , 0 , 1 , 0 ) } U { ( x ^ + 3 + / , x, y, y2-x2^-xy\ THAS-PAYNE OVOIDS T7 7 [ 1 9 0 ] .
1) || x, j, G GFfa)}.
Put q = 3h, h > 3, and let n be a non-
square of GF(q). Then T7? is given by TV = {(0,0,0,1,0)} U {{xn + (xn" 1 ) 1 7 9 + y 1 / 3 , x, y, y 2 - xixn + (xn" 1 ) 1 7 9 + t/ 1 / 3 ), 1) G Q(4, q) || x, y G GF(g)}. If ft is odd, - 1 is a nonsquare, and then the ovoid can be written as TV = {(0,0,0,1,0)} U { ( - x - x 1 / 9 + 2/1/3, x, y, y2 + x2 + x 1 0 / 9 - xy 1 / 3 ,1) GQ(4,g)||x,yGGF(g)}. THE PENTTILA-WILLIAMS OVOID
VW [131].
The Penttila-Williams ovoid
PWofQ(4,35)isgivenby P W = {(0,0,0, l,0)}U{(x 9 + y 8 \ x , y , y 2 - x 1 0 - x y 8 1 , l ) || x,y G GF(3 5 )}.
Chapter 7. Ovoids and Subquadrangles
150
7.5
Subquadrangles of T(£>), with O Good: the Even Case
First we discuss briefly the connection between GQs with a regular point and covers of the net associated with the regular point. In particular, we shall consider subGQs of order s of GQs of order (s, s2) in this context. Let TV = (V, B, I) be a net of order k and degree r. An m-fold cover of A/" is a geometry M = (V,B,I) such that there exists a surjective map
satisfying: (i) for any x e V, x1
consists of m pairwise noncollinear points;
(ii) for any pair {x, y} of collinear points of 77, {x, y}1 disjoint pairs of collinear points;
consists of m
(iii) for any pair {x,y} of noncollinear points of N, {^,y}7 pairwise noncollinear points of Af; (iv) for any line L of M, if VL = {% € V || xlL), then VI points of m disjoint lines of A/".
is a set of consists of the
The map 7 is called a covering map. A point or line of M is said to be a cover of its image under 7. Now let 5 = (V, B, I) be a GQ of order (s, t) with a regular point x, and let Afx be the associated net of order s and degree t + l. The geometry 77^ with • POINT SET V
\x^,
• LINE SET B \ {L e B |j xlL}
and
• INCIDENCE inherited from S, is a i-fold cover of A/^. The covering map 7 maps the point y e V \ x1to the point {x, y}1- of Afx. The cover has the additional property that for a nonincident point-line pair {y, L} of J7^, either y1 = z1 with zlL and z £ {x,y}~LJ~, or 2/is collinear with a unique point of 772 incident with L. In general, given a net of order s and degree t + l with such a i-fold cover,
7.5. Subquadrangles of T(0), with O Good: the Even Case
151
it is possible to construct a GQ of order (s, t), which in the above case is a reconstruction of <S. Suppose that Mx has a proper subnet J\fx of order s', with s' > 2, and degree t + 1. Then the subset V' of V consisting of x, the lines of J\f'x and j e P \ i i such that {x, y}1- is a point of M'x, is the point set of a subGQ 5 ' of order (s', t) of 5 . The lines of S' are the lines of S incident with at least one (and hence s' + 1) points of V. The point x is a regular point of S1 with associated net A^. It follows that t = s',t2 = s and so the subnet Mx is an affine plane of order t. For more details see K. Thas [200]. Let S be a GQ of order (q2, q) with a regular point x such that the associated dual net Mx is isomorphic to H^. In Theorems 5.1.2, 6.2.1, 6.3.3 and 6.3.4 examples were given. In the next results, due to Brown and Thas [28], subGQs of such GQs will be considered. Theorem 7.5.1 Let S = (P,B,I) be a GQ of order (q2,q), 9 / 1, with a regular point x such that the associated dual net N® is isomorphic to H3. Then x is contained in the maximal number q3 + q2 subGQs of order q. If S' = (V, B', I') is any subGQ of order q not containing x, then S' is isomorphic to the point-line dual ofT2(0)forsome oval O o / P G ( 2 , q). Proof. Suppose that the net (H^) D = J\fx is constructed from the line L of P G ( 3 , q); that is, by taking as POINTS the lines of P G ( 3 , q) not meeting L, as LINES the points of P G ( 3 , q) \L, and as INCIDENCE the incidence from P G ( 3 , q). Then each plane n of P G ( 3 , q) not containing L gives rise to an affine plane subnet of (H^) D , which is the dual of IT, with the point TT n L and the lines of n on n n L removed. From the foregoing we see that TT gives rise to a subGQ of order q. This gives q3 + q2 distinct subGQs of order q containing x, the maximal number by Lemma 5.1.8. Now suppose that S' = (V, B\ I') is a subGQ of order q of S, not containing x. The set 'P\a;-L is the point set of a g-fold cover of the net J\fx = (H^) with covering map 7 taking the point y e V \ x1- to the point {x^}1- of A4 = (H^) . The subGQ S' contains a unique line, say M, incident with x; see Section 2.2 of Payne and Thas [128]. The points of S incident with M, but distinct from x, define a parallel class of (H^) the elements of which are contained in a plane of P G ( 3 , q) containing L which we will denote by M 7 . The subGQ S' contains q + 1 points of M. These points define a set O of size q + 1 in the plane M 7 , and O n L = 0. Consider a line TV of S' not concurrent with M. If PN = {z e V \\ zlN}, then Vjj defines a point of P G ( 3 , q) not in the plane M 7 . Further, since no two lines of S' are incident
152
Chapter 7. Ovoids and Subquadrangles
with a common point of xL not incident with M, it follows that 7 gives a one-to-one correspondence between the q3 lines of S' not concurrent with M and the q3 points of PG(3, q) \ M 7 . Each point of S' not incident with M is collinear with a unique point of M and so, under the map 7, defines a line of PG(3, q) meeting M 7 in a unique point of O. Since no two lines of S' are concurrent in a point of x1- not incident with M, it must also be the case that no two points of S', not incident with M, correspond under 7 to the same line of PG(3, q). Thus 7 gives a one-to-one correspondence between the set V \VM, with VM = {z € V \\ zlM}, and the lines of PG(3, q) not in M 7 meeting M 7 in a point of O. It is now a straightforward exercise to verify that O is an oval and that S' is isomorphic to the point-line dual of T 2 (£>). • We have an immediate corollary. Corollary 7.5.2 Let S = (V,B, I) be a GQ of order (q2,q), q odd, with a regular point x such that the associated dual net Af® is isomorphic to H3. If S' = (V, B', I') is any subGQ of order q not containing x, then S' is isomorphic to the classical GQ W(g). Proof. This follows from Theorem 7.5.1, Theorem 1.5.2(f), the remark which follows Theorem 1.5.2, and Theorem 1.4.1. • Suppose that S = (V, B, I) is a TGQ of order (s, s2), s ^ 1, with S = T(0), where O is good at its element -K. By Theorem 6.2.1 the dual net N® is isomorphic to Hf. Suppose that H^ is constructed from the line L in PG(3, s). We identify H^ and J\f®, and assume that 7 is the covering map from V \ x1- to the point set of (H^) D . Let G be the translation group of S about (00). Then G has order s4, fixes each line incident with (00) and acts regularly on the elements of V \ ar1. The group G induces an automorphism group G' of the dual net A/jf = H^. The identity mapping of J\f® is induced by all symmetries of S about n, and so G' has size s3. The group G' is induced by a collineation group of PG(3, s) fixing the line L; this collineation group will also be denoted by G'. If we denote the plane of PG(3, s) corresponding to the elements of O \ {71-} by (oo)7, then each element of G' fixes all points of (00)7, so is an axial collineation with axis (oo)7. If 9 e G fixes a line of ir1- not incident with (00), then it follows that 9 is a symmetry about n. Hence G' acts regularly on the points of PG(3, s) \ (oo)7 and must be the group of elations of PG(3, s) with axis (oo)7.
7.5. Subquadrangles ofT(C), with O Good: the Even Case
153
By Theorem 5.1.9 <S = T(0) contains exactly s 3 + s2 subGQs of order s containing the line n. Now suppose that S' = (V',B', I') is a subGQ of order s of S not containing the point (oo); then n cannot be a line of S'. Consequently, by applying Theorem 7.5.1 it follows that S' = T2(0'), for some oval O' of PG(2, s), where PG(2, s) corresponds to the point m of S' which is incident with w; this plane PG(2, s) will be denoted by vrC. Lemma 7.5.3 Let S = (V, B, I) be a TGQ of order (s, s2), s ^ 1, with S = T(0), where O is good at its element TT. Suppose that S' — (V',B', I') is a subGQ of order s ofS not containing the point (oo). Then S' is isomorphic to the classical GQ Q(4, s). Proof. By the foregoing we have S' = T 2 (£>')• N o w w e show that O' is a conic in m 7 , with m the unique point of S' incident with the line w. Let y be a point of (oo)7 not on the line L and let
154
Chapter 7. Ovoids and Subquadrangles
Theorem 7.5.4 Let S = (V, B, I) be a GQ of order (s, s2), s even, having a subGQ S' isomorphic to Q(4, s). If in S' each ovoid Ox consisting of all the points collinear with a given point xofS\S'isan elliptic quadric, then S is isomorphic to Q(5, s). • Remark 7.5.5 The equivalent theorem was proved in the odd case by Brown [22] and a proof valid for both s odd and even may be found in Brouns, Thas and Van Maldeghem [19] and in Brown [24]. We are now equipped to prove the classification theorem. Theorem 7.5.6 Let S = (V,B,I) be a TGQ of order {s,s2), s even, with S = T(O), where O is good at the element IT. Suppose that S' = (V, B', I') is a subGQ of order s ofS not containing the point (oo). Then S is isomorphic to the classical GQ Q(5, s). Proof. By Lemma 7.5.3, if m is the point incident with the line IT and contained in S', then S' = T 2 (C) where C is a conic in the plane rrC of PG(3,s). Let y be a point oiT)\V' such that y ^ m, and let Oy be the ovoid of S' subtended by y. We now show that Oy corresponds to an elliptic quadric of Q(4, s) ^ T 2 (C). Let tp be a nontrivial elation of P G ( 3 , s) with axis (oo) 7 and center on the line L of P G ( 3 , s). Let Tp be an automorphism of <S that induces ip in P G ( 3 , s). By Sections 2 ^ and 2.3 of Payne and Thas [128] there are three possibilities for S' n S'v: (i) a point u, the s + 1 lines of S' and S^ incident with u and the points incident with these lines; (ii) an ovoid of S' and S'v, and (iii) a grid. Hence the subGQs S' and S^ can share either 0,1,2 or s + 1 lines incident with the point m. Consequently, C and Cv may share 0,1,2 or s + 1 points. However, ip cannot fix C so the possible intersection numbers are 0,1 and 2. As any nontrivial elation of P G ( 3 , s) has order 2, we may choose
7.5. Subquadrangles ofT(O), with O Good: the Even Case
155
the subGQs S' and S'v share exactly two lines incident with m, and so S' n S'^ is a grid. Under the covering map 7, with the lines of the grid correspond z, z' and the 2s lines on z or z', not in m 7 , in a fixed plane 5 of PG(3, s). Since the point y of <S is not contained in 5', but is contained in S'v, it follows that y1 fi 5. Each of the s + 1 lines of <S'^ incident with y is incident with a unique point of 5 ' n S^. Looking in PG(3, s) at the T2(C) and T2(CV) models for 5' and <S'^, respectively, we see that this set V of s + 1 points must contain the s - 1 points of the projection of Cv \ {z, z'} from y7 onto 5, that is, s - 1 points of a conic (completed to a conic by adding z and z')- The remaining two points of V are the planes spanned by y1 and the tangent lines of C at z and z'\ these tangent lines are also the tangent lines of Cv at z and z'. Under the isomorphism from T2(C) to Q(4, s) the images of these s + 1 points of T2(C) are the s + 1 points of a conic, so under the isomorphism from Q(4, s) to W(s) the corresponding s + 1 points of W(s) are also the s + 1 points of a conic C; see Brown [23] for an explicit isomorphism from T2(C) to W(s). To Oy corresponds an ovoid 0'y ofW(s). By Thas [158] 0'y is also an ovoid of the projective space PG(3, s) which contains W(s). As Oy contains a conic, by Brown [23], Oy is an elliptic quadric of PG(3,s). Hence to Oy corresponds an elliptic quadric on Q(4, s). Now suppose that y e V \T" such that y ~ m with y ^ m. Again let Oy be the ovoid of S' subtended by y. Let TV be any line of S incident with y but not with m. Then N meets S' in a unique point, n say. Each of the s - 1 points incident with N, but distinct from y and n, subtends an ovoid of S' which corresponds to an elliptic quadric on Q(4, s). Since the s points incident with N, but distinct from n, subtend s ovoids that partition the points of S' not collinear with n, it follows that the ovoid Oy is determined by the s - 1 ovoids subtended by the points incident with N, but distinct from y and n. Looking in the Q(4, s) or T2(C) model of S', we see that with Oy corresponds an elliptic quadric of Q(4, s). Since each point oiV\T" subtends an ovoid in S', which defines an elliptic quadric in Q(4, s), it follows from Theorem 7.5.4 that S = Q(5, s). • Note that a similar proof may be possible in the case when s is odd. However, given that the characterizations of an elliptic quadric of Q(4, s) (as ovoid of the latter) are more involved for s odd than in the case where s is even, any such proof will be more complicated and probably much longer than that given in Theorem 7.6.2. Now we will consider subGQs of order s of TGQs of order (s, s2), for any
156
Chapter 7. Ovoids and Subquadrangles
s ^ 1, where S' does contain the point (oo). The following theorem is folklore. Theorem 7.5.7 Let S = {V,B,I) be a TGQ of order (s,s2), s ^ I, with O = G(n,2n,q), and assume that S' = (P',B',I') is a subGQ of order s containing the point (oo) of S. Then S' = 1(0'), with O' = 0'{n,n,q) a pseudo-oval on O. Proof. Assume S admits a subquadrangle S' = (V, B', I') of order s containing the point (oo). Let L0, Li,..., Ls be the s + 1 lines of <S incident with (oo) and contained in <S', say Li = 71-j with 7r; e O, i = 0 , 1 , . . . , s. If PG(n, q) = M is a line of S' concurrent with L 0 and not containing (oo), then (PG(n,q),iti) \ PG(4n - l,q), with (O) = P G ( 4 n - l,q) and i e { 1 , 2 , . . . , s}, is a 2n-dimensional affine space AG ( -^(2n, q) consisting entirely of points of <S'. In this way arise the s3-s2 points of S' not collinear with (oo) or with L0 n M, together with the s points of M not incident with L0. Remark that the points of A G ^ (2n, q), i E { 1 , 2 , . . . , s}, are the points not collinear with (oo) of a grid
7.6
Subquadrangles of T(0), with O Good: the Odd Case
The theorems of this section are due to Brown and Thas [29]. Theorem 7.6.1 Let S = (V, B, I) he a TGQ of order (s, s2), s odd and s ^ 1, with S = T(O) and O good at some element IT. If S' is a subGQ of order s of S containing the point (oo), then S' = Q(4, s) and either S' is one of the s3+s2 (classical) subGQs of order s containing the line n ofS, orS = Q(5, s).
7.7. Subquadrangles: Remaining Cases and Some Applications
157
Proof. By Theorem 5.1.9 the GQ contains exactly s 3 + s 2 subGQs of order s containing the line -K of S; each of them is isomorphic to the classical GQ Q(4,s). Assume <S admits a subGQ S' = (V, B', I') of order s containing the point (oo) but not containing the line n. If <S = T(O), with O = 0(n,2n,q), then 5 ' = T(O'), with O' = {TT0, v n , . . . , TTJ, TT, e O for » = 0 , 1 , . . . , s, and 7ro, 7ri,..., 7rs contained in a common P G ( 3 n - 1, q); see the proof of Theorem 7.5.7. Now by Theorem 5.2.3 and the uniqueness of the GQ of order (3,9) the egg O is classical, and so <S = Q(5, s). • Theorem 7.6.2 Let S = (V, B, I) be a TGQ of order (s, s2), s odd and s=£l, with S = T(O) and O good at some element z.IfS'isa subGQ of order s of S not containing the point (oo), then S = Q(5, s). Proof. By Corollary 5.1.4, T ( 0 ) is the translation dual of the point-line dual of a flock GQ. If -K is the good element of O, then, by Section 3 of K. Thas [205] (see also K. Thas [206] or Chapter 10) every point of the line 7r is a translation point of the TGQ S. One of these translation points, say m, belongs to S'. As there is an automorphism of <S mapping m onto (oo), there is a subGQ S" of order s which contains the point (oo). Hence, by Theorem 7.6.1, S ^ Q(5, s). •
7.7
Subquadrangles: Remaining Cases and Some Applications
Each known GQ of order (s, s 2 ), s ^ 1, is either the point-line dual of a flock GQ, or a TGQ T(O), with either O or O* good. In this section we will determine the subquadrangles of order s of flock GQs and of TGQs T ( 0 ) , with O* good. As most of the proofs are quite long, just a few of them will actually be given. Finally, we will mention some interesting characterization theorems, which are related to the determination of these subquadrangles. The next theorem is due to O'Keefe and Penttila [92], but we give here another and very short proof. Theorem 7.7.1 If S(F) is a flock GQ of order (q2,q), q even, with center (oo), then S(J7) contains exactly q3 + q2 subGQs of order q containing the point (oo).
158
Chapter 7. Ovoids and Subquadrangles
Proof. Payne [118] proved that | { L , M , i V } ± x | = q + 1 for all triads {L, M, N}, with (oo)lL, for which there is a line U such that (oo)lU and {L,M,N} c UL; see also Theorem 4.10.1. That is, the set {L,M,N} is 3-regular. Now let {L, M, N} be such a triad. If B' is the set of lines of S {J7) incident with points of the form X nY, with X e {L, M, N}1- and Y e {L, M, N}-11-, and "P' is the set of points which are incident with at least two distinct lines of £>', then endowed with the induced incidence I', S' = (V',B', I') is a subGQ of order q of S{F), see Theorem 2.6.2. Now a standard counting argument yields q3 + q2 of such subGQs. By Lemma 5.1.8, 5(J - ) contains exactly q3 + q2 subGQs of order q containing the point (oo). • Now we mention, without proof, a strong characterization theorem for Q(5, s) on which we will rely in the proof of Theorem 7.7.3. Theorem 7.7.2 and 7.7.3 are due to Brown and Thas [28]. Theorem 7.7.2 Let S = {V, B, I) he a GQ of order (s, s2), s + 1, and assume one of the following conditions is satisfied: (i) s is odd and S satisfies Property (G) at the pair {x, y} with i ~ t / / i and L = xy, or (ii) s is even, S is the point-line dual of a flock GQ and L is the corresponding base-line of S. If at least one triad \z\,zi,z), with zlL and where {zi,z2,z}1contain a point incident with L, is 3-regular, then S is classical.
does not •
Theorem 7.7.3 Let Sbea nonclassical flock GQ of order (q2,q), q even, with base-point (oo). Then any subGQ of S of order q contains (oo). Proof. Let SD be the point-line dual of a nonclassical flock GQ of order (q2,q), q even. The base-line of SD will be denoted by L. Suppose, by way of contradiction, that S' is a subGQ of order q of SD which does not contain L. Let x be the point of <S' incident with L an let ylL, with x ^ y. Further, let Oy be the ovoid of S' consisting of the q2 + 1 points of S' collinear with y. Also, let z e Oy\ {x}, let M be a line of S' incident with x, let N be a line of S' incident with z, and let M / N. Let Mlu ~ z and Nlw ~ x. Then by Theorem 4.10.1 and Theorem 2.6.2 there is a subGQ S" of order q of SD containing y, u, w, x, z. From Lemma 1.1.1 easily follows that <S' n <S" is
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a grid containing the lines M, N. Hence the pair {M, N} is regular for any line AT of S' not concurrent with M. Consequently M is a regular line of <S', so in S' any line incident with x is regular. Now by Theorem 2.1.5 the point x is regular in S'. If r is a point of S' not collinear with x, then {x, r}- 1 ' and {x, r}-1-'1-' contain q + 1 points of S'. Let {x, r}1-' = A and {x, r}1-'x' = A. If r' e A\{x, r}, then {x, r, r'} is 3-regular in 5 and {x, r, r'}-1 = A does not contain a point incident with L. By Theorem 7.7.2, the GQ SD is classical, a contradiction. • Theorem 7.7.4 Let S fee a flock GQ of order (q2, q), q odd, with base-point (oo). If S' is a subGQ of order q of S containing the point (oo), then S is a Kantor-Knuth semifield flock GQ. Hence S' ^ W(g) and either S ^ H(3, q2) or S' is one of the q3 + q2 subGQs of order q containing the point (oo). • Remark 7.7.5 The proof of this theorem is quite long; the last part of the statement follows from Theorem 7.6.1. Corollary 7.7.6 Let S be a flock GQ of order (q2,q), q odd, with base-point (oo). If the net M^oo) defined by the regular point (oo) of S contains at least one proper subnet having the same degree as M^), then S is a Kantor-Knuth semifield flock GQ. Proof. In K. Thas [200] it is shown that the subnet is an affine plane of order q. From the proof of Theorem 6.5.1 then follows that <S has a subGQ of order q containing the point (oo), and so by Theorem 7.7.4 we are done.
• Theorem 7.7.7 Let S be a nonclassical flock GQ of order (q2, q), q odd, with base-point (oo). Then any subGQ ofS of order q contains (oo). • Remark 7.7.8 The proof of Theorem 1.1.1 is long and technical. Central in this proof is the construction of Knarr; see Section 4.9. As a byproduct of the proof of Theorem 1.1.1 we obtain the counterpart of Corollary 6.6.8; no proof will be given. Theorem 7.7.9 If S = (V,B,I) is a flock GQ of order (q2,q), with q odd, then the point-line dual of S is a TGQ if and only if S has a regular point x, with (oo) ^ x ~ (oo), where (oo) is the elation point of S. •
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Next we will consider subGQs of TGQs T(0*) of order (s, s2), where the translation dual O* of O is good. If s is even, then the egg 0(n, 2n, q) is good if and only if 0*(n, 2n, q) is good; see Corollary 5.1.6. If 0*(n,2n,q), q odd, is good, then, by Corollary 5.1.4 the TGQ T(C) is isomorphic to the point-line dual of a flock GQ. So from Theorems 7.5.6, 7.5.7, 7.7.4 and 7.7.7 we have the following result. Theorem 7.7.10 Let S = T(0), with O = 0{n, 2n, q), be a TGQ such that the translation dual 0*ofOis good at the tangent space TofOatit&O, and assume that S contains a subGQ S' of order qn = s. If q is even and S' does not contain the translation point (oo), then S = Q(5, s); if q is even and S' contains the translation point (oo), then S' = T(0'), with O' = 0'(n,n,q) a pseudo-oval on O. If qis odd and S' does not contain the translation point (oo), then S = Q(5, s); if q is odd and S' contains the translation point (oo), then S is a Kantor-Knuth semifield flock GQ, S' = Q(4, s) and either S = Q(5, s) or S' is one of the s3 + s2 subGQs of order s containing the point (oo). •
7.8 Translation Generalized Quadrangles with One Classical Subquadrangle Recall Brown's result on ovoids in PG(3, q) and its equivalent statement in the theory of TGQs. Theorem 7.8.1 (Brown [23]) (i) If an ovoid o/PG(3,?), q even, has a conic section, the ovoid must be an elliptic quadric. (ii) If a T 3 (C) of order (q, q2), where Oisan ovoid o/PG(3, q), q even, has a classical subGQ of order q containing the point (oo), then T 3 (O) = Q(5, q), that is, T 3 (C) arises from a nonsingular elliptic quadric in PG(5,g). Proof. See Brown [23].
•
We note that the only classical GQs of order q are the GQ Q(4,q), which arises from a nonsingular quadric in PG(4, q), and the symplectic quadrangle W(q). However, if q is even, then by Theorem 1.4.1, Q(4, q) = W(g).
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Brown and Lavrauw [26] generalized Theorem 7.8.1 by obtaining a similar result for generalized ovoids. We first prove a lemma. Lemma 7.8.2 Every (2n~l)-space a in PG(3n— 1, q), q even, which is skew to all elements of a pseudo-conic, is the span of two elements of the regular (n — l)-spread induced in a by the pseudo-conic. Proof. Let a be a (2ra — l)-space skew to all elements of the pseudoconic O in w = P G ( 3 n - l,q). After dualizing, we obtain an (n - 1)space a* that is disjoint from all elements of a dual pseudo-conic O*. Embed 7r in a 3n-dimensional space PG(3n, q), and embed a* in an n-space P G ( n , q) C PG(3n, q), where P G ( n , q)
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(xnPG(3n, g ) ) \ P G ( 3 n - 1 , q), with P G ( 3 n - 1 , q) the space containing O'. The (2n-l)-dimensional space x n P G ( 3 n - l , g ) = P G ( 2 n - l , q ) is skew to the pseudo-conic O', so applying Lemma 7.8.2, we have that it is the span of two elements of the regular (n - 1)-spread of P G ( 2 n -l,q) induced by the pseudo-conic. Now represent S' over QF(qn), so as a T 2 (C) = Q(4, qn) with C a conic of P G ( 2 , q n ) . Then the ovoids of T 2 (C) consisting of the point (oo) of T 2 (C) together with the affine points of a plane skew to C correspond to the elliptic quadrics of Q(4, qn) containing a fixed point. As a subtended ovoid can be subtended by at most two distinct points (by Section 2.5), and as there are q2n(qn - l ) / 2 elliptic quadrics on Q(4,g n ) containing a given point, it follows that each of the q2n(qn - 1) points of S \ S' of Type (ii) subtends a doubly subtended classical ovoid. Now let y be a point of <S \ S' not collinear with (oo), and let Oy be the corresponding subtended ovoid of S'. We now work in the T 2 (C) model of S'; assume T 2 (C) is contained in P G ( 3 , qn). As y ^ (oo), we have that Oy = A U {irp || p £ C}, where A is a set of q2n - qn points of Type (i) of T 2 (C) and ITP is a point of Type (ii) of T 2 (C) (a plane containing p e C). If a plane •K of P G ( 3 , qn) contains no point of C, then ( T T \ P G ( 2 , qn)) U {(oo)} is a classical ovoid subtended by two points, x and x' of S \ S', by the first part of the proof. If y is collinear with x or x', then ir n A is a single point. If y is collinear with neither x nor x', then {x, x', y} is a triad of S. Hence |7rn„4| = qn + l, as this is the number of centers of {x, x', y}. Next, suppose that 7T contains a unique point p of C. If -K = ixp C Oy, then ir contains no point of A. If -K ^ irp, then the qn lines of -K incident with p and not in the plane of C are lines of T 2 (C), and so contain precisely one point of A. Whence J7r n A\ = qn. Next, suppose that n contains two distinct points, p and p', of C. Of the qn + 1 projective lines in n incident with p, one is contained in PG(2, qn), one in ixp and qn - 1 are lines of T 2 (C) containing a unique point of A Hence J7rn^| = qn - 1. Finally, if 7r = PG(2, g"), then •K contains no point of A. Consider the set of points of PG(3,g™) defined byOy = AllC. Bythe above, the plane intersections with Oy have size 1 or qn + 1, and a straightforward count shows that Oy is an ovoid of P G ( 3 , qn). Further, since Oy contains C, it is an elliptic quadric by Theorem 7.8.1. Hence Oy is a classical ovoid of S' = Q(4, qn). So every point of S outside S' subtends an elliptic quadric in S'. By Theorem 7.5.4 the theorem now follows. • Corollary 7.8.4 An egg in P G ( 4 n - l,q), q even, contains a pseudo-conic if and only if it is classical. •
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Further on in this chapter, we will improve these results by showing that an elation generalized quadrangle of order (q,q2), q even, with a subGQ Q(4,g) containing the elation point, arises from a nonsingular elliptic quadric in PG(5,q). This is taken from K. Thas [208]. The theorem is a corollary of a structural result (cf. Proposition 7.9.2 and Corollary 7.9.3) independent of the characteristic. Another corollary generalizes also a result of Brown and Lavrauw [27].
7.9 Elation Generalized Quadrangles with a Subquadrangle Recall that a center of transitivity a; of a GQ S is a point for which there is an automorphism group of S fixing x linewise and acting transitively on the points noncollinear with x. The next lemma is well-known (it can, for instance, be found in [197]). Lemma 7.9.1 Let S be a GQ of order (s, s2), s > 1, and suppose that S' is a subGQ of order s. Furthermore, assume that x e S' is a center of transitivity of S. Then x is also a center of transitivity of S', and there are s distinct subGQs of order s which all contain the same lines through x, and of which the sets of points not collinear with x partition the set of points of S not collinear with x. Proof. Let y, y', y ^ y', in S' both be not collinear with x. Let 6 be a whorl of S about x mapping y onto y'. If S'e ^ S', then S'C\S'e must be a subGQ of order (s, 1) (by Lemma 1.1.1), a contradiction. Whence S' n S'e = S' and it follows that a; is a center of transitivity for S'. The lemma now easily follows. • Proposition 7.9.2 (i) Let S^ be an EGQ of order (q, q2), q > 1, containing a subGQ S' of order q which has an axis of symmetry L incident with x. Then L is an axis of symmetry in S^x\ (ii) Let S^ be a GQ of order (q,q2), q > 1, with center of transitivity x, containing a subGQ S' of order q which has an axis of symmetry L incident with x. Then L is an axis of symmetry in S^. Proof, (i) Let S' be as in the statement of the proposition, and suppose G is the group of elations of S about x. Then there are q elements in the
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orbit S'G by Lemma 7.9.1, and by that same lemma they precisely intersect in the q + 1 lines of <S' which are incident with x, together with the points on these lines. Whence the elements of S' partition the lines of S not incident with x and concurrent with one of the q + 1 lines on x in S', and also the points of S not collinear with x. Let L be as in the statement of the proposition, and suppose M ~ L ^ M is not incident with x. Then there is precisely one element, denoted S'M, in S'G which contains M, and no other element of S'G contains a point of M\{LnM}. Take an arbitrary element 8 of GM- Then 8 fixes S'M. By Theorem 3.2.1, 8 induces a symmetry about L in S'M, so that 8 fixes all lines of S'M through each point of L. We obtain the following crucial property for G: (S) For each point ylL, y j^ x, each element of G fixes a constant number c = aq, a G N, of lines incident with y and different from L. We emphasize the fact that this constant is the same for each point on L different from x. Suppose that 8 ^ 1 is as above (8 fixes M ) . Suppose that 8 fixes some point z ~ x which is not on L. Then Theorem 3.2.1 implies that the fixed elements structure of 8 is a thick subGQ of order (q,q2), and so 8 = 1, contradiction. So 8 fixes / = q + 1 points, namely the points of L. Suppose that z is a point not collinear with x for which z9 ~ z ^ z6; then zze ~ L (otherwise z, z e ,proj L z are the points of a triangle). It follows that the number of points of S which are mapped by 8 onto a collinear point different from itself, is g — q3 + q2c. Applying Theorem 3.3.1, we obtain (g2 + l){q + l) + q3 + q3a = q3 + 1 mod q2 + q. Hence a + 1 = 0 mod q+ 1, so that c = aq > q2, hence c = q2. It follows that 8 is a symmetry about L of <S. (ii) By Bose and Shrikhande [18], see also Section 2.5, we have that for each z • x, {z,x}±± = {z,x}. Then by Theorem 3.2.2(4), there is a normal subgroup G'
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(ii) Let <S(x) be a GQ of order {q,q2), q > 1, with center of transitivity x, containing a subGQ S' of order q which has at least one axis of symmetry L incident with x. Then S^ is a TGQ for the translation point x. Proof. By the proof of Proposition 7.9.2, we only have to obtain (i). Suppose that <S(Z) is as assumed in (i). By Lemma 7.9.1, one derives that <S' is an EGQ with elation point x (the stabilizer of S' in the elation group of S^ is an elation group of S'). Say this elation group of S' is G, and suppose Llx is an axis of symmetry in S'. Then clearly the full group of symmetries (of S') about L is a (normal) subgroup of G. By Theorem 3.3.18, it follows that S' is a TGQ with translation point x, so that each line incident with x in S' is an axis of symmetry of S'. By Proposition 7.9.2, each of these lines is also an axis of symmetry of S. Now the theorem follows from Theorem 3.3.17. • Remark 7.9.4 (a) Let <S(x) be as in (i) or (ii) of Proposition 7.9.2, and suppose there is a subGQ S' of order q which has a regular line L incident with x. Then by Theorem 3.3.13, L is an axis of symmetry of S'. So by Proposition 7.9.2, L is an axis of symmetry of S^. Whence Corollary 7.9.3 can also be restated in this way (and the same holds for some of the following results). (b) Let S^ be an EGQ of order (s, s2), s > 1, containing the s distinct subGQs Si, S2, • •., Ss of order s which share a line L incident with some point x of S, and which also share the same s +1 lines incident with x. Suppose L is regular in all these subGQs. Suppose that there is some line M ~ L ^ M which is not incident with x so that there is a group of automorphisms H of S which fixes x linewise, and acts transitively on the points of M different from L n M . Then there is an i e {1,2,..., s} so that M e <%. By Theorem 2.3.10 of [206], L is an axis of symmetry in «S», and a similar argument as in the proof of Proposition 7.9.2 leads to the fact that L is an axis of symmetry in S. Not much is known about EGQs which are also dual EGQs, and which are nonclassical. We will prove some specific results in Section 12.2. In O'Keefe and Penttila [90], some results on certain TGQs which are dual EGQs can be found. In Thas and K. Thas [192], see Chapter 8, a general theory is initiated about TGQs which are also dual TGQs. Also, if S is a thick nonclassical GQ with elation point x and elation line L, then by K. Thas
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and Van Maldeghem [215], or Van Maldeghem [231], one may assume that xlL. (If x IL, it follows easily that each point of <S is an elation point, and each line an elation line, and then [231] — see also Corollary 11.7.3 — implies that the GQ is classical or dual classical.) Here, we obtain a result on flock GQs which are also dual EGQs. Theorem 7.9.5 Let S = <S(.F) be a flock generalized quadrangle of order (q2,q)> 1 even, which is abo a dual EGQ. Then S is classical, i.e., S(T) = H(3,g 2 ). Proof. We look at the situation in the point-line dual 5(^ 7 ) r> of S (T). Then the line [oo] which corresponds to the point (oo) of S{F) is contained in q3 + q2 subGQs of order q; see Theorem 7.7.1. Suppose S' is such a subGQ. We may assume that the line [oo] is incident with an elation point x, so that by Lemma 7.9.1, <S' is an EGQ with elation point x. The line [oo] is an axis of symmetry of S(T)D, cf. Section 4.6, so that it also is an axis of symmetry of S'. Now by Corollary 7.9.3 we have that S is also a TGQ with translation point x. The conclusion follows from Theorem 4.5.1. • The following theorem is a nice corollary of Corollary 7.9.3. Theorem 7.9.6 (i) Let S^ be an EGQ of order (q,q2), q > 1, containing a subGQ S' of order q which is a TGQ with translation point x. Then is a TGQ with translation point x. (ii) Let S^ be a GQ of order (q,q2), q > 1, with center of transitivity x, containing a subGQ S' of order q which is a TGQ with translation point x. Then S^ is a TGQ with translation point x. Proof.
Each line of <S' incident with x is an axis of symmetry of S'.
•
The next theorem generalizes the result of Brown and Lavrauw to EGQs. Theorem 7.9.7 Let S^ be an EGQ of order (q,q2), q even, containing a classical subGQ S' of order q containing x. Then S^ = Q(5, q). Proof. The only classical GQ of order q, q even, is Q(4, q) = W(q), and this is a TGQ for each of its points. By Theorem 7.9.6, S^ is a TGQ. Now apply Theorem 7.8.3. •
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Let O be a generalized oval in PG(3n - l,q) = (O), q even. Then by Theorem 3.9.1 the qn + 1 tangent spaces to the elements of O meet in an (n - 1)-dimensional subspace 77 of P G ( 3 n - 1, q). It is easily seen that for any IT e O, O' = (O \ {ir}) U {77} is again a generalized oval. If O is a generalized conic, that is, if T(£>) = Q(4, qn), then O' is called a pointed generalized conic or a generalized pointed conic. We will not prove the next result in the present monograph. Theorem 7.9.8 Let S ^ = T(O) be a TGQ of order (qn,q2n), q even, containing a subGQ T ( C ) through (00), where O' is a pointed generalized conic in PG(3n - l,q). Then either qn = 4 and S ^ ^ Q(5,4), or qn = 8 and 5(00) ^ T 3 ( C " ) with O" a Suzuki-Tits ovoid o/PG(3,8). Proof.
See Brown and Lavrauw [27].
•
We now generalize to EGQs. Theorem 7.9.9 Let <S(°°) be an EGQ of order (qn,q2n), q even, containing a subGQT(0), where O is a pointed generalized conic in P G ( 3 n - l , q), having (00) as translation point. Then either qn — 4 and <S^°°) = Q(5,4), or qn = 8 and <S(°°) ^ T 3 ( 0 ' ) with O' a Suzuki-Tits ovoid o/PG(3,8). Proof. This theorem follows immediately from Theorem 7.9.6 and Theorem 7.9.8. •
Chapter 8
Translation Generalized Ovals In this chapter, we first introduce new objects called "translation generalized ovals" and "translation generalized ovoids", and make a thorough study of these objects. We then obtain several new characterizations of the T 2 (C) of Tits and the classical generalized quadrangle Q(4, q) in even characteristic, including the complete classification of 2-transitive generalized ovals for the even case. We also prove that translation generalized ovoids do not exist.
8.1 Translation Generalized Ovoids and Translation Generalized Ovals Let 6 be an involution in the projective general linear automorphism group PGLfc + i(q) of the projective space PG(/c, q), with q even. Then the set of all fixed points of 9 is a subspace PG(r, q) of PG(fc, q), called the axis of 8. The fixed hyperplanes for 9 are all hyperplanes containing some subspace PG(fc - r - 1, q) of PG(fc, q), which is called the center of 8. Also, PG(fc r — l,q) Q PG(r,q) and so A: < 2r + 1. For any point x $ PG(r,q) we have xx9 n PG(fc - r - l,q) ^ 0. If TTI and 7r2 are mutually skew (k - r + 1)-dimensional subspaces of PG(fc, q), which are skew to a given PG(r,q) C PG(k,q), then there is exactly one involution 9 of PG(k,q) with axis P G ( r , q) mapping -K\ onto 7r2; the center of 9 is the subspace (7ri,7T2) n P G ( r , q ) . Further, all involutions of PG(k,q) with given axis PG(r, q) form an elementary abelian group. For more details on involutions of finite projective spaces we refer to Chapter 16 of [139]. Notice that in [139] the term "fundamental space" is used 169
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instead of "axis", and the term "fundamental axis" instead of "center". Let O be a generalized ovoid in PG(2n + m - l,q), n ^ m, respectively a generalized oval in PG(3n - 1, q), with q even. Then O is a translation generalized ovoid, respectively a translation generalized oval, with axis the tangent space P G w ( n + m - l , ? ) at P G w ( n - l , g ) e O, if there is a group of involutions of PGL 2 n +m(g) with axis P G ^ ( n + m-l,q), fixing O and acting regularly on O \ { P G ( i ) (n-l,q)}. If n = m = 1, then a translation generalized oval is just called a translation oval; if 2n = m = 2, it is called a translation ovoid. All translation ovals of PG(2, q), q = 2h, were determined by Payne in [107]; up to recoordinatization, they are always of the form { ( l , M 2 i ) || t £ G F ( g ) } U {(0,0,1)}, where i is fixed in { 1 , 2 , . . . , h - 1} and (h, i) = 1. Suppose S = (P,B, I) is a TGQ of order q n with translation point (oo). Let x T^ (oo) be a regular point. Then each point of V \ (oo)-1- is regular. Whence each point of <S is regular, and S = W(g ra ) by Theorem 2.2.2. Theorem 8.1.1 (Structure of Translation Generalized Ovoids and Ovals) (i) A translation generalized ovoid in PG(2n + m — l,q), q even, does not exist. (ii) Let O be a nonclassical generalized oval in P G ( 3 n — 1, q), with q even. Then the following are equivalent. (a) S = T ( 0 ) contains a regular point x different from the translation point (which is necessarily collinear with (oo),); (b) the point-line dual SD of S is a TGQ (with as base-point the line x(oo) ofS); (c) O is a translation generalized oval with axis the tangent space of O at its element x(oo). Proof. Let O be a nonclassical generalized oval in P G ( 3 n - 1 , q). Suppose a; is a regular point different from (oo). Then by the observation preceding the theorem, x is collinear with (oo). Hence each point yla;(oo), y ^ (oo), is regular. By Corollary 3.3.9, also the point (oo) is regular. The translations of T(C) fixing the point x, induce elations, with as axis the set of points
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of x(oo), of the projective plane TTX defined by the regular point x of S. Hence TTX is a translation plane having as translation line the set of points on x(oo). Then by Thas [185], SD is a TGQ where the base-point is the line x(oo) of 5. So S contains a regular point x different from the translation point if and only x ~ (oo) and SD is a TGQ with as base-point the line x(cx>) of <S.
Suppose that SD is a TGQ with as base-point the line x(oo) of S. Then the translation group of SD has a subgroup of size qn acting regularly on the lines, distinct from a;(oo), incident with (oo). Remark that all translations of SD are involutions. Further, let x(oo) = TT and let r be the tangent space of O at TT. We have TT0 = TT, T0 = T and rf = r\, with r? the kernel of O. Clearly the axis of 9 is contained in T, as otherwise elements of O \ {TT} would be fixed by 9. If TT\ e O \ {TT}, then (TTI, TT0) PIT = TT' belongs to the center of 9, so is a subspace of the axis of 6. Let u e TT'. The unique line containing u and intersecting r\ and TT is fixed by 9, and it follows that also TT and T] belong to the axis of 9. Hence T is the axis of 9 and TT' is the center of 9. Hence O is a translation generalized oval with as axis the tangent space of O at its element x(oo). Finally, suppose that O is a translation generalized oval with axis the tangent space T of O at TT, where TT is the line x(oo). Let G(0) be the group of involutions with axis T which acts regularly on O \ {TT}. Embed P G ( 3 n — l,q) = (O) in PG(3n, q). Let 77 be an arbitrary 2n-dimensional subspace of PG(3n, q) which is not contained in PG(3n - 1, q), and which contains r. Then there is a group G' of involutions of PG(3n, q) with axis r\ which induces G(G) in PG(3n - 1, q); the corresponding automorphism group of T(O) will also be denoted by G'. The axis 7 is a point of T(O) and G' fixes 71- pointwise. As \G'\ = qn, the point 7 is a center of symmetry, and so is a regular point by Theorem 3.3.4. Part (ii) of the theorem is proved. Now suppose that O is a translation generalized ovoid in P G ( 2 n + m — 1 , q), so n < m, with axis T (T being the tangent space of O at the element IT). Then by the final part of the proof of Part (ii), each (n + m)-dimensional subspace of PG(2n + m,q) D P G ( 2 n + m — l,q) which is not contained in PG(2n + m — l,q) and which contains r, is a regular point of T(O). Hence s > t, so n > m, contradiction. The theorem is proved. • Remark 8.1.2 (On the Notion of "Translation Generalized Oval/Ovoid") Let O be a generalized ovoid, respectively generalized oval, in PG(2n + m - l,q), and let TT G O. Suppose r is the tangent space of O at TT. Suppose there is a subgroup G of P r L 2 n + m (q) that fixes r pointwise and
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that acts regularly on O \ {K}. We will call this Property (*) for the moment. Then by an argument similar to the one in the final part of the proof of Theorem 8.1.1, each (n + m)-dimensional subspace of PG(2n + m,q) 3 PG(2n + m-l,q) which is not contained in PG(2n + m — l,q) and which contains T, is a regular point of T(O). So n = m. It easily follows that also (oo) is regular, and so, by Theorem 2.1.5, q is even. (Alternatively by 1.5.2(f) of [128] it follows that q is even and so, by Theorem 2.1.5, (oo) is regular.) Let Ll(oo) be the line of regular points of T(0) that corresponds to the space r of PG(3n - 1, q). Then in the same way as in the first part of the proof of Theorem 8.1.1, the dual T(0)D of T(0) is a TGQ with as base-point the line L. It follows that G is an elementary abelian group, so that each nontrivial element of G is an involution. Hence a generalized oval, respectively ovoid, satisfies Property (*) if and only if it is a translation generalized oval, respectively ovoid. This is the reason why we only defined translation generalized ovals and ovoids in even characteristic. Corollary 8.1.3 (see also Chapter 12 of [128] and K. Thas [206]) Let O be a nonclassical oval in PG(2,g), q even. Then S = T 2 (C) contains a regular point x different from the translation point if and only if' x ~ (oo) and SD is a T 2 (C) with base-point the point corresponding to x(oo), if and only if O is a translation oval with axis the tangent space of O at its element a;(oo). • Remark 8.1.4 Tits [221] defines a translation ovoid ("ovoide a translation") O of P G ( n , I ) , n = 2,3 and K a (not necessarily finite) field, in a slightly different way; here O is a set of points such that for each point x e O the union of the tangent lines of O at x is a hyperplane of PG(n, K), called the tangent space of O at x. He demands that for each tangent space T to O, there is a group of elations with axis T, fixing O globally, and acting regularly on the points of O not incident with T. Amongst many other results, he then proves that PG(n, K), n = 2,3, contains a translation ovoid if and only if the characteristic of K is 2 and there is a subfield L of K for which (n - 1)[K : L] < [L : L2]. Suppose that K is finite, i.e., K is isomorphic to GF(2 r ) for some r. If n = 3, then clearly no such subfield exists, so PG(3,2 r ) does not contain translation ovoids in the sense of Tits. If n = 2, then the only possibility is that L = K = GF(2 r ), and in this case the translation ovoid is a conic of PG(2, K). Let S = T(0) be a TGQ of order s, with s = qn even. Let O C PG(3n 1, q), and write O = {it, Tt\,..., wqn.}. Suppose n is the nucleus of O, that is,
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the common (n -1)-dimensional space of all tangent spaces of O. Consider the (2n - l)-space r = (IT, 77) (that is, the tangent space of O at 71-). For each i e {1,2,..., qn}, the set { 7 r,r ? }U{{7r i ,7r i )nr||i^j} is an (n - 1)-spread of r, denoted 7^. Note that if O is a translation generalized oval with axis T, then all the %'s coincide. It is also important to note that if all the 7^'s coincide, we have that if 7 is an element of that spread distinct from it and 77, and if j G {1,2,..., qn}, then (7,itj) contains precisely one other element irr of O (and hence is disjoint from itu if k ^ j , r). Suppose that all the (n — 1)spreads 7^ coincide. We say that O is projective atr if the following property holds: Let 7 be an element of T = % (for all i), where it ^ 7 ^ 77, and let j •£ k be in {l,2,...,qn} such that (7,^} ^ (l,itk)- By the above, there are elements ity, j ' ^ j , and -K^, k! ^ k, so that 7iy C (7,^) and 7Tfc' c (7,7Tfc). Then (itj, itk) n (7Tfe/, ity) is an element of T. Remark 8.1.5 Each translation oval O of PG(2, s) with axis L, s even, has the (in that case trivial) property that the %s coincide, and also is projective at L. It should be noticed that for a generalized oval the fact that the %s coincide, does not imply that O is projective; this follows from Theorem 8.1.6 below, and the existence of ovals in PG(2, q), q even, which are not translation ovals. Suppose all the %s coincide, and that the generalized oval O is projective at (ir, 77) = r. Let 7r, and Wk be arbitrary distinct elements of O \ {it}, and suppose that (TTJ, irk) HT = r ^ . Let 6jk be the involution with axis r, center Tjk, and which maps TTJ onto irk- Then 9jk fixes O globally, and sends TTJ onto Tifc. If then G is the group consisting of all the involutions 6jk, j ^ k, and j , k € {1,2,..., qn}, together with the identity, then it is clear that G is a group of involutions with axis T, acting regularly on O \ {-K}. Whence O is a translation generalized oval. So we have the following theorem. Theorem 8.1.6 Let O be a generalized oval in PG(3n - l,q), q even, and use the above notation. Then O is a translation generalized oval with axis r if and only if all the %s coincide and O is projective at r. •
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Note that for n = 1, this is the well-known geometrical characterization of translation ovals in PG(2, q) {q even).
8.2
Note on the Definition of Translation Generalized Oval/Ovoid
In this section, we make the following useful observation. Theorem 8.2.1 A generalized oval O o / P G ( 3 n - 1, q), q even, is a translation generalized oval with axis T, where r is the tangent space ofOatireO, if and only if there is a subgroup of involutions o / P G L 3 n ( g ) which fixes n and acts regularly on the remaining elements of O. Proof. Let G be a group of involutions as stated. Let 9 be a nontrivial element of G. Suppose r\ is the kernel of O. Let n be an arbitrary ndimensional subspace of P G ( 3 n - 1, q) containing IT but not contained in T, and suppose x is a (2n - 1)-dimensional subspace of PG(3n -l,q) containing r\ and meeting T in r\. First of all, one notes that Ue ^ II. For, suppose that this is not the case. Let IT' be the unique element of O \ {IT} intersecting II. Then n' = TT', clearly a contradiction. For each p e x, put pa = (p, ir)e Hx- Then a is a linear involutory automorphism of x- As a does not fix any point of x \ V> it follows that a must fix each point of r\ (as the dimension of r\ is n - 1). But 0 and a have the same action on n, so that 9 must fix r/ pointwise. By the same argument, 9 also fixes IT pointwise, so that r is fixed pointwise by 6. Now suppose ( e 0 \ {IT}. Then (CC") n (7r,??) belongs to the center, so that the center has dimension at least n - 1. So the dimension of the axis is at most 2n - 1, from which it follows that r is the axis of 9. The theorem readily follows. • From the proof of Theorem 8.2.1 we have the following useful observation. Observation 8.2.2 Let Obea generalized oval o / P G ( 3 n — l,q),q even, and let T be the tangent space of O at the element IT G O. Suppose that there is an involution in PGL 3 n (q) which stabilizes O, which fixes n and does not fix any of the remaining elements of O. Then 9 is an involution with axis r. • Remark 8.2.3 It should be remarked that Theorem 8.2.1 is not true for generalized ovoids. For, consider the GQ Q(5, qn), q even, and suppose p
8.3. Characterizations oftheT2(0)
of Tits
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is a point of Q(5,g ,n ). Then p is a translation point, and Aut(Q(5,g n )) p contains a subgroup H which acts naturally as PSL2(g2™) on the set of lines incident with p. In particular, for any line Lip, the stabilizer of L in H contains a normal elementary abelian 2-subgroup which acts regularly on the lines incident with p and distinct from L. So, if O is the generalized ovoid in P G ( 4 n — l,q) corresponding to the TGQ Q(5, qn) , and r is the tangent space of O at IT e O, where IT corresponds to L, then there is a subgroup G of involutions of PGL 4 „(q) which fixes IT and acts regularly on the remaining elements of O. But G can not fix r pointwise by Theorem 8.1.1.
8.3
Characterizations of the T 2 (O) of Tits
Not many characterizations of the T2(£>) of Tits, O an oval in PG(2,s) with s even, are known. It is the goal of this section to provide new characterizations of these objects relying on Theorem 3.9.7 and the theory of translation generalized ovals. Theorem 8.3.1 Let <S(°°) = T(O) be a TGQ of order s with base-point (oo). Suppose there is some line Ll(oo) for which the associated projective plane irL is Desarguesian. Suppose that x ^ (oo) is a regular point, x I L. Then O is a regular translation generalized oval, i.e. S = T2(0') of Tits for some translation oval O' o / P G ( 2 , s), and conversely. Proof. First suppose that x is not collinear with (oo). Then since <S(°°) is a TGQ, each point not collinear with (oo) is regular. It follows that each point is regular, and then <S(°°) = W(s) by Theorem 2.2.2. Also, s is even by Theorem 2.1.5. The theorem follows. Now suppose x ~ (oo). By 1.5.2(f) of [128] s is even (alternatively, all points of the line x(oo) are regular, so s is even by Theorem 2.1.5). Then Theorem 8.1.1 implies that O is a translation generalized oval with axis the tangent of O at x(oo). The result now follows from Theorems 8.1.6 and 3.9.7. The converse is obvious. The following theorem is very general.
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Theorem 8.3.2 Let S = <S(°°) = T(£>) be a TGQ of order s, s even, with base-point (oo). Suppose there is a line LI (oo) for which the associated projective plane TTL is Desarguesian. Suppose abo that there is a line M ^ L which is incident with (oo), and an involution 8 of Aut(<S)(oo) which fixes M pointwise and which does not fix any of the lines incident with (oo) and different from M. Then S £* T 2 (C) of Tits for some oval O' o/PG(2, s). Proof. Let it, respectively n', be the element of O corresponding to L, respectively M, and let r' be the tangent space of O at -IT'. Suppose 8 is as in the theorem. Then by Observation 8.2.2, 8 fixes r' pointwise and does not fix any element of 0\{ir'}. So n and ire induce the same spread T in T'. Moreover, since TTL is Desarguesian, we have that T is regular. Now apply Theorem 3.9.7. • Corollary 8.3.3 Let S = <S(°°) = T(£>) be a TGQ of order s, s even, with base-point (oo). Suppose there is a line LI (oo) for which the associated projective plane -KL is Desarguesian. Suppose that there is a nontrivial symmetry about some point x I L. Then S = T 2 (C) of Tits for some oval O' of PG(2,s). Proof. If x is not collinear with (oo) then one observes that each point of S is a translation point, so that each point is regular. The corollary then follows from Theorem 2.2.2 and Theorem 1.5.2(i). If x ~ (oo), we consider a nontrivial involution in the group of symmetries generated by the given one, and then the corollary follows from Theorem 8.3.2. •
8.4 A Characterization of Translation Generalized Ovals LetO = {71-, 7ri,... ,7r9n}be a generalized oval in PG(3n-l,g), with q — 2h even, and let r be the tangent of O at n. Now we define a point-line incidence structure A(0) as follows: • POINTS are the elements of O \ {n}; • LINES are the pairs {7r,,7rj}, with i, j e {1,2,... ,qn} and i ^ j ; • INCIDENCE is containment.
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Hence A(O) is the complete graph with vertex set O \ {IT}. Further, two lines {-Ki, -Kj) and {71-fc, 17} are called parallel if either {71-j, TTJ} = {-Kk, -K{\ or (•Ki.TTj) DT = (lTk,1Tl) PIT.
Theorem 8.4.1 The incidence structure A((D) provided with parallelism is isomorphic to the hn-dimensional ajfine space AG(hn, 2) over GF(2) if and only if O is a translation generalized oval with axis r. Proof. Assume that A(O) provided with parallelism is isomorphic to AG(hn, 2). If {wi,TTj} is a line of A(0) and -Kk $ {TT^TTJ}, then there is exactly one line of A(0) containing irk and parallel to {71^,7^}. Also, if the distinct lines {7^,77,} and {71-fc, 717} are parallel, then also {71^, 71-fc} and {ITJ,TTI} are parallel. Hence, by Theorem 8.1.6, O is a translation generalized oval with axis r. Conversely, assume that O is a translation generalized oval with axis r. Then, by Theorem 8.1.6, the qn (n - l)-spreads of r defined by the respective elements of O \ {-K} coincide, and further O is projective at TT. NOW by [98] the point-line incidence structure A(O) provided with parallelism is isomorphic to AG(hn, 2). • Remark 8.4.2 Let P G ( k , 2) be the projective completion of the affine space A(0). Then the points at infinity of A ( 0 ) , that is, the points of PG(/m, 2) not in A(0), can be identified with the elements of T \ {n, 77}, where T is the common (n - 1)-spread of r defined by the elements of O \ {TT} and 7? is the kernel of O. Any line of P G ( k - 1 , 2 ) = PG(hn, 2)\ A(C) is of the form {a,/?, 7}, with a = (iri,irj) n (7^,7^), (3 = (7ri,7rfc) n {TTJ,TTI}, 7 = (-Ki,-KI) n (7Tj, 7Tfc) and 7T,, 7Tj, 7Tfe, 7T; distinct elements of 0 \ {7r}.
8.5
Classification of 2-Transitive Generalized Ovals in Even Characteristic
We call a generalized oval O in P G ( 3 n -l,q) 2-transitive if there is a subgroup of PTL3n(q) which stabilizes 0 and acts 2-transitively on its elements. Note that by Thas and K. Thas [191], cf. Theorem 3.7.2, this is equivalent to asking that the automorphism group of T(0), which we may demand to fix the translation point (00) (as otherwise T(O) = Q(4,
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at the equivalent problem for the corresponding TGQs. From the theorem will follow which groups arise. Theorem 8.5.1 Let S = S ^ ) = T(O) be the TGQ of order {qn, qn), q even, which arises from a 2-transitive generalized oval O in PG(3n — l,q). Then S = Q(4, qn), and so O is classical. Proof. Since q is even, the point (oo) of T(O) is regular. The plane # , , which is the dual of TT^, is a translation plane of order qn, where the parallel classes correspond to the lines of S incident with (oo). We suppose that Aut(S) fixes (oo) for reasons of conveniency (otherwise S = Q(4, qn)). Then Aut(<S) acts 2-transitively on the parallel classes of 7I-P*. We have the following possibilities by Schulz [136] and Czerwinsky [42], see also Liineburg [101] (Chapter VI): (a) TTP, is Desarguesian, and Aut(S) has a subgroup K which induces PSL 2 ((7 n ) on the line at infinity of nPy (b) nP, is a Liineburg plane (so that n is even, and h odd, with q = 2h — cf. also the next chapter), and Aut(<S) has a subgroup K which induces the Suzuki group Sz( v /g") on the line at infinity of irPy We take K in such a way that the translation group G of S is contained in K. Let a be an involution of K which fixes the element n e O but not all elements of O; then -K is the only element of O which is fixed by a. We regard a as being an element of ~PVL5n(q). Since h is odd, a is a linear involution. By Observation 8.2.2, a is an involution with axis r, the latter being the tangent space of O at n. As PSL 2 (g n ) and Sz( v / q") both are generated by their involutions (a fact that can be proved, for instance, by using their representations given in the next chapter), we have proved that the nucleus 77 of O is fixed pointwise by K (regarded as a subgroup of PGL 3 „(g)). Let z' be a point of <S not collinear with (00). Then since S is a TGQ, it is easy to see that Kz< = K' still induces PSL 2 (g"), respectively Sz( v /g") J on the lines incident with (00). Let LI(00) be arbitrary, and consider a Sylow 2-subgroup 5 2 of K'L. As 5 2 — seen as a subgroup of PGL 3 n ((/) — fixes 77 pointwise, S2 fixes each line of S incident with projL2;' (and 5 2 acts sharply transitively on the set of lines incident with (00) and different from
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L). If we interpret 5*2 as an automorphism group of nL, it thus follows that 5*2 also must fix L pointwise (since in a finite projective plane any central collineation is also axial, and conversely). The group Gp ro j z, has order q2n and fixes L pointwise, so together with 5*2, it generates a group H such that, if N is the group of symmetries about L, H/N acts sharply transitively on the set of spans {[/, V} ±J - of nonconcurrent lines U,V e L-1, and fixes L pointwise. So H/N is a translation group of 7r£ of order q2n, and hence it is elementary abelian. In particular S2 = S2N/N also is. So PGL3n(g)e> has a subgroup of involutions that fixes 7r and acts sharply transitively on O \ {n}. By Theorem 8.2.1, we conclude that O is a translation generalized oval with axis r. So any point incident with L is regular. As L was an arbitrary line incident with (00), it follows that each point of (oo)-1- is regular, so that each point of S is regular (cf. 1.3.6(iv) of [128]). It follows that S ^ Q(4,qn) by Theorem 2.2.2, so that Case (b) cannot occur. Hence by Theorem 1.5.2 and §3.6, O is classical. •
Chapter 9
Moufang Sets and Translation Moufang Sets
In this chapter, we gather some results from group theory, that we will need later on. The central notion is the notion of a Moufang set. This can be viewed as a generalization of a one-dimensional translation group. We begin without any finiteness restriction, but later on, when dealing with classification, we will have to consider only finite Moufang sets.
9.1 Definition and General Results A Moufang set (on the set X) is a system (X, (Ux)xex) consisting of a set X,\X\ > 2, and a family of groups Ux of permutations of X indexed by X itself and satisfying the following conditions. (MSI) Ux fixes x € X and is sharply transitive on
X\{x}.
(MS2) In the full permutation group of X, each Ux normalizes the set of subgroups {Uv || y £ X} (that is, it acts on the latter by conjugation). The groups Ux will be called root groups. The elements of Ux are often called root elations. If Ux is abelian for some x G X, then for all x e X and we call the Moufang set a translation Moufang set. If (X, (Ux)x€X) is a Moufang set, and Y c X, then Y, \Y\ > 2, induces a sub Moufang set if for each y eY, the stabilizer (Uy)y acts sharply transitively on Y \ {y}. In this case (Y, ((Uy)Y)yeY) is a Moufang set. 181
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Chapter 9. Moufang Sets and Translation Moufang Sets
The group S generated by the Ux, for all x e X, is called the little projective group of the Moufang set. A permutation of X that normalizes the set of subgroups {Uy || y e X}, is called an automorphism of the Moufang set. The set S of all automorphisms of the Moufang set, is a group G, called the full projective group of the Moufang set. Any group H, with S < H < G, is called a projective group of the Moufang set. We have the following easy lemma. Lemma 9.1.1 (i) The little projective group S of a Moufang set (X, (Ux)X£x) acts doubly transitively on the set X. (ii) A permutation group H (acting on X) is a projective group if and only if Ux
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A group element g is said to act freely on a set S, if the group generated by g does not contain any nontrivial element that fixes x G S. A group acts freely on 5 if only the identity fixes some x e S. Proposition 9.1.2 Let (G, X) be a sharply transitive group. (i) Then we can identify X with G and the action of g G G on the element x G X = G is given by right multiplication xg. (ii) Suppose some permutation u commutes with every element of G = X. Then u acts freely on X and the action ofu is given by left multiplication with some element h G G (so u maps every x G X to hx). (iii) If a transitive permutation group H centralizes G, then H acts sharply transitively on X and there is an isomorphism ip : H —» G such that the action ofheHonG = Xis given by h : x \-> h~vx. If in particular G is abelian, then H = G. (iv) Suppose the sharply transitive permutation groups H and G normalize each other. Then either G and H have a nontrivial intersection, or G and H centralize each other. Proof, (i) Take any fixed element x e X and identify each element g e G with xs G X. Then (x9)h gets identified with gh, and (i) follows. (ii) Now put X = G. If [u, G] = {id}, then every element of (u) centralizes G, so we only have to show that u acts fixed point freely if it is nontrivial. Let e be the identity in G and suppose that u maps e to ft G G. If g G G is arbitrary, then gu = (eg)u = (eu)g = hg (the second equality uses the fact that u commutes with g), hence u maps g to hg. (iii) By (ii), H acts freely on X and hence sharply transitively, and there is a bijection 9 : H —> G such that the action of ft G H on X = G is given by left multiplication with he. If fti, ft2 G H, and g e G, then (h1h2)eg = ghlh2 = (ghi)h2 = heheg> s o (hlh2)e = he2h{. Putting y : H -» G : h .-> (ft 6 )" 1 , we see that (p is an isomorphism and so we can write he = (h~ly = h~v. (iv) If G and H intersect trivially, then from [g,ft]= g~1h~1gh e H9h C H and g~1h~1gh G g~lGh = G we deduce that [g,ft]is trivial, for all g e G and all h G H. •
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Chapter 9. Moufang Sets and Translation Moufang Sets
By (iii) of the previous proposition, and its proof, it follows that for every sharply transitive permutation group G acting on a set X, there is a unique sharply transitive permutation group H acting on X and centralizing G in Sym(X). We say that the action of H is opposite the action of G and denote H by G°PP. Note that, if G is abelian, then the two actions are obviously the same. We now get back to the Moufang sets. A first question that can be asked is: do we really need all the root groups in the definition? This is one answer. Proposition 9.1.3 Let X (\X\ > 2) be a set and let a,b G X be distinct. Let Ua and Ub be two permutation groups acting on X such that Ua fixes a and acts sharply transitively on X \ {a}, while Ub fixes b and acts sharply transitively on X \ {b}. Then Ua and Ub are root groups of at most one common Moufang set. Also, they are root groups of a Moufang set if and only if for each u G U£, u ^ id, there exists v e Ub such that U% = Ujf, if and only if Ua is normal in Ga, Ub is normal in Gb, and Ua is conjugate to Ub in G:=(Ua,Ub). Proof. Suppose first that (X, (Ux)xex) is a Moufang set. For given u € Ua \ {id}, define v e Ub by av = bu. Then (MS2) implies that Uva = Ua« = Ub* =Ubu.By the doubly transitivity of G, it maps Ua under conjugation to Ub- Also, since G normalizes the Ux, the stabilizer Ga of a must normalize Ua, hence Ua < Ga and likewise Ub
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and so, by assumption, U™ lg 1 = Ua, hence Uva = U%u = U%. The previous paragraph completes the proof of the proposition. • The condition that for each u e C/ax there exists v G Ub such that U% = Ug is used by Timmesfeld [216] to define rank one groups. But a rank one group has the additional condition that all root groups are nilpotent, and we do not want this restriction. For a set X, elements a, b e X (with \X\ > 2), and permutation groups Ua, Ub of X such that Ua (respectively, Ub) fixes o (respectively, b) and acts sharply transitively o n I \ {a} (respectively, X \ {&}), we call (X, Ua, Ub) a Moufang triple if, for G = (Ua, Ub), Ua is normal in Ga, Ub is normal in Gb, and Ua is conjugate to Ub in G. We can use the previous proposition to recognize easily sub Moufang sets. Corollary 9.1.4 Let M = (X,(Ux)xex) be a Moufang set. Suppose that Y C X, with \Y\ > 3, and consider two distinct elements oo,0 e Y. Let Voc and Vo be the permutation groups induced on Y by the stabilizers (£/oo) y and (UO)Y, respectively. Then (Y, V^, Vo) is a Moufang triple — and hence induces a sub Moufang set of M — if and only if Vx and Vo act transitively on Y \{oo} and Y \{0}, respectively. Proof. It is clear that the stated transitivity condition is necessary. We now show that it is also sufficient. First we observe that the group H generated by V^ and Vo acts doubly transitively on Y. Let h e H be such that oo'' = O and Oh = oo. Then h can be written as product of elements of Voo U Vo, and reading these elements as belonging to U^ U [/"o, we see that there is some element g e (f/oo, Uo) such that G/Y = h. It follows that v£ is the restriction to Y of (C/oo)y, and hence this must be equal to Vo- Hence V^ and Vo are conjugate in H. Similarly one shows that V^ and Vo are normal in Hoc and Ho, respectively. The assertion now follows directly from Proposition 9.1.3. • Proposition 9.1.3 tells us something about the possibility of two root groups to be contained in the same Moufang set. Now we want to look at the situation where we have two Moufang sets acting on the same set X and sharing at least one root group, and we want to find conditions under which these Moufang sets are the same. More precisely, we have the following result.
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Chapter 9. Moufang Sets and Translation Moufang Sets
Theorem 9.1.5 Let (X,(Ux)xGx) be a Moufang set, and let a, b £ X be distinct. Moreover, let Wb be a permutation group acting on X, fixing b and acting sharply transitively on X\ {b}. Suppose that (X, Ua, Wb) is a Moufang triple. If UbC\Wb is nontrivial, or if Uh and Wb normalize each other, then Wb = Ub- If Ub and Wb centralize each other, then Ub = Wb and the Moufang set is a translation Moufang set. Proof. Let M be the Moufang set determined by Ua and Ub, and let Af = (X, (Wx)x€x) be the one determined by Ua and Wb, with Ua = Wa. First suppose that v e Ub n Wb is nontrivial. Then Wav = Wl = Uva = Uav is a root group in both M and Af, and since these Moufang sets are both determined by Ua and !/ 0 ., they must coincide. Hence Ub = Wb. Now suppose that Ub centralizes Wb. Then the action of Wb is opposite the action Ub. Take arbitrary elements u e Ua, v e Ub and w e Wb. Then [v,w] = id, hence [vu ,wu] = id and we see that the action of Wb* is opposite the action of Ubu. li\X\ = 3, then clearly Ub = Wb is commutative and the result follows. Hence we may assume |X| > 4, so that we can take two distinct elements c, d in X \ {a, b}. We then see that jV is determined by the permutation groups Wc and Wd the actions of which are opposite those of Uc and Ud, respectively. By the doubly transitivity of the little projective group G(M) of M, there is a permutation g e G(M) interchanging a with b. Since [Ub, Wb] = {id}, we also have [U^Wj]} = {id}. Hence, if we denote Ua by Va, then, since Ujj = Ub9 = Ua, and since W{] acts sharply transitively o n I \ { a } , w e obtain Wb9 = Va.lfc9 = c' and d9 = d!, then similarly, one easily shows 9 that W C, = Wc and W|, = Wd- Hence the conjugate Af9 of M contains the root groups Wc and Wd and hence coincides with Af. Comparing the root groups fixing b, we see that Ub = Wb, and [Ub, Wb] = {id} implies that Ub is abelian. Hence M = N is a translation Moufang set. Finally suppose that Ub and Wb normalize each other. If they share a nontrivial permutation of X, then by the first part of the proof Ub = Wh. If they intersect trivially, then by Proposition 9.1.2(iv) they centralize each other, and the second part of our proof implies again Ub = Wb. •
9.2 Finite Moufang Sets In this section, our aim is to comment on the following classification result of finite Moufang sets.
9.2. Finite Moufang Sets
187
Theorem 9.2.1 Let M = (X, (Ux)xeX) be a finite Moufang set with little projective group S and full projective group G. Then one of the following cases occurs. (2T) S acts sharply doubly transitively on X, there is a prime number p and a positive integer n such that \X\ = pn, and S contains a normal sharply transitive subgroup N, which is an elementary abelian p-group. (Ch) S is a Chevalley group, there exists a prime number p and a positive integer n such that \X\ =pn + l, and Uxis a p-group ofnilpotency class at most 3. We have one of the following cases: - S = PSL2(p"), pn > 4, is simple and X is the projective line
PG(l,p"); - nisa multiple of3,S^ is the Hermitian unital
PSU 3 (p™/ 3 ), pn > 21, is simple and X Un(pn^3);
- p = 2, n = 2n' is even, n' is odd, S = Sz(2 n ), n' > 3, is simple and X is the Suzuki-Tits ovoid C ST (2™'); -
= 3, n = 3n', n' is odd, S ^ R(3™'), n' > 1, is simple for n' > 3, and X is the Ree-Tits unital UR(3n'); if n' = 1, then R(3) = P r L 2 (8) has a simple subgroup of index 3, which coincides with the commutator subgroup of S.
p
In all cases, we have that G is the full automorphism group of S. Also, the root groups are precisely the Sylow p-subgroups of S.
We will not prove this theorem here, but the interested reader can find proofs in Shult [143] and Hering, Kantor and Seitz [69]. We will content ourselves with some comments on this theorem, and with an explicit definition of the Moufang sets appearing in (Ch) above. It should be noted that in the papers [143, 69], the term "split BN-pair of rank 1" is used instead of "Moufang set". Concerning the sharply doubly transitive case, it follows from Frobenius' Theorem that there is a normal sharply transitive subgroup N acting on X. The group Ux, for arbitrary x e X, clearly acts transitively on iV x by conjugation, hence all elements in N have the same order and N is a p-group. Since every p-group has a nontrivial center, and since Ux acts transitively on Nx ,we see that ,/V must be elementary abelian. It follows
188
Chapter 9. Moufang Sets and Translation Moufang Sets
that |X| is a prime power. The standard example here acts on a finite field GF(g), for some prime power q, and the actions are given by GF(g) -> GF(g) : x >-> ax + b, with a e GF(q) \ {0} and b e GF(q). The normal sharply transitive subgroup consists of the maps with a = 1. A second class of examples is given by a slight modification of the previous example. If q is odd and a perfect square, then we retain the previous maps with a a square in GF(q) \ {0}, and substitute the other maps with the maps GF(q) —• GF(q) : x i-> ax^ + b, with a a nonsquare in GF(g). We call it the nonstandard example related to GF(q). We now turn to the examples under (Ch), the Chevalley type groups. As a general remark, we would like to point out that in the following descriptions we are primarily interested in the finite case, but we choose the notation in such a way that also the general (infinite) case is covered, disregarding nevertheless noncommutative fields.
9.2.1 The Case PSL2(g) This is the prototype of all Moufang sets. Let K be any (finite) field, and denote by PG(1, K) the projective line over K. So we may identify PG(1,K) with the set of all 1-dimensional subspaces of a given 2-dimensional vector space I x l . The elements of PG(1, K) can be written as (a, b), with a,b e K, (a,b) ^ (0,0) and (a,b) identified with (ca,cb) for all c e Kx. We set O = (0,1) and oo = (1,0). We define U0 as the (multiplicative) group of matrices (k)0 := (
), k e K, Uoo consists of the matrices
(fc)oo := ( , 1 ), for all fc e K, and the action of a matrix M on an element (a, b) is by right multiplication (a, b)M (conceiving (a, b) as a (2 x 1)matrix). It is easy to calculate that U^}° = UQ ) O °, hence by Proposition 9.1.3, we indeed have a Moufang set acting on PG(1,K). Since l ^ is abelian, we have a translation Moufang set. We denote this Moufang set by M(PG(1,I)). The little projective group is equal to PSL 2 (K) and is simple if |K| > 4. For |K| = 2, we have the unique Moufang set on 3 points, and for |K| = 3, we obtain the unique Moufang set on 4 points. Both are improper Moufang sets, isomorphic to the standard examples related to GF(3) and GF(4), respectively.
9-2. Finite Moufang Sets
9.2.2
189
The Case PSU3(g)
Let K be a field having some quadratic Galois extension F. This means that F admits some field involution a and the elements of F fixed by a are precisely those of the subfield K. In order to treat all characteristics at the same time, we introduce the following notation. For an arbitrary i G F \ K, we can write any element k € F as
a
a
a
k"i -ki" —. :
k- ka + -. —i,
l — la
i — ia
_1
with both (ki - k i)(i - i ) and (k - ka)(i - i")"1 in K. So F can be regarded as a 2-dimensional vector space over K and the map F —> K : x i-> x + x" is K-linear, nontrivial, hence surjective with 1 -dimensional kernel. We denote the inverse image of the element k G K by K^. Now we define our set X. It is the set of all projective points (x0,xi,X2), up to a multiplicative nonzero factor, of the projective plane PG(2,F) satisfying the equation XQX^ + x^x2 = x\x\. We can write X = {(1,0,0)} U {(jfc, x, 1) j| x e F, k G K ^ " ) } , and also X = {(0,0,1)} U {(1, x, k) || x G F, k e K^*")}. We set O = (0,0,1) and oo = (1,0,0). The group of collineations Uoo = {{x,k)oo || x GF,fcGK( xx ' T )},where(x,fc) 00 isthecollineationofPG(2,F) defined by the linear transformation with matrix
acts sharply transitively on X \ {oo} (on the right!) and fixes oo. Likewise, the group Uo = {(x,k)o || x G F, k G K ^ " ) } , where (x,k)o is the collineation of PG(2, F) defined by the linear transformation with matrix
fixes O and acts sharply transitively o n I \ {O}. It is also easy to calculate that, if k G K^'K then k~x e K«x/fc)(x/fc)a>, and also that rrC^/rOoo _ O —
u
Tr(x/k,l/k)o
u
oo
190
Chapter 9. Moufang Sets and Translation Moufang Sets
By Proposition 9.1.3, we obtain a Moufang set, which we denote by M{H(2,F,a)), or when |K| = q, briefly by M(H{2,q2)), since in this case a is uniquely determined and given by a : x \-^ xq. We call it a Hermitian Moufang set. The little projective group of M(H(2,¥,a)) is the unitary group P S U 3 ( F , a), denoted just PSU 3 (g) when |K| = q. It is a simple group whenever |K| > 3 and has order q3(q3 + \){q2 - \)/{q + 1,3). Note that \X\ = q3 +1 in this case. For |K| = 2, PSU 3 (2) is isomorphic to the sharply doubly transitive nonstandard example related to GF(9), and hence is not simple (but solvable). There is a little proposition that can be proved now. Proposition 9.2.2 Let U^ = {(a;, k)x \\ x e F, k e K^ 35 ")} be, as above, a root group of the Moufang set M(H(2,¥,a)), with |F| > 4. Then U^ is nilpotent of class 2 and the following subgroups of U^, coincide: (i) the center Z(Uoo); (ii) the derived group [[7^, [7^]; (iii) the subgroup of elements in Uoo that induce a central collineation in PG(2, F) (necessarily with center (1,0,0)). If \K\ = q, then {U^ = q3 and |Z(t/oo)| = q. Proof. It is easy to calculate that all these subgroups coincide with {(0,fc)oo|| k G K(°)}. The other assertions are obvious. •
9.2.3 The Case Sz(q) Let K be a field of characteristic 2, and denote by K 2 its subfield of all squares. Suppose that K admits some Tits endomorphism 9, i.e., the endomorphism 6 is such that it maps x9 to x2, for all x e K. If |K| = 2 n , then n must be odd, say n = 2e + 1, and 8 maps x to xT , for all x e K. Let Ke denote the image of K under 9. In the finite case K6 = K. Let L be a subspace of the vector space K over Ke, such that tfcl (this implies that L \ {0} is closed under taking multiplicative inverse). In the finite case L = K. We also assume that L generates IK as a ring. We now define the Suzuki-Tits Moufang set M(Sz(K, L, 9)).
9.2. Finite Moufang Sets
191
Let X be the following set of points of PG(3,K), where the coordinates with respect to some given basis are: X = {(1,0,0,0)} U {(a2+e + aa' + a'6,1, a', a) || a, a' e L}. One now calculates that X = {(0,1,0,0)} U {(1, a2+e + aa' + a'6, a, a') || a, a' £ L}. Wesetoo = (1,0,0,0) and O = (0,1,0,0). Let (x, a;')oo be the collineation of P G ( 3 , K) determined by / (x0 xi x2 x3)
H^
(x0 xi x2 x3)
V
1 0 0 0\ x2+9 + xx' + x' 1 x' x x 0 10 1+e x + x' 0 xe 1 /
and let (x, x')o be the collineation of P G ( 3 , K) determined by
(x0
XI X2 X3) H^ (X 0 XI X 2 X 3 )
/lx2+^+xx' + x ' % x ' \ 0 1 0 0 0 x1+e + x' 1 x6 \0 x 0 1/
Define the groups )oo || x,x' G L} and C/0 = {(x,x')o
|| x,x' G L}.
Both groups [/oo and [/o act on X, as an easy computation shows (for Uo use the second description of X above), and they act sharply transitively o n I \ {(1,0,0,0)} and X \ {(0,1,0,0)}, respectively. Moreover, one can check that (U0){x'x')a° = ([/ 00 ) (!/ ' J/ ' )c, ) with _./ 2-2+61 _)_ xxi
_ _|_ xit
and y'
c2+e + xx,
+ x,b
It follows from Proposition 9.1.3 that we obtain a Moufang set M(Sz(K,L,6)), called a Suzuki-Tits Moufang set. The group Sz(K,L,0) is the Suzuki group generated by Uoo and Ifo. If |K| = g, then we denote it by Sz(g). One has |Sz(g)| = q2(q2 + l)(g - 1) and \X\ = q2 + l. All Suzuki groups are simple groups unless |K| = 2. The group Sz(2) is a sharply doubly transitive standard Moufang set related to GF(5).
Chapter 9. Moufang Sets and Translation Moufang Sets
192
In order to understand better the structure of Uoo, we can identify each point (a2+e + aa' + a'e, 1, a', a) of X \ {oo} with the ordered pair (a, a'). Then the unique element of U^ that maps (0,0) to (b, b') is given by {b,6')oo : {a,a') ^{a + b,a' + b' + abe). The root group Uoo is hence given abstractly by the set {(a,a')oo II a, a' G £} with operation (a, o')oo © {b, &)'<x> = (a + 6, a' + 6' + ab 6 )^. We can now state the following proposition. Proposition 9.2.3 Let U^ be as above (a root group of the Moufang set M{Sz(K,L,9))), and suppose \Uoo\ > 4. Then Uoo is nilpotent of class 2 and the following subgroups of Uoo coincide: (i) the center Z(Uoo); (ii) the derived group [Uoo, Uoo}; (hi) the subset of elements in U^ which fix at least two points o/PG(3, K); (iv) the subset of elements of order 1 and 2 of UooIf\K\ = q, then \Uoo\ = q2 and \Z(Uoo)\ = q. Proof. The equality of the subgroups under (i), (ii) and (iv) follows from an elementary and easy computation, using the abstract description of Uoo above. For (iii) we consider the matrix representation of U^o and one easily calculates that (0, o')oo fixes the line spanned by (1,0,0,0) and (0,0,1,0) pointwise. If a ± 0, then (a, a')oo only fixes (1,0,0,0), as the reader can verify for himself. • Remark 9.2.4 The Moufang set .M(Sz(K, L, 9)) can also be defined as the set of absolute points of a polarity in a certain generalized quadrangle. We explain this in detail further on. 9.2.4
The Case R(g)
Let K be a field of characteristic 3, and denote by K3 its subfield of all third powers. In the finite case, K3 is just K. Suppose that K admits some Tits endomorphism 9, i.e., the endomorphism 9 is such that it maps xe to x 3 , for all x G K. In the finite case, |K| is a power of 3 with odd exponent, say
9.2. Finite Moufang Sets
193
|K| = 3 2 e + 1 , and 0 maps x to a;3 . Let Ke denote the image of K under 0. In the finite case, again necessarily Ke = K. We now define the Ree-Tits Moufang set M(R(K, 0)). For a, a', a" e K, we put „"\ - _„4+20 fi(a,a ' ,a")
f2(a,a
',an"\) =- -_ ^ a3+9 ,
/ 3 (a, a', a") = - a
3+2e
aa
+a
/0
a
T •
a
+a
+a
-a a
- a a ,
2 I
•a — aa • a a , - a"6 + aea'9 + a'a" + aa'2.
Let X be the following set of points of PG(6,K), where the coordinates with respect to some given basis are: X = {(1,0,0,0,0,0,0)} U{(f1(a,a',a"),-a',-a,-a",l,f2{a,a',a"),f3{a,a',a"))
\\ a,a',a"
A tedious calculation shows that X = {(0,0,0,0,1,0,0)} U{(1, f2(a, a', a"),h{a,
a', a"), a", fx{a, a', a"), - a ' , - a ) || a, a', a" e
We set oo = (1,0,0,0,0,0,0) and O = (0,0,0,0,1,0,0). Let (x, x', x")oo be the collineation of PG(6, K) determined by (XQ XI
(
x2 x 3 x 4 x5 xe) i-> (x 0 xi x 2 x 3 x 4 x 5 x6)
1 0 0 0 1 0 —x p X6 1 x' - x 1 + e a 1 0 0 x" fi(x,x' x") -x' —x -x" x' — xi+e 0 0 0 X 0 0 0
0 0 0 \ 2 —x" — xx' 0 x r s 0 0 X -x1 1 hi* , X x")f3(x,x',x") 0 1 -xe 0 0 1 )
with p = x3+e - x'6 q = x"6 + xex
•f x'x" — xx'
r = x"~ xx' + x „ /
X
1+(
2
r
2+e
V _TV
x2+ex>
_
xwx»
„3+2e
GK}.
Chapter 9. Moufang Sets and Translation Moufang Sets
194
and let, with the same notation, (x, x', x")o be the collineation of PG(6,1 determined by (xo xi x-i X3 X4 x5 xe) "-> (xo xi x2 x3 x\ x5 XQ) ^ 1 f2(x,x',x")
V
0 0 0 0 0 0
1 0 —x 0 X2
r
f3(x,x',x")
x"
-X6
0 0 1 0
1 x' 0 —x" — xx' X i+e_ x s
fi(x,x',x") x' _xi+e X
-x" 1 p
x'
Q
—x' — X '
0 0 0 0 1 xe
0 0 0 0 0
w
For the computations, it can be interesting to note that (x,x',x")o (x, x', x")^, with g the collineation of PG(6, K) determined by (x0,Xi,X2,X3,X4,X5,Xe)
H->
=
(xi,X5,Xe,-X3,X0,Xl,X2)-
Define the groups ^oo = {{x,x'',a;")oo || x,x'7x" e K} and U0 = {(x,x',x")o
\\ x,x',x" € K}.
The groups U<x, and Uo both act on X, as an easy computation shows (for Uo use the second description of X above), and they act sharply transitively o n I \ {(1,0,0,0,0,0,0)} and X \ {(0,0,0,0,1,0,0)}, respectively. Moreover, it can be checked that (t/0)(*.*V)oo = (jjoo^v,y',y")o> w j t h
y' =
y" =
f3(x,x',x") fi{x,x',x"y f2(x,x',x") fi(x,x',x"Y x" fi(x,x',x")'
Hence, by Proposition 9.1.3 we obtain a Moufang set, which we denote by A/((R(K, 6)) and call a Ree-Tits Moufang set. Its little projective group, generated by Ux and Uo, is the Ree group R(K, 9). If |K| = q, then we denote the corresponding Ree group by R(g). In this case we have |R(g) | = q3(q3 + l)(q - 1). All Ree groups are simple groups unless q = 3, in which case R(3) = PrL 2 (8). Hence [R(3),R(3)] ^ PSL2(8) is simple, but not sharply doubly transitive, because it has order 504.
9.2. Finite Moufang Sets
195
Following 1.1.1 of [232], we can now define U^ abstractly as follows. Define (a, a', a") := (/i(a, a', a"), -a', -a, -a", 1, / 2 (a, a', a"), /3(a, a', a")). Then the unique element of Uoo that maps (0,0,0) to (x, x', x") is given by (x,x',x")oo
'• (a,a',a")
t—y (a + x, a' + x'+ axe ,a"+ x" + ax' — ax — axl+6).
The root group Uoo can now be defined as the set {(a, a', a")^ || a, a', a" e K} with operation (a,a',a")oo®{b,b',b")oo
= (a + b,a' + b' + ab0,a" + b" +
ab'-a'b-ab1+e)oo.
We have the following proposition. Proposition 9.2.5 Let U^ be as above (a root group of the Moufang set M(R(K,9))). Then Uoo is nilpotent of class 3 and the center Z(JJoo) coincides with the second central derived group [Uoo,Uoo][2]- Also the following subgroups of Uoo coincide: (i) the derived group [Uoo, Uoo]; (ii) the subset of elements in U^ which fix at least two points o / P G ( 6 , K); (iii) the subset of elements of order 3 of UooAlso, the derived group [Uoo, Uoo] is abelian, hence Uoo is solvable of length 2. Finally, if\K\ = q, then \UX\ = q3, {[U^, Uoo]\ = q2 and \Z{Uoo)\ = q. Proof. The coinciding of Z(Uoo) and [Uoo, E^»][2] is readily verified using the above abstract description of Uoo- Likewise, the equality of the subgroups under (i) and (iii) follows from an elementary and easy computation. For (ii) we consider the matrix representation of Uoo and one easily calculates that (0, x', x")<x> fixes the plane spanned by (1,0,0,0,0,0,0), (0,0,0,0,0,0,1) and (0, x'2,0, -x'x", 0, x"2 - x'l+e, 0) pointwise. If x ^ 0, then {x,x',x")oo only fixes (1,0,0,0,0,0,0), as the reader can verify for himself. • Remark 9.2.6 The Moufang set A4(R(K, 9)) can also be defined as the set of absolute points of a polarity in a certain generalized hexagon, see 1.1 of [232].
196
9.2.5
Chapter 9. Moufang Sets and Translation Moufang Sets
Sub Moufang Sets
We end this chapter with describing which finite Moufang sets are sub Moufang sets of other finite ones. First consider M(PG(1, q)). Clearly if GF(q') is a subfield of GF(q), then by considering a projective subline over GF(q') of PG(l,g), we obtain a sub Moufang set isomorphic to M(PG(l,q')). Up to isomorphism there are no other sub Moufang sets of M(PG(l,q)). Note that the improper standard Moufang sets related to any field cannot be isomorphic to sub Moufang sets of M(PG(l, q)) unless they are isomorphic to M(PG(1,2)) or M{PG(1,3)), although they are somehow related to affine lines, which can be viewed as subgeometries of projective lines. Next consider M{H(2, q2)). Again, M(H.(2, q'2)) is isomorphic to a sub Moufang set if and only if GF(g'2) is a subfield of GF(g2) but not of GF(g). But here, also M{PG(1, q)) is a sub Moufang set, together with all its sub Moufang sets mentioned in the previous paragraph. Indeed, with the notation of Subsection 9.2.2, we restrict X to the projective points with second coordinate 0, i.e., to the set {(1,0,0)} U {(fc,0,1) || k + ka = 0}. Also, we restrict [Too to the elements (0, k)^, with k + k" = 0 and with Uo to the elements (0, k)o, with k + k? = 0. There are no other sub Moufang sets up to isomorphism. The only sub Moufang sets of the Suzuki-Tits Moufang set M(Sz(q)) are isomorphic to the Suzuki-Tits Moufang sets M{Sz(q')), where GF(q') is a subfield of GF(q). No projective lines to be found here! Finally, concerning the Ree-Tits Moufang set M(R(q)), we have the sub Moufang sets over a subfield, M(R(q')), with GF(g') a subfield of GF(q), and also M(PG(l, q)), together with all its sub Moufang sets. This can be seen as follows. With the notation of Subsection 9.2.4, we restrict X to the point set Y = {(x'1+$ ,-x',0,0,1, x'e ,0) \\ x' G K} U {(1,0,0,0,0,0,0)}. Then one easily sees that the commutative groups {(0,x',0)oo || x' e K} and {(0,x',0)o || x' e K} act sharply transitively on Y \ {oo} and Y \ {O}, respectively. Corollary 9.1.4 now implies that we indeed obtain a sub Moufang set, which is readily seen to be isomorphic to M{PG(l,q)). There are no other sub Moufang sets up to isomorphism. Remark 9.2.7 Let M(R(q)) be a Ree-Tits Moufang set on the set X. Recall that |X| = q3 + 1. Let B be the collection of subsets of size q + 1 on which a sub Moufang set isomorphic to M(PG(1, q)) exists. Then the space UR = (X, B) is a unital, i.e., a system of points and blocks such that every
9.2. Finite Moufang Sets
197
two points are contained in a unique block, and, for some natural number q, the number of points is equal to q3 +1, and the number of points on each block is q + 1. This unital is called the Ree-Tits unital. The Hermitian curve H(2, q2) is also a unital, if endowed with all secants. In fact, this Hermitian unital can also be constructed as follows. Let M{H(2, q2)) be a Hermitian Moufang set on the set X. Let B be the collection of subsets of size q + 1 on which a sub Moufang set isomorphic to M(PG(l,q)) exists. ThenUH = (X,B) is that unital.
Chapter 10
Configurations of Translation Points Let S^ be a TGQ with translation point p. Then an easy exercise yields that we have the following possibilities for its configuration of translation points: (a) p is the only translation point; (b) there is a line of translation points (incident with p); (c) every point is a translation point. If we are in Case (c), then by Theorems 10.5.3 and 10.7.1 below, and Theorem 2.2.2, the TGQ is isomorphic to one of Q(4, q), Q(5, q) for some q. Often in the existing literature, one speaks of "the translation point" of a given nonclassical TGQ S. So in view of the hypothetical Class (b), the question arises whether there are nonclassical examples with a configuration of translation points as described in (b). In the papers [86, 201, 202] and the monograph [206], a more general class of GQs was studied, namely the "span-symmetric generalized quadrangles" ("SPGQs"). The defining property is that they have nonconcurrent axes of symmetry. The main result of Kantor [86] and K. Thas [201] is that an SPGQ of order s (s > 1) necessarily is isomorphic to Q(4, s) (and then s is a prime power). So since each SPGQ of order s (s > 1) is classical, each nonclassical TGQ of order s has a unique, and whence well-defined, translation point. Surprisingly there are nonclassical examples of TGQs of order (s, s2), s > 1 and s an odd prime power, with distinct translation points, so TGQs for which the notion "translation point" is not uniquely defined! This was first 199
200
Chapter 10. Configurations of Translation Points
noted in [205]. From the paper [202], a classification of all GQs in the Class (b) could then be deduced. In this chapter, we will describe several aspects of the theory of SPGQs (which have direct implications for TGQs), state the main results, along with a number of lemmas (without proofs) by which we will point out the general ideas of the proofs of these main results. Some lemmas have an elementary (combinatorial) proof, and sometimes a sketch of a proof will be given. We will then elaborate on the consequences for the theory of TGQs. Also, we will comment on the classification of TGQs which lie in (c), in particular on the proof of K. Thas [199] which does not use the deep group theory of Fong and Seitz [59, 60].
10.1 Span-Symmetric Generalized Quadrangles Suppose S is a GQ of order (s, t), s,t^\, and suppose L and M are distinct nonconcurrent axes of symmetry; then it is easy to see by transitivity that every line of {L, M}-11- is an axis of symmetry, and <S is called a spansymmetric generalized quadrangle (SPGQ) with base-span {L, M}^^. Note that \{L, M}-1-1] = s + 1, as L and M are regular lines by Theorem 3.3.4. Let 5 be a span-symmetric generalized quadrangle of order (s,t), s,t ^ 1, with base-span {L, M}-1-1. Throughout this chapter, we will use the following notation. First of all, the base-span will always be denoted by £. The group which is generated by all the symmetries about the lines of £ is G, and sometimes we will call it the base-group. This group clearly acts 2-transitively on the lines of £, and fixes every line of £-*-. The set of all the points which are on lines of {L, M} XJ - is denoted by Q (of course, Q, is also the set of points on the lines of {L, M } x ; we have that \{L, M } x | = \{L, M}^\ = s + 1). We will refer to r = (fi, £ U £-•-, I'), with I' being the restriction of I to (fi x (£ U £-L)) U ((£ U £ ± ) x n), as being the base-grid. The substructure of fixed elements of an element of the base-group is given by the following theorem. Recall that a partial spread of a GQ is a set of mutually nonconcurrent lines. Theorem 10.1.1 Let S be an SPGQ of order (s,t), s ^ 1 ^ t, with basespan £ and base-group G. If 0 / 1 is an element of G, then the substructure Se = (Pe, Be,I$) of elements fixed by 0 must be given by one of the following.
10.2. Groups Admitting a 4-Gonal Basis
201
(i) Ve = 0 and Be is a partial spread containing C1-. (ii) There is a line L e Cfor which Ve is the set of points incident with L, and M~Lfor each M e B9 (C1- C B6). (iii) Be consists of C1- together with a subset B' of C; Ve consists of those points incident with lines of B'. (iv) Se is a subGQ of order (s, t') with s < t' < t. This forces t' = s and t = s2.
Proof. See 10.7.1 of [128].
•
We also mention the next result. Theorem 10.1.2 If S is an SPGQ of order (s,t), s,t ^ 1 and t < s2, with base-group G, then G acts regularly on the set of (s + l)s(t - 1) points of S which are not on any line of C. Proof. See 10.7.2 of [128].
•
10.2 Groups Admitting a 4-Gonal Basis Let S be an SPGQ of order s ^ 1 with base-span C, and put £ — {Lo,Li,...,Ls}. The group of symmetries about Li is denoted by d, i = 0 , 1 , . . . , s. One notes the following properties (see [114] and 10.7.3 of [128]): (1) the groups Go, Gi, • • •, Gs form a complete conjugacy class in G, and are all of order s, s > 2; (2) GinNG(Gj)
=
{l}foii^j;
(3) dG-j n Gk = {1} for i,j, h distinct, and (4) \G\ =
s3-s.
We say that G is a group with a A-gonal basis 55 = {G0, G\,..., Gs} if these four conditions are satisfied. It is possible to recover the SPGQ <S of order s, s > 1, from the base-group G starting from 4-gonal bases, in the following way, see [114, 128].
202
Chapter 10. Configurations of Translation Points
Suppose G is a group of order s 3 - s with a 4-gonal basis 03 = {G 0 , Gi, • . ., Gs}, and let G* = NG(Gi) for i = 0 , 1 , . . . , s. Define a point-line incidence structure S
POINTS.
(a) Elements of G. (b) Right cosets of the G*s. •
B
of three kinds of
LINES.
(i) Right cosets of Gi, 0 < i < s. (ii) Sets Mi = {G*g \\ g e G}, 0 < i < s. (iii) Sets U = {G*g || G* n G* 3 = 0,0 < j < s,j ^ i } U {G*}, 0 < i < s. •
INCIDENCE. I«B is the natural incidence: a line dg of Type (i) is incident with the s points of Type (a) contained in it, together with that point G*g of Type (b) containing it. The lines of Types (ii) and (iii) are already described as sets of those points with which they are to be incident.
Then 5® = (V^s, 6
See 10.7.8 of [128].
•
It is also important to recall the following from [128]. Theorem 10.2.2 Let S be an SPGQ of order s =£ 1, with base-span £. Then every line o / £ x is an axis of symmetry.
10.3. SPGQs and Moufang Sets
Proof. See 10.7.9 of [128].
203
•
So for any two distinct lines U and V of £-*-, S is also an SPGQ with basespan {£/, V}-1-1. The corresponding base-group will be denoted by G-1. It should be emphasized that this property only holds for SPGQs of order s, s > 1, as will follow from later observations. Now suppose G is a group of order s3 - s, s > 1, where s is a power of the prime p, and suppose that G has a 4-gonal basis 05 = {Go, G\,..., Gs}. Since the groups Gi all have order s, all these groups are Sylow p-subgroups in G. Since 03 is a complete conjugacy class, this means that every Sylow p-subgroup of G is contained in 03, and hence G has exactly s + 1 Sylow p-subgroups. We have proved the following theorem. Theorem 10.2.3 Suppose G is a group of order s3 — s, s > 1, with s a prime power. Then G can have at most one A-gonal basis. In particular, if G has a A-gonal basis, then it is unique. • As a corollary we obtain the following fundamental result, see [206]. Theorem 10.2.4 Suppose Sis an SPGQ of order (s,t),l<s
10.3 SPGQs and Moufang Sets By the following theorem, the base-group of an SPGQ is essentially known.
Chapter 10. Configurations of Translation Points
204
Theorem 10.3.1 Suppose S is an SPGQ of order (s,i), s,t ^ I, with basespan £ and base-group G. Let N be the kernel of the action of G on the lines of £. Then G/N acts as a sharply 2-transitive group on £, or is isomorphic (as a permutation group) to one of the following: (i) PSL 2 (s); (ii) the Ree group R( ^fs); (iii) the Suzuki group Sz(y / s); (iv) the unitary group PSXJ3(
10.4 Basic Structural Lemmas Let S be an SPGQ of order (s, t), with s ^ 1 ^ t. We keep using the standard notation introduced earlier. Suppose that s
The following set of lemmas can be found in Kantor [86] and K. Thas [201]. One can also use [206] as a general reference. Lemma 10.4.1 The group G is a perfect group if (G/N,£) 2-transitive, and if G/N ¥ R(3). Proof.
See Lemma 7.6.2 of [206].
Lemma 10.4.2 N is a subgroup of the center ofG.
is not sharply
•
10.4. Basic Structural Lemmas
205
Proof. Let H be a full group of symmetries about any line L of £. Then H and N normalize each other, so N n H = {1} implies that they centralize each other. The lemma now follows as G is generated by the symmetries about the lines of C. • Lemma 10.4.3 We have that G/N either acts naturally as PSL 2 (s) on C, or sharply 2-transitively. Proof. See Lemma 7.6.5 of [206].
•
The proof of the following lemma uses the theory of universal central extensions of (perfect) groups. Lemma 10.4.4 If G/N acts as PSL 2 (s), then G ** SL2(s) and S ^ Q(4, s). Proof. See Lemma 7.6.6 of [206].
•
It remains to consider the case where G/N acts sharply 2-transitively. First suppose that G/N acts sharply 2-transitively on the lines of C, and that s = t. Every line of £-L is an axis of symmetry by Theorem 10.2.2. Since the lines of C1- are also axes of symmetry, we can assume that the base-group G1- corresponding to these lines also acts sharply 2-transitively on C1 — otherwise G1- is isomorphic to SL2(s), and then <S is classical by Theorem 10.2.4. Hence G and G1- contain normal central subgroups N and N1- which act trivially on the points of tt, both of order s - 1 (where fi is the set of points on the lines of the base-span). Also, G and G1- act regularly on the points of S not in £1 by Theorem 10.1.2. Let a; be a point and L a line of a projective plane II. Then II is said to be (x, L)-transitive if the group of all collineations of II with center x and axis L acts transitively on the points, distinct from x and not on L, of any line through x. The following theorem is a step in the Lenz-Barlotti classification of finite projective planes, see e.g. [45,145]; it states that the Lenz-Barlotti Class III.2 is empty. Theorem 10.4.5 (Yaqub [239]) Let II be a finite projective plane, containing a nonincident point-line pair (x, L) for which II is (x, L)-transitive, and
206
Chapter 10. Configurations of Translation Points
assume that II is (y, xy)-transitive for every point y on L. Desarguesian.
Then II is •
As every axis of symmetry L of a GQ is a regular line, there is a projective plane irL canonically associated with L (by Theorem 2.2.1) if the GQ has order s (s ^ 1). Using Theorem 10.4.5, one then obtains the following result. Theorem 10.4.6 Suppose that S is an SPGQ of order s, where s / 1 , with base-group G and base-span C. Also, let N be the kernel of the action of G on the lines of L, and suppose that G/N acts as a sharply 2-transitive group on the lines of C. Then S is isomorphic to Q(4,2) or Q(4,3). Proof.
See Theorem 7.7.2 of [206].
•
Now suppose that s
10.4. Basic Structural Lemmas
207
the stabilizer in G of a line M which does not contain a point of Y has size at least (s + l)/2. Since s + 1 = ph with p odd, it follows that the size of the stabilizer is exactly s + 1 (recall that G acts regularly on the points of S \ fi), and that the action of the stabilizer on the points of M is regular. As A' is a Sylow p-subgroup of G, we have that GM < A', and \MA' \ = s + 1. As N < Z(G) acts transitively on the lines of M A , we have that GM fixes each line in MA . If two distinct lines of MA' would intersect, then any element of GM fixes their common point, clearly a contradiction. Hence MA is a partial spread of the GQ. The set of all points incident with lines of MA is an orbit R of A'. If mlM and M is the second line incident with m which does not contain a point of Q,, then similarly R is the set of all points incident with lines of MA'. So R is the point set of a grid having as lines MA' U MA . It easily follows that MG is a partial spread of S and that MGUCJ- is a spread of S. Consider any i e {0,1,..., s}, take any point mlM and suppose O is the unique line on m meeting L» e £. As any orbit R' of A' is the point set of a grid, it contains at most one point of O. Hence O has a unique point in common with every orbit of A1. Take any 9 £ Gt and j3 e GM- Then Me/3 is concurrent with O13 and belongs to the spread MG U £-"-, so M9/3 = Me. It easily follows that GM fixes every line of the spread, while acting sharply transitively on the points of each of these lines. By definition, this means that C1- U MG is a spread of symmetry. Before proceeding, we need a nice construction, first observed by Payne, of GQs of order (s - 1, s + 1) from (thick) GQs of order s with a regular point.
The Payne Derivative Let s be a regular point of the GQ S = (V, B, I) of order s, s > 1. Define an incidence structure V{S, x) = S' = {V, B', I') as follows: • The POINT SET V is the set V \
x^.
• The LINES of B' are of two types: - the elements of Type (a) are the lines of B which are not incident with x; - the elements of Type (b) are the hyperbolic lines {x, y}1-1- where y i- x. • INCIDENCE I' is containment (regarding a line of S as a set of points).
208
Chapter 10. Configurations of Translation Points
Then S' is a GQ of order (s - 1, s + 1), which we call the Payne derivative of <S with respect to x. Theorem 10.4.7 (De Soete and Thas [49]) Let S be a GQ of order (s, s + 2), with s > 1, and suppose S has a spread of symmetry T. Let H be the group of automorphisms of S that fix T linewise. Then there exists a GQ S' of order s + 1 with center of symmetry x, such that S^V{S',x). Moreover, T arises from the set of hyperbolic lines containing x, and H is induced by the group of symmetries about x. • So there exists a GQ S' of order s + 1 with center of symmetry (oo) such that V{S', (oo)) = S, and £ x U MG = V arises from the hyperbolic lines containing (oo), while GM is induced by the symmetries about (oo). Moreover, as G preserves T', a result of De Winter and K. Thas [50] states that G is induced by an automorphism group of S' that fixes (oo). Now consider a nontrivial symmetry <j> of S about some fixed line Lj of the base-span C Then Lj is a line of Type (a) in S', and
10.5 Classification of SPGQs of Order (s,t),l < s
10.6. SPGQs of Order
(s,s2)
209
Theorem 10.5.2 A finite group is isomorphic to SL2(s) for some s if and only if it has a 4-gonal basis. • It also follows that the order of an SPGQ is essentially known. Theorem 10.5.3 Let S be an SPGQ of order (s, t), s,t^l.
Then t e {s, s2}.
10.6 SPGQs of Order (s,s2) In [202], the author proved that if S is a span-symmetric generalized quadrangle of order (s, s2), s / 1 and s odd, with base-span £, then <S contains s + 1 subquadrangles, all isomorphic to the classical GQ Q(4, s), which mutually intersect in the base-grid T = (Q,C U £-L, I'). Also, the basegroup G acts freely on the points of S \ £1 and G = SL2(s). (Note that |G| = |SL 2 (s)| = s 3 - S . ) Already since the beginning of the study of SPGQs (late 1970's), it was believed (and posed as a problem to prove) that SPGQs of order (s, s2), s > 1, always have ("many") classical subGQs of order s, all passing through the base-grid. The solution of that problem in the odd case by K. Thas [202] involved a proof that was largely valid in the even case. In Chapter 7 of [206], the problem was completely solved for the general case. The fact that an SPGQ of order (s, s2) has s +1 subGQs isomorphic to Q(4, s) in this position endows the GQ with a lot of structure, and this observation is crucial for many applications in theory of SPGQs. For instance, Kantor used this to classify what he calls "grid-symmetric generalized quadrangles" — these are SPGQs for which the perp of the base-span also is a base-span, see [87]. Other applications will be obtained further on. Here, we give a brief account of the proof. The reader is referred to [206] for further details. For the rest of this section, let S be an SPGQ of order (s, t), s ^ 1 ^ t, with base-grid T = (fi,£ U £-*-, I') and base-group G, and let Li,Gj, etc. be as before. If p is a point which is not an element of f2, and U is a line through p which meets fl in a certain point q!Lk of Y, then clearly every point on U which
Chapter 10. Configurations of Translation Points
210
is different from q is a point of the G-orbit which contains p. We will use this observation freely throughout this chapter. The proof of the next lemma was first obtained in K. Thas [202]. Lemma 10.6.1 Suppose S is an SPGQ of order (s,t), s / 1 / t, with basegrid T and base-group G. Then G has size at least s 3 - s. Proof. If s = t, then we already know that G has order s 3 — s, so suppose s f t (so that t = s2 by Theorem 10.5.3). Set C = {L0, LI, . . . , Ls} and suppose Gj is the group of symmetries about Li for all feasible i. Suppose p is a point of S not in Q, and consider the following s + 1 lines JVj := proj p L;. If A is the G-orbit which contains p, then every point of Nt not in fl is also a point of A. Now consider an arbitrary point q ^ p on N0 which is not on a line of C. Then again every point of proj^Li, i =£ 0, not in Q, is a point of A. Hence we have the following inequality: |A| > l + (s + l ) ( s - l ) + ( s - l ) 2 s , (10.1) from which it follows that |A| > s3 - s2 + s. Now fix a line U of £-"-. Every line of S which meets this line and which contains a point of A is completely contained in A U Q,. Also, G acts transitively on the points of U. Suppose k is the number of lines through a (= every) point of U that are completely contained in A U SI (as point sets). If we count in two ways the number of point-line pairs (u, M) for which u G A, M ~ U and ulM, then it follows that k(s + l)s
= |A| > s(s2 - s + 1)
3 ~+ n —— = s - 2 + — - , s+1 s+1 and hence, since k e N, we have that k > s - 1. Thus |A| > s 3 - s and so also \G\. M
, ^ ^ k >
S2 S 1
In the rest of this section, we can suppose that s > 3; if s = 2, then S = Q(5,2) by Section 1.8. In that case, there are s + 1 = 3 subGQs of order 2, isomorphic to Q(4,2) and mutually intersecting in the base-grid. The basegroup stabilizes each of these subGQs (as it is generated by symmetries), and acts freely on the points of S not in f2; whence G = SL 2 (2). Similar properties hold for the case s = 3.
10.6. SPGQs of Order {s,s2)
211
For y/i = s > 3, the strategy of handling the situation is essentially the same as that followed in Section 10.4. The aim again is to prove that G is a perfect central extension of G/N, when (G/N, C) is not sharply 2-transitive. The latter case has to be handled separately (in an entirely different way than in the case t < s2). Then, one eventually comes to the following result, using Lemma 10.6.1. Proposition 10.6.2 Suppose that S is an SPGQ of order (s, s2), s > 3, with base-span C = {U, V}1-1, base-group G and base-grid T = (fi, £ U Cx, I'). Furthermore, let N be the kernel of the action of G on C. Then G/N is isomorphic to PSL 2 (s). Also, G acts freely on S \ fi. Proof. See Section 7.10 of [206], in particular the Lemmas 7.10.2 — 7.10.7. • We know that G acts freely on the points of <S \ fl. Assume that G has order (s + l)s(s - 1). Let A be an arbitrary G-orbit in S \ f2, and fix a line W of £-"-. By the fact that G acts freely on the points of S \ tt, that \G\ = (s + l)s(s — 1) and that G acts transitively on the points incident with W, we have that any point on W is incident with exactly s - 1 lines of S which are completely contained in A except for the point on W which is in Q., and that every point of A is incident with a line which meets W (recall that G is generated by groups of symmetries). Now define the following incidence structure S' = S'(T) = (V, B', I'). • LINES. The elements of B' are the lines of S' and they are essentially of two types: (1) the lines of
C^uL^;
(2) the lines of <S which contain a point of A and a point of Q,. • POINTS . The elements of V are the points of the incidence structure and they are just the points of f2 U A. • INCIDENCE. Incidence I' is the induced incidence. Then S' is an SPGQ of order s. By Theorem 10.5.1, S' is isomorphic to the GQQ(4,s).
212
Chapter 10. Configurations of Translation Points
Theorem 10.6.3 Suppose S is an SPGQ with base-span C and of order (s, s2), s ^ 1. Assume that G has size s3 — s. Then there exist s + 1 subquadrangles of order s which are all isomorphic to Q(4, s), so that they mutually intersect in the base-grid F. Proof. From each G-orbit in S \ fl arises a subGQ of order s which is isomorphic to Q(4, s), and there are exactly s + 1 such distinct G-orbits. •
The Main Structure Theorem for SPGQs of Order (s, s2) We are ready to prove the main result for SPGQs of order (s,s2) [206, Chapter 7]. Theorem 10.6.4 Suppose S is an SPGQ of order (s, s2), s ^ 1. Then S contains s + 1 subquadrangles, all isomorphic to the classical GQ Q(4, s), which mutually intersect in the base-grid T. Also, the base-group G acts freely onS\Q, \G\ = s3-sandG^ SL2(s). Proof. The fact that G acts freely on S \ Q, implies that G is a perfect group, and also that G/N ^ PSL 2 (s) if ./V is the kernel of the action of G on £; see [206]. Moreover, ./V is contained in the center of G (recall the proof of Lemma 10.4.2). Thus, G is a perfect central extension of PSL 2 (s), and this leads to the fact that G = SL2(s) and that \G\ = s3 - s; see again [206]. The result follows from Theorem 10.6.3. •
10.7 Translation Generalized Quadrangles with a Line of Translation Points Although it was conjectured in 1983 by Payne that every SPGQ of order (s, s 2 ), s > 1, is isomorphic to Q(5, s), see PROBLEM 26 of [115], such a similar result as in the case t < s2 cannot hold. For, let /C be the quadratic cone with equation X0Xi = X2 of PG(3, q), q odd. Then the q planes irt with equation tX0 - mfXx + X3 = 0, where t £ GF(q), m a given nonsquare in GF(q) and a a given field automorphism of GF(g), define a Kantor-Knuth semifield flock T of K; see
10.7. Generalized Quadrangles witii a Line of Translation Points
213
Section 4.5. In [116], Payne noted that the dual Kantor-Knuth flock generalized quadrangles are span-symmetric (this observation is rather hidden in [116]), and this infinite class of generalized quadrangles contains nonclassical examples; T is not a linear flock (and then the GQ is nonclassical) if and only if a is not the identity. Moreover, every nonclassical dual Kantor-Knuth flock GQ contains a line L for which every line which meets L is an axis of symmetry (see [206]). Hence there is some line each point of which is a translation point. In order to initialize a theory for thick span-symmetric generalized quadrangles of order (s, s2) with s ^ 1, and to answer several important questions in a broader context, K. Thas studied the following problem in [202, 205, 206]. PROBLEM.
Classify all thick generalized quadrangles with distinct transla-
tion points. In the papers [202], [205] and [206, Chapter 7], this program was carried out completely. Before coming to a synthesis of the main result, cf. Theorem 10.7.2 below, we need the next theorem. Theorem 10.7.1 (Fong and Seitz [59, 60]; Kantor [85]; K. Thas [199]) 7/<S is a GQ of order (s, s2), s > 1, each line of which is an axis of symmetry, then S is isomorphic to Q(5, s). Proof (Taken from K. Thas [199]). Let U and V be any two nonconcurrent lines of S. Then S is an SPGQ with base-span {U, V}-11- and by Theorem 10.6.4, {U, V} is contained in s + 1 subGQs of order s. Now 5.3.5 of [128] implies the theorem. • Theorem 10.7.2 Suppose S is a generalized quadrangle of order (s,t), s ^ l ^ t , with two distinct collinear translation points. Then we have the following possibilities: (i) s = t, sis a prime power and S = Q(4, s); (ii) t — s2, sis even, sis a prime power and S = Q(5, s); (iii) t = s2, s = qn with q odd, where GF(q) is the kernel of the TGQ S = <S(°°) with (oo) an arbitrary translation point of S, q > 4n 2 -8n+2 and S is the point-line dual of a flock GQ S^) where T is a KantorKnuth flock;
214
Chapter 10. Configurations of Translation Points
(iv) t = s2, s = qn with q odd, where GF(q) is the kernel of the TGQ S = <S(°°) with (oo) an arbitrary translation point ofS, q < 4 n 2 - 8 n + 2 and S is the translation dual of the point-line dual of a flock GQ S(F) for some flock T. If a thick GQ S has two noncollinear translation points, then S is isomorphic to one of Q(4, s), Q(5,s). Conversely, if S = T(O) is a good thick TGQ of order (s, s 2 ) in odd characteristic, then S has a line of translation points. Proof. Let S be as assumed. If S contains two noncollinear translation points, then by Theorems 10.5.1, 10.5.3 and 10.7.1 S is isomorphic to one ofQ(4,s), Q(5,s). Suppose that S contains two collinear distinct translation points u and v. Then for s = t, the statement follows immediately from Theorem 10.5.1. If s ^ t, then t = s2 by Theorem 10.5.3. By transitivity, every point of L -.— uv is a translation point, and hence every line of L1- is an axis of symmetry. If we fix some base-span C for which L e £-*-, then by Theorem 10.6.4 there are s + 1 classical subGQs of order s which mutually intersect precisely in the base-grid r = (f2, £ U £-"-, I') (with the usual notation). Fix an arbitrary translation point (oo)lL, and consider the TGQ S(°°) = T(O) with base-point (oo). Then by Theorem 7.5.7, O is good at its element TX which corresponds to L. If s is even, the result follows by Theorem 7.8.3. Now let s be odd. Then Theorem 5.1.2 implies that the translation dual T(0*) of T(O) satisfies Property (G) at the flag ((oo)',L'), where (oo)' corresponds to (oo), and V to L. If we now apply Theorem 4.10.6, then it follows that T(0*) is the point-line dual of a flock generalized quadrangle S^J7). The theorem now follows from Theorem 5.1.14. Conversely, suppose that S = T ( 0 ) is a good thick TGQ of order (s, s 2 ), s odd, with TT e O a good element. Let x be any point of T ( 0 ) incident with •K. Then by Theorem 6.2.1 and Theorem 6.3.5, x is coregular. By Corollary 6.6.9, we conclude that a; is a translation point. The theorem is proved.
•
Corollary 10.7.3 A GQ of order (s,t), s =£ 1 ^ t and s even, has two distinct collinear translation points if and only if S is isomorphic to Q(4, s) or Q(5,s). •
10.7. Generalized Quadrangles with a Line of Translation Points
215
Corollary 10.7.4 Suppose S = T(£>) is a TGQ of order (qn, qm), where q is odd ifn = m. Suppose (oo) is the base-point of the TGQ. Let S* = T{0*) be its translation dual with base-point (oo)'. Then we have one of the following possibilities: (i) |Aut(5)| = |Aut(«S*)|; (ii) |Aut(<S)| = (qn + l)|Aut(<S*)| with n ^ TO; (iii) |Aut(5*)| = (qn + l)|Aut(5)| with
n^m.
Proof. The Aut(5)-orbit of (oo), with S = {V, B, I), is either V, a line, or {(oo)}. If n = TO with q odd and if the Aut(5)-orbit of (oo) is not {(oo)}, then by 1.3.6 (iv) of [128] each line of S is regular, so 5 ^ Q(4,g n ) by Theorem 2.2.2, hence S = S*. If every point of S is a translation point, then S is classical, S = S*, and so Aut(<S) = Aut(<S*). In all other cases we have one of (i), (ii), (iii) by Theorem 3.11.1. • The following corollary, which shows that all possibilities in Corollary 10.7.4 occur, was deduced in K. Thas [205] in the context of TGQs arising from flocks. Corollary 10.7.5 (1) Suppose S = T(O) is a TGQ of order (s, s 2 ), s even, and let (oo) be the base-point. Let S* = T(0*) be the translation dual with base-point (oo)'. Then |Aut(<S)| ^ |Aut(5*)|. (2) Suppose S = T(0) is a TGQ of order (s, s2), s odd, which arises from a flock but which is not the point-line dual of a Kantor-Knuth semifield flock GQ. Let (oo) be the base-point of T(0). Let S* = T(0*) be the translation dual ofT(0) with base-point (oo)'. Then |Aut(<S*)| = (s+l)|Aut(5)|. Proof. Suppose we are in Case (1). Then by Theorem 10.7.2, both <S and S* either have precisely one translation point, or all their points are as such. The result follows from Theorem 3.11.1. If we are in Case (2), then S* is the translation dual of the point-line dual of a flock GQ — hence S* is good, and so by Theorem 10.7.2 it has a line of translation points. By Theorem 10.7.2 and Theorem 6.4.4, <S has only one translation point, and the result follows from Theorem 3.11.1. •
Chapter 10. Configurations of Translation Points
216
10.8 On the Classification of Translation Generalized Quadrangles Combining the obtained results, we arrive at the following theorem. Theorem 10.8.1 Let S = T(0) be a TGQ of order (qn,qm) with translation point (oo). Then we have the following possibilities. (i) n = m, and either S = Q(4, qn), or (a) S has one translation point, q is even, and Aut(<S) = Aut(<S)(oo); (b) S has one translation point, q is odd, and if u and v are arbitrary points ofT(0) and T(0*), not collinear with (oo) and (oo)' (where (oo)' is the translation point of S* = T(0*)), respectively, then Aut(<S)„ = [Aut(<S)(oo)]„ 3* Aut(5*)„ = [Aut(S*)(oo)»]„. Also, |Aut(5)| = |Aut(5*)|. (ii) n < m, and then we have one of the following cases. (a)
S^Q(5,qn).
(b) S and S* are not classical, have two distinct collinear translation points, and then q is odd, m = 2n, and S = S* = S{F)D, where T is a Kantor-Knuth semifield flock. (c) S and S* are not classical, S is not the point-line dual of a KantorKnuth semifield flock GQ, has a line of translation points, and then m = 2n, q is odd, and S is the translation dual of the point-line dual of a flock GQ. In this case we have that |Aut(<S)| = (qn + l)|Aut(5*)|. Moreover, we have that [Aut(5) ( o o ) ] u = Aut(S*)„ = [Aut(S*) (oo) ,]„, where (oo) is an arbitrary translation point on the line of translation points [oo] of S, where (oo)' is the translation point of S*, and where u and v are arbitrary points o/T(£>) and T(0*), not collinear with (oo) and (oo)', respectively. (d) S andS* are not classical, S* is not the point-line dual of a KantorKnuth semifieldflock GQ, has a line of translation points, and then
10.8. On the Classification of Translation Generalized Quadrangles
217
m = 2n, qis odd, and S* is the translation dual of the point-line dual of a flock GQ. Similarly as in Case (ii) (c), we have that |Aut(5*)| = (qn + l)|Aut(5)|. Also, [Aut(S*) (oo) ,]„ = Aut(5) u = [Aut(5) ( o o ) ] u , where (oo)' is an arbitrary translation point on the line of translation points [oo]' of S*, where (oo) is the translation point of S, and where u and v are arbitrary points ofT(0) and T(0*), not collinear with (oo) and (oo)', respectively. (e) S and S* both have precisely one translation point, and then Aut(S) = Aut(<S)(oo) and Aut(<S*) = Aut(<S*)(ooy, where the notation is obvious. Furthermore, |Aut(5)| = |Aut(<S*)|, and Aut(S)u^Aut(S*)v, where u and v are arbitrary points of T(O) and T(0*), collinear with (oo) and (oo)', respectively.
not
If H — Aut(<S)(00) is the stabilizer of (oo) in Aut(«S), then we have that H = T xi Hx, with T the translation group of <S(°°) and x any point not collinear with (oo). Proof. First suppose that n = vn. Then by Theorem 10.7.2, S has one translation point if S is not classical. Part (i) then follows from Theorem 3.11.1. Next suppose that n < m, and that S is not isomorphic to Q(5,
218
Chapter 10. Configurations of Translation Points
In the next chapter, we will consider GQs which are TGQs with respect to every point without any finiteness assumption. We will develop our ideas in the more general context of Moufang GQs.
Chapter 11
Moufang Quadrangles with a Translation Point In this chapter, we study generalized quadrangles which are translation with respect to every point. In fact, these are a special case of the class of so-called Moufang quadrangles. We will formulate and prove a lot of equivalent, but seemingly much weaker, conditions than the Moufang Condition. From time to time, and when appropriate, we apply these results to translation Moufang GQs. We also show a basic correspondence between general Moufang GQs and translation Moufang GQs, namely, that every Moufang GQ contains a full or ideal thick translation Moufang GQ. All these results will be shown without any finiteness assumption. This guarantees that all arguments be purely geometric, or combinatorial group theoretic. In the finite case, other proofs exist (but not necessarily shorter ones!) making use of deeper group theoretic theorems such as the classification of Moufang sets and of irreducible split BN-pairs of rank 2 (which are not available in the infinite case).
11.1 Notation Two nonintersecting lines L, M will sometimes be called opposite; likewise two noncollinear points x, y will sometimes be called opposite. For every point x and line L, we denote the set of elements incident with x and L by [x\ and [L], respectively. With this notation, the mapping pL>M • [L] —» [M] : x H-> projMx is a bijection for every line M opposite L. We shall use this map below, so we treat pL,M as standard notation. 219
220
Chapter 11. Moufang Quadrangles with a Translation Point
Let <S be a finite GQ of order (s,t), s,t > 2. A subquadrangle of order (s,t'), for some t' e N, will be called a full subGQ; a subquadrangle of order (s',t), for some s' e N, will be called an ideal subGQ. In general, a full subquadrangle <S' of a (not necessarily finite) generalized quadrangle S is a subquadrangle with the property that every point of S incident with some line of S' belongs to S'. Dually, one defines in general an ideal subquadrangle. Now let <S be a thick GQ of order (s, t), let S' be a full subGQ of order (s, t'), t' < t, and let S" be a full subGQ of S' of order (s, t"), t" < t'. Then t = s2, t' = s and t" = 1; see Lemma 1.1.1. In particular, S" does not admit a proper full subGQ. We start with some general lemmas on subGQs and coUineation groups, some of which are trivial in the finite case. We gather these in the next section. In Section 11.3 we introduce a lot of conditions on the coUineation group of a GQ S and its action on S. All these conditions assume a transitive coUineation group with a so-called Tits system, and therefore we study this situation in Section 11.4. In the subsequent sections, we prove the promised equivalences of conditions we introduced, and show that every Moufang GQ contains a subGQ which is a TGQ.
11.2 Some General Elementary Lemmas The first lemma is a characterization of GQs, grids and dual grids. Lemma 11.2.1 Let S = (V,B,I) be an incidence structure satisfying the following properties. (i) Each element is incident with at least 2 elements. (ii) There exist a point x and a line L not incident with x. (iii) For each x e V and each L e B, with x not incident with L, there exist a unique point y and a unique line M for which xlMIylL. Then S is either a generalized quadrangle, or a grid, or a dual grid. Proof. Suppose, by way of contradiction, that there are two points x\, x2 both incident with two distinct lines Li, L2. Then every line incident with x\ must be incident with x2, as otherwise we contradict (iii). Dually, every
11.2. Some General Elementary Lemmas
221
point on any line incident with both xi,x2 must be incident with every line through both xi, x2. Conditions (i) and (ii) imply that there exists a line L not incident with any of the points on L\. Then the line through x\ meeting L must also be incident with x2, but it contains a point — namely, the point on L — not incident with L\, a contradiction. Suppose now that S is not a grid, nor a dual grid. Left to show is that there are a constant number of points on a line and, dually, a constant number of lines through a point. If all lines are incident with exactly two points, then it is easy to see that we obtain a dual grid; likewise, if all points are incident with exactly two lines, then we have a grid. So we may suppose that there is a line L with three points, and a point x incident with three lines. Condition (iii) readily implies that opposite lines carry the same number of points, and opposite points are incident with the same number of lines. We claim that every line M ~ L, M ^ L, is opposite some line K ^ L. Indeed, let x\,X2, X3 be three points incident with L, and let Mtlxi be arbitrary but distinct from L, i = 1,2,3. We show that there is a line opposite both L and Mi. Let yjlMj, j = 2,3, yj ^ Xj. It is clear that x cannot be collinear with all four of x2, £3,2/2,2/3- But since x2 and x3 are opposite 2/3 and y2, respectively, at least one of y2,2/3 is incident with at least three lines. One of these three lines does not meet L nor Mi. The claim is proved. Now Mi is incident with the same number of points as L, and the lemma is proved.
We refer to Condition (iii) of the previous lemma as the Main Axiom of GQs. For thick GQs, we have the following alternative definition. Lemma 11.2.2 Let S = (V,B,I) be an incidence structure satisfying the following properties. (i') Some point is incident with at least three lines, and some line is incident with at least three points. (ii') There exists an ordinary quadrangle in S, i.e., there are four points xi,x2,X3,Xi G V with x\ ~ x2 ~ £3 ~ £4 ~ x\, where xi 7^ £3 and X2 ^ £4.
(iii) For each x e V and each L e B, with x not incident with L, there exist a unique point y and a unique line M for which xlMlylL. Then S is a thick generalized quadrangle.
222
Chapter 11. Moufang Quadrangles with a Translation Point
Proof. By (iii), the lines x\x-i and £3x4, and the lines £2X3 and x4xi in (ii') do not meet. Let L be any line of S. If L is incident with one of the points Xi, i e {1,2,3,4}, say with x\, then we can apply (iii) to the pair {x3,L} and obtain a second point on L. If L is not incident with one of xi,...,X4, then we can apply (iii) to L and each of xi,..., 24. It is easy to see that this gives rise to at least two distinct points on L. Similarly, one shows that each point of S is incident with at least two lines. Hence we have proved (i) of the previous lemma. As (ii) of that lemma is an immediate consequence of (ii'), we conclude that S is either a thick GQ, or a grid or dual grid. But the latter two do not satisfy (i')The lemma is proved.
•
The next lemma is evidently true in the finite case, comparing orders. Proposition 11.2.3 Let S = (V, B, I) be a thick GQ and let S' be a thick subGQ of S. If S' contains at least one line L such that every point inS on L is also a point of S', then S' is a full subGQ. Also, if S' is both full and ideal in S, then it coincides with S. Proof. If M is a line in S' opposite L, then the above map PL,M is a bijection in both S and S'; hence all points of M in S belong to <S'. The first result now follows by repeatedly applying this observation and noting that the graph (B, rf) is connected (it even has diameter 2). For the second assertion, let z be an arbitrary point of S, and let L be a line of S'. If zlL, then z belongs to S' since S' is a full subGQ. Suppose now z IL. Since S' is full in S, the point projL2; belongs to S'. Since S' is ideal in S, the line proj z Llproj L ^ belongs to S'. Since S' is full in S, the point z must now belong to S'. Similarly all lines of S belong to S'. The proposition is proved.
•
The connection between coUineations and subGQs will become apparent from the following proposition. Proposition 11.2.4 Let ipbea collineation of the GQ <S. Let V (B') be the set of invariant points (lines) of S under the action of ip. If V contains two opposite points of S, and B' contains two opposite lines of S, then V' and B' are the point and line set, respectively, of either a subGQ S' of S, or a grid or dual grid.
11.2. Some General Elementary Lemmas
223
Proof. Let I' be the restriction of I to the sets V and B'. According to Lemma 11.2.1, it suffices to show that S' = (V',B',I') is an incidence structure in which every element is incident with at least two elements, in which there is a point of V not incident with some line of B', and in which the Main Axiom of GQs holds. The latter is clear since, if a point x is fixed, and a line L not through x is fixed, then proj^L and projLa; are uniquely determined by x and L and therefore fixed. We now show that every line of <S' is incident with at least two points of S'. Let L be any line of S'. By assumption, there are two opposite points xi, x 2 in <S'. Hence the points p r o j ^ , i = 1,2, belong to S'. If proj L xi ^ proj L x 2 , then we are done. Suppose now that proj L xi = projLa:2. By assumption there are also two opposite lines in <S'. Consequently not all elements of B' contain the point proj L a;i. Let M e B' be such that it is not incident with proj L a;i. If M meets L in another point, then we are done, so we assume that M is opposite LinS'. Then the line M cannot be concurrent with both proj^L and proj^L, since a triangle would appear. We may assume without loss of generality that M is not concurrent with proj L. Since x\ and proj^rri are collinear, the Main Axiom of GQs implies that y\ := proj M ;n and 2/2 := P r oj M (P r o JL : r i) a r e different points, belonging to V. Hence also proj L yi and proj L xi = proj L (proj M (proj L a;i)) are two distinct points on L belonging to S'. Hence every line of S' is incident with at least two points of <S' and dually, every point of S' is on at least two lines of S'. Now it is also clear that S' contains a point x and a line L which are not incident. • Here is an immediate useful corollary. Corollary 11.2.5 Let ipbe a collineation in a thick GQ S fixing an ordinary quadrangle, fixing all points on a certain line of that quadrangle, and fixing all lines through a certain point (vertex) of that quadrangle. Then ip is the identity. Proof. By the previous proposition, ip fixes pointwise a subGQ which is by assumption thick. By Proposition 11.2.3, this subGQ coincides with S. Hence ip fixes everything and is the identity. • We end this section with two general combinatorial results. The first one is folklore, the second one is an observation of Cuypers, see [21].
224
Chapter 11. Moufang Quadrangles with a Translation Point
Lemma 11.2.6 Let S = (V, B, I) be a thick GQ and let x e V be arbitrary. Then the graph (T, E) with vertices the points ofS opposite x, and adjacency given by collinearity is connected. Proof. Let y, z be two points opposite x. If some point in {y, z}-1 is not coUinear with x, then y and z belong to the same connected component of T. If all points of {y, z}1- belong to Xs-, then replacing y by any point y' coUinear with y and opposite x yields a pair {y1', z} for which {y1', z}1- is not contained in x1-. Now y is adjacent to y', which is in the same connected component as z. • Lemma 11.2.7 Let S = (V, B, I) be a thick GQ not of order (2,2). Let {x, L} be a flag ofS, with x e V and L e B. Define the following graph (T, E). The vertex set T is the union of the set of points of S opposite x and the set of lines ofS opposite L. The edge set E is the set offlags ofS whose elements are both contained in T. Then the graph (F, E) is connected. Proof. It is clearly enough to show that every two vertices of (r, E) that correspond with points of <S can be connected by a path in (F, E). So let y, z be two points of S opposite x. Since the order of <S is unequal (2,2), we may assume, without loss of generality, that every point of S is incident with at least four lines. Since there is exactly one line through each point opposite x which is not opposite L, we see that the valency of y in (r, E) is at least 3. Similarly for z. We may obviously assume that y and z cannot be joined by a path of length 2 or 4 in (r, E). This means that y and z are not incident with a line opposite L, and that the intersection point (if it exists) of any line through y opposite L with any line through z opposite L is coUinear with x. We now assume first that projLy = yro)Lz, but proj^L ^ projzL. Since there is one line through each of y, z not opposite L, we see that there are at least three points xi, X2, ^3 not on L coUinear with all of x, y, z. Now let y' be a point coUinear with y, y' ^ y, such that x3Iyy' and such that y' is opposite x (hence belonging to T). Clearly, y and y' can be connected inside (r, E) (by a path of length 2). If y' is coUinear with z, then y',z,xz are the vertices of a triangle, contradiction. So y' is not coUinear with z, and at least one of the lines Li := projyZXj, i = 1,2, say L\, is not concurrent with L (since they cannot be equal). Also, the intersection point z' of Li and zx\ is opposite x. We now have the path (z,zz',z',Li,y',yy',y) of length 6 connecting z with y.
11.3. The Moufang Property and Analogues
225
If, still under the assumption that pro]Ly = projLz, we have proj^L = projzL, and so yz is a line, then we apply the previous paragraph to a point y* opposite x, collinear with projLy, and not on yz. We obtain that y and y* are connected in (T,E), and y* and z are connected in (T,E), hence y and z are connected by a path in (r, E). We now assume that projLy ^ projLz. Then the line M := projz(projJ/L) is opposite L and the point projMy is opposite x. Hence we have a path from z to projM,z inside (F, E). But by the foregoing, we also have a path from projMy to y, hence from z to y. The lemma is proved. • If S is a GQ of order (2,2), then it is easy to show that (T, E) of the previous lemma has exactly two connected components, each being an ordinary quadrangle in S.
11.3 The Moufang Property and Analogues We introduce some more notation. Let <S = (P,B, I) be a GQ. An apartment is an ordinary quadrangle in S. It consists of four points and four lines. A root is a set of five "consecutive" elements of an apartment. Hence a root contains either two points xi,x2 and three lines Li,L2,L3, with L1IX1IL2IX2IL3, or dually, it contains three points and two lines. It will be useful to distinguish between these dual notions. Therefore, we will call the former a root, and the latter (dually) a dual root. A root without its extremal lines will be called a panel, more precisely, the interior (panel) of the root. Similar but dual definitions for dual panels. So a panel is a set of two distinct collinear points, together with the joining line. Note that the Main Axiom of GQs implies that every pair of flags (a flag being an incident point-line pair) of <S is contained in at least one apartment of S. We are now ready to define the Moufang Property for generalized quadrangles. Let S be a generalized quadrangle with full collineation group G = Aut(S). Let 7r be a panel. Then we say that TT has the Moufang Property (or, equivalently, n is a Moufang panel) if for some root a with interior IT, the group G^l of collineations (called root elations and, dually, dual root elations) fixing every element incident with any element of n acts transitively on the set
226
Chapter 11. Moufang Quadrangles with a Translation Point
of apartments containing a. Let n = {x, y, L}, with x, y e V and L £ B. Let M be any line through x distinct from L. Suppose IT is a Moufang panel, and let a be a root with interior ix such that GH acts transitively on the set of apartments containing a. Let u, v! be two points on M distinct from x. Suppose the unique line of a incident with y and unequal L is N. Then N is opposite M and we may consider the distinct points u\ = proj^u and u[ = pro)Nu'. Using the Main Axiom of GQs, it is quite easy to see that MI and u[ determine unique apartments £ and £', respectively, containing a. Hence there is a collineation 9 e G^l mapping S to £', and hence mapping u\ to u[. But since M is fixed under 9 by assumption, 9 maps u to w', and hence, for every line M' through y distinct from L, 9 maps the (unique) apartment containing TT, U and M' to the (unique) apartment containing TY, v! and M'. We have shown that the definition of Moufang panel is independent of the root involved in that definition. We can even say a little more. Suppose that 9 e G^', with TT as above. Suppose that 9 leaves an apartment through n invariant. Then by Corollary 11.2.5, 9 is the identity. Hence, in general, G ^ acts freely on the set of apartments through a, with a a root with interior ir, and it acts sharply transitively on that set precisely when TT is a Moufang panel. If every panel and every dual panel of the GQ S has the Moufang Property, then we say that S has the Moufang Property, or that S is a Moufang GQ. If every panel is a Moufang panel, or if every dual panel is a Moufang dual panel, then we say that <S is a half Moufang GQ. If S is a Moufang GQ, then the group generated by all root elations and dual root elations is called the little projective group ofS. Note that a symmetry about a line L is automatically a root elation, and it will sometimes be called an axial root elation (with axis L). Dually we define central root elations (with center some point x). It is now clear that every line of a GQ in which every point is a translation point, is an axis of symmetry, and hence the GQ is half Moufang. Let {x, L} be a flag of the GQ S. Let Mix and ylL such that y is not incident with M. As above, it is easy to see with the aid of Corollary 11.2.5, that the group G^X'L^ of collineations fixing all lines through x and fixing all points on L acts freely on the set of apartments containing {x,y,L, M}. We call the flag {x, L} a Moufang flag if the group G^x^ acts transitively (and hence sharply transitively) on the set of apartments containing {x, y, L, M}. As before, one shows easily that this definition is independent of the chosen line M through x, M =£ L, and of the chosen point y on L, y ^ x.
23.3. The Moufang Property and Analogues
227
We see that the definition of a Moufang flag is a self-dual one, hence there is no need to introduce something like a "dual Moufang flag". If every flag of the GQ <S is a Moufang flag, then we call S 3-Moufang, where the number 3 refers to the length of the sequence (y, L, x, M) as a path in the incidence graph of S (which is the graph (V U B, I)). The 3-Moufang Condition is a completely self-dual one, and so one might think that a notion of "half 3-Moufang" cannot be defined. But actually, one can take a kind of greatest common divisor of the 3-Moufang Condition and the half Moufang Condition to obtain the following definition of a half 3-Moufang generalized quadrangle. Let {x, L} be a flag of the GQ S. Let zlMlx and KlylL such that z ^ x, K 7^ L and y is not incident with M. As before, the group df acts freely on the set of apartments containing {x, y, z,L,M}, and the group GK acts freely on the set of apartments containing {x, y, K, L, M). When we have transitivity for each z, then in the first case, we say that the flag {x,L}is half 3-Moufang at x, while in the second case we talk about half 3Moufang at L. The GQ S is called half 3-Moufang if either every flag {x, L} of S is half 3-Moufang at x, or every flag {x, L} of <S is half 3-Moufang at L. Now let x be any point of the GQ S. Recall that if the group G ^ of collineations fixing all lines through x acts transitively on the set of points of S opposite x (here, this is not necessarily a regular action!), then we say that x is a center of transitivity. Dually, we define an axis of transitivity. If all points are centers of transitivity, and all lines are axes of transitivity, then we say that S is a 2-Moufang GQ. If either all points are centers of transitivity, or all lines are axes of transitivity, then we call S a half 2-Moufang GQ. Let {x,L} again be a flag of the GQ S, with x a point and L a line. Another flag {y, M} is called opposite {x, L} if the point y is opposite x and the line M is opposite L. If the group GX,L of all collineations of S fixing the flag {x, L} acts transitively on the set of flags opposite {x, L}, then we say that {x, L} is a transitive flag. If all flags are transitive, then we say that S is a 1-Moufang GQ, or, equivalently, that <S satisfies the Tits Condition, or that <S is a Tits GQ. The last two names are motivated by their group theoretic counterparts, the Tits systems, see below. If S satisfies the Tits Condition, then the corresponding coUineation group G acts transitively on the set of ordered pairs of opposite flags of S. For convenience, we shall sometimes also refer to the (half) 4-Moufang Condition for a (half) Moufang GQ.
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Chapter 11. Moufang Quadrangles with a Translation Point
Let S = (V, B, I) be a Tits GQ with full collineation group G. Let {x, L} be any flag in S, with x e V and L e B. Set B := GXjL. Let £ be an apartment containing {x, L}. Define iV := Gs, the (setwise) stabilizer of £, and set H = Br\N. Then iJ is precisely the group of collineations fixing X elementwise. Hence H < N. Now, if there exists a nilpotent normal subgroup U of B such that f/i? = B, then we say that S has the FongSeitz Property. A Fong-Seitz GQ is a Tits GQ with this additional group theoretic property. From group-theoretic point of view this is a natural condition. Note that, a priori, nothing guarantees uniqueness of U, but one can actually show that it is unique (see De Medts, Haot, Tent and Van Maldeghem [48]); in the finite case it is not so difficult to see that U is the unique Sylow p-subgroup of B, when G contains the little projective group of the corresponding Moufang GQ — as we will later prove, every Fong-Seitz GQ is a Moufang GQ. Hence B is always a split extension of the nilpotent group U, and B = U:H. All Moufang GQs have been classified by Tits and Weiss in the monograph Moufang Polygons [229]. Technically, the classification was completed in 1997. In the finite case, the classification of Moufang GQs follows from the classification of Fong-Seitz GQs by Fong and Seitz [59, 60]. They show that the Moufang Condition implies the Fong-Seitz Property, and then they use entirely group theoretic methods to classify the Fong-Seitz GQs (in fact, they classify all Fong-Seitz generalized polygons, but the part of quadrangles is the largest by far). The proof is highly nontransparent from a geometric point of view, and it is approximately 100 pages long. Corollary 11.6.7 and Theorem 11.7.4 reduce in a completely geometric and elementary combinatorial group theoretic way the Fong-Seitz Condition to the Moufang Condition (including the infinite case!), and hence combined with the classification theorem of Tits and Weiss (mentioned above), this classifies all Fong-Seitz GQs. In the finite case, one can also appeal to the alternative geometric classification of Moufang quadrangles by Payne and Thas [128, Chapter 9], Kantor [85] and K. Thas [199], see also Thas [188]. All this replaces the bulk of the papers of Fong and Seitz [59, 60] by elementary methods, and generalizes it at the same time to the infinite case. In this chapter, we essentially prove that all the conditions introduced above, except for the Tits Condition and the half 2-Moufang Condition, are equivalent. In particular, this will imply that a GQ in which every point is a translation point is a Moufang GQ. Also, we will show that every Moufang GQ has a thick subGQ which is a TGQ. For all this, we will not need finiteness, so we prove things in the most general setting.
11.4. Tits Generalized Quadrangles and Tits Systems
229
Concerning the Tits Condition, it can be proved, using the classification of finite simple groups, that also this condition is, in the finite case, equivalent with the Moufang Condition; see Buekenhout and Van Maldeghem [33]. In the infinite case, there are a lot of counter examples, the most striking ones being without doubt those constructed by Tent [150]. But we will not use the classification of finite simple groups in this book, and so we will not prove the result of Buekenhout and Van Maldeghem mentioned above. We refer to work of K. Thas [209, 210, 211] on the classification of finite Tits GQs without the classification of finite simple groups. Concerning the half 2-Moufang Condition, it can be shown that it is also equivalent to the Moufang Condition, at least in the finite case. This has been done by K. Thas and Van Maldeghem [215], but the proof of that result is beyond the scope of this book.
11.4 Tits Generalized Quadrangles and Tits Systems As we saw in Chapter 3, a 4-gonal family is the group theoretic equivalent of the notion of an EGQ. Likewise, we will axiomatize group theoretically the notion of a Tits GQ. This gives rise to groups with a BN-pair, also called Tits systems. This notion was introduced by Tits [218, 220] for spherical buildings, and we specialize it here to the class of GQs. At this stage, it is convenient to introduce a generalized distance between flags of S as follows. Let F = {x, L} and F' = {x', L'} with x, x' e V and L,L' e B be two flags of S. Then 5(F,F') = z £ {-3,-2,-1,0,1,2,3,4}, where \z\ = \{d(x,x') + d{L,L')) and if \z\ ^ 4, then z has the same sign as d(x', x) + d(L', x) - d(x',L) - d(L', L) (distances are measured in the incidence graph of <S and the sign of 0 is "+")• Note that 8(F, F') = -5(F', F) if 5(F, F') is even but not equal to 4; otherwise 8{F, F') = S(F', F). In the next proposition we will be dealing with double cosets. It is convenient to think of these as unions of ordinary cosets. For instance, the double coset HgK for subgroups H, K of some group G and g e G consists of the union of all left cosets hgK for h ranging over H, or of all right cosets Hgk for k ranging over K. Note that, in contrast with ordinary cosets, double cosets need not contain an equal number of elements. For instance, \HgH\ > \HH\ whenever g does not normalize H. We will also consider a quotient group G/N as consisting of cosets of TV in G. With this convention, it makes sense to multiply elements of the
Chapter 11. Moufang Quadrangles with a Translation Point
230
quotient group with elements or subgroups of G; it is just the ordinary multiplication of subsets in a group. Proposition 11.4.1 Let S = (V, B, I) be a thick Tits generalized quadrangle and let G be a collineation group of S such that G acts transitively on the set of ordered pairs of opposite flags of S. Let F = {x,L} be any flag in S, with x e V and L e B, and let E be an apartment containing F. Define B := GxM N := G E and H = B n N as above. Then N n (Gx\ GL) and N n (GL \ Gx) are nonempty. Choose arbitrarily sx and sL, respectively, in these sets. Then (G, B, N) and SX,SL satisfy the following properties. (BN1) G=
(B,N).
(BN2) H < N andW := N/H = (sxH, sLH) is isomorphic to Dih 8 , the dihedral group of order 8. (BN3) BsBwB
C BwB U BswB, for allw EN and s e {sx, sL}.
(BN4)
sBs^B,fors€{sx,sL}.
Also, B does not contain any nontrivial normal subgroup ofG and H coincides with
P| w~lBw.
Proof. Let F' = {x1, L'}, x' e V, V e B, be the flag in E that is opposite F. Also, let M ^ L and M' / L' be the "second" lines in E incident with x and x', respectively. By the Tits Property there is some element sx e Gx mapping the pair ({x, L}, {x', L'}) to the pair ({x, M}, {x\ M'}). Clearly sx stabilizes E and belongs toJVfl (Gx \ GL). Similarly we find sL e N n (GL \ Gx). We may treat the elements sx and SL as arbitrary elements in N n (Gx \ GL) and sL e N n (GL \ Gx), respectively. Remark that the Tits Property is equivalent with G being transitive on the set of flags of S, together with requiring that B acts transitively on the set of apartments containing F. We now show that N acts transitively on the flags of E. By flag transitivity, there is a collineation g mapping any flag F' of E onto F, and by our previous remark, there is a collineation g' e B mapping E 9 back to E. Then gg' e N and maps F' onto F. We now show (BN1). Let g e Gbe arbitrary. Let E' be an apartment of S containing the flags F and F9. By the Tits Property there is some gx e B
11.4. Tits Generalized Quadrangles and Tits Systems
231
mapping E' to E. Since N acts transitively on the flags of E, we can find w£N with {F991 ) w = F. Hence ggxw € B and so G = BNB. We turn to (BN2). The action of N on E is clearly isomorphic to a subgroup of the natural action of Dih 8 on a square. By our remark above, this action in fact coincides with that natural action. Since Dih 8 acts sharply transitively on the "flags" of the corresponding square, we see that H = B n N is the elementwise stabilizer of E, and hence a normal subgroup of the setwise stabilizer N. Moreover, H is the kernel of the permutation group (N, E) (S seen as a set of flags) and as such N/H ^ Dih 8 . Returning to our proof above of (BN1), one sees that, if we define wz e N/H as the coset of H in N whose elements map F onto the flag of E at generalized distance z, then an arbitrary element g e G that maps F onto a flag at generalized distance z from F belongs to BwzB (which defines a unique double coset of B since H < B). Conversely, it is easy to see that every element of BwzB maps F to a flag F' at generalized distance S(F, F') — z from F. Hence G is the disjoint union of the double cosets BwzB, for z £ {-3, - 2 , - 1 , 0 , 1 , 2 , 3 , 4 } . Now let F' be a flag with 6(F, F') = i. If i < 0, then one easily sees that S(F, F'SL) = 1- i. If i > 0, then either 5{F,F'SL) = 1- i, or S(F, F'SL) = i. Now, one checks that wi-i = W%SL = WiW\. This shows BWBSLB C BWB U BWSLB, for all w e W. Taking inverses, (BN3) for s = SL follows. Similarly, (BN3) is also true for s = sx (and sxH = w-\). Regarding (BN4), we may choose a g e B which does not fix the unique point incident with L contained in E and distinct from x. This is possible by the thickness assumption. For this g, the image of F under SLgsL clearly differs from F (and lies at generalized distance 1 from F). Hence sLBsL ^ B. Similarly sxBsx ^ B. If K < G is contained in B, then K = K9 < B n B3, for every g e G. Hence K stabilizes every flag F9, for g ranging over G, and so K fixes every flag (by flag transitivity of G) and must therefore be trivial. Since w~1Bw is the stabilizer of the flag Fw, and Fw ranges over all flags of E as w ranges over N, every element of f]weN w~1Bw belongs to N and hence to H. Conversely, every element of H fixes every flag of E and hence belongs to w~1Bw, for all w e N. The proposition is proved.
•
The group W is called the Weyl group of (G,B,N), and the elements sx and SL representatives of the standard generators ofW.
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Chapter 11. Moufang Quadrangles with a Translation Point
We now would like to prove a converse of the previous proposition. To that aim, we define the notion of a group with a BN-pair of Type B 2 , also called a Tits system of Type B 2 . Since we will not encounter any other types of such groups, we will shortly talk about groups with a BN-pair and Tits systems. Let G be a group, and let B and N be two subgroups of G. Set H = B n N. Then (G, B, N) is called a Tits system, or (B, N) is called a BN-pair in G, or G is called a group with BN-pair (B, N), if there exist elements sx and SL of G, with {s2, s |} C if, such that (BN1) up to (BN4) of Proposition 11.4.1 hold. The group W in (BN2) will be called the Weyl group and the cosets sxH and s^H the standard generators ofW. If S is a Tits GQ, and (G,B,N) is as in Theorem 11.4.1, then we call (G, B, N) a natural Tits system associated with S. If (G, P, AT) is a Tits system with Weyl group W and if s^ and sL are corresponding representatives of the standard generators of W, then we define the following incidence structure SG,B,N = (P, &, I)- Define Px = (B, BSx) andP L = {B,BSL). • POINTS. The elements of V are the right cosets of Px. • LINES. The elements of B are the right cosets of PL. • INCIDENCE. For g,h e G, the point Pxg is incident with the line PLh if Pxg fl P^h ^ 0 (and in this case we may choose g = h). The group G acts (on the right) as a collineation group on SG,B,N, and it is flag transitive since every flag can be written as {Pxg, Ptg} (and is the image under g of the "standard" flag {PXl PL}). If we denote the point Px by x, then the point Pxg can be written as x9. Similarly, we write PL as E, and every line can be written as L9, for some g e G. Before continuing, we prove a lemma, which will later lead to the Bruhat Decomposition. First a definition. Let W be the Weyl group of some Tits system, and let sx,sL be representatives of the standard generators. Then the minimum number of factors needed to write an element w e W as a product of sxH and SLH is called the length of w, and denoted £(w). Since W is a dihedral group of order 8, we have £(w) e {0,1,2,3,4}, for an arbitrary element w oiW.
11.4. Tits Generalized Quadrangles and Tits Systems
233
Lemma 11.4.2 If (G,B,N) is a Tits system with Weyl group W, then for w, w' G W, one has BwB = Bw'B if and only ifw = w'. Proof. Clearly, if w = w', then BwB = Bw'B. Now assume that w ^ w' and assume that BwB = Bw'B. First suppose that w = s is a standard generator of W. Note that B C BsBsB, since sBsB contains sHsH = H < B. Then (BN3) tells us that B C BsBsB = BsBw'B C Bsw'B U Bw'B. Hence either sw' C B or w' C B, implying that either w' = s or w' is the identity. The former contradicts our hypothesis, the latter implies BsB = B, implying s is the identity, again a contradiction. So we may assume that 1 < £(w) < £(w'). Let w = sw", with s a standard generator, and £{w") = £(w) - 1. Then Bsw"B = Bw'B implies the existence of b, b' G B with sw" = bw'b', hence w" = s~1bw'b' G BsBw'B C Bsw'BuBw'B. So either Bw"B = Bsw'B, or Bw"B = Bw'B. An easy induction on the length of w implies now that either w" = sw', contradicting w = sw" as s is an involution, or w" = w', contradicting £{w') > £(w) > £(w"). The lemma is proved.
•
Now let s e {sx,sL} and note that (BN3) implies BsBsB c BsBUB. Since (BN4) says that BsBsB ^ B, we deduce that at least one element w of BsB belongs to BsBsB, and hence multiplying w at the left and at the right with B, we obtain BsB C BsBsB. We now see that BsBsB = BsBUB. It follows that BSXB\JB = BsxBsxB = Px and BsLBuB = BsLBsLB = PL. We now claim that G is generated by Px and PL. Indeed, by the previous paragraph, sxH c Px and sLH c PL. Also, (BN2) implies that N is generated by sxH and sLH, and so N < (PX,PL)- Hence, since B < Px n PL, we obtain (B, N) < (Px, PL} and (BN1) completes the proof of our claim. Next we claim that Px n PL = B. Indeed, since sx £ B, the double cosets BsxB and B are distinct, and likewise BsLB and B are distinct. Also, Lemma 11.4.2 implies that BsxB ^ Bs^B. The claim is now clear. As a consequence, we may identify the flags of SatB,N with the right cosets of B. Denote by £ the subgeometry induced by the set of flags {B, Bsx,
BSL,BSXSL,
BSLSX,
BSXSLSX,
BSLSXSL,BS
SL}-
Just as we showed BsxB ^ BSLB above using (BN2) and (BN3), the same conditions also guarantee that those eight flags are different. Notice that
Chapter 11. Moufang Quadrangles with a Translation Point
234
(using the equality Dih 8 )
BSXSLSXSL
=
which follows from W =
BSLSXSLSX,
B
QPxsLnPL, C PxsL
BSLSXSL BsxsLsxsL BSXSLSX
DPLsxsL,
Q PXSLSXSL
n
C PXSLSXSL
n PL
PLSXSL,
C PX ^x)
BsLsx Bsx
C Px
nPL CPXDPLSX.
Hence E induces a quadrangle in SG,B,N, if we can show that the point x is not collinear with X3LSXSL and dually, that the line L is not concurrent with t 8 ' 8 1 8 ' (and then it follows that xSL is not collinear with xSLSx by applying SL to both points and remarking that BSLSXSLSL C BSLSXH = BHSLSX = Sx BSLSX, and likewise for the lines i and LS:cSL). Now PXSLSXSL
= BSXBSLSXSL
U
C BsxsLsxsLB
U
BSLSXSL BSLSXSLB,
which follows from multiplying the right hand side of the first line at the right with B and then applying (BN3) with s = sx and w = sLsxsL. Any line P^g incident with Px shares a coset of B with Px. The cosets of B in Pj; are of the form B and Ps^fe, with b e P . Hence we may put # = s x 6 (leaving the more trivial case of g e P to the reader). Hence the coset P L # is equal to Bsxb U BsLBsxb, and hence belongs to the union of the double cosets BsxB\JBsLsxB (indeed, (BN3) implies that BsLBsxB C BsLsxBU BsxB; taking inverses of the left hand side, applying (BN3) again, and then again taking inverses, we deduce BSLBSXB C BSLSXB U BSLB. SO BsLBsxB c BsLsxB by taking intersections, using Lemma 11.4.2). This union should now contain a right coset of P that is also contained in one of the double cosets BSXSLSXSLB or BSLSXSLB. This implies, however, that one of these double cosets coincides with one of P s ^ P or BSLSXB, contradicting Lemma 11.4.2. The dual argument is similar, and so we conclude that E is a quadrangle. It is easy to see that sx and sL preserve E and act as generating reflections of the dihedral action of N on E.
11.4. Tits Generalized Quadrangles and Tits Systems
235
Now we claim that G = BNB. Since G = (B, N), it suffices to show that every element bnb'n'b", with b, b', b" G B and n, n' e N, belongs to BNB. Noticing that every element of N can be written as a finite product of sx and SL times an element of H < B, the claim easily follows from (BN3) using induction on the length £(n) of n. Together with Lemma 11.4.2, this gives rise to the Bruhat Decomposition, which states that
G = (j BwB. wGW 9
Consider any flag F . This flag corresponds to the coset Bg. By the foregoing paragraphs we can write g = b'nb, with b,b' G B and n G N. Hence Bg = Bnb and so b^1 fixes F and maps F9 into E. It now follows rather easily that the geometry SG,B,N does not contain any digon {xi,x2,Li,L2}, with xi,x2 G V and Li,L2 G B with XilLj, i,j G {1,2}. Indeed, by flag transitivity, we may assume x\ = xaxAL\ = L. By the previous paragraph, we may also assume that the flag {x2,L2} is contained in E, a contradiction. Now suppose that SQ,B,N contains a triangle xilL3Ix2lLi!x3lL2Ixi, with the x% points and the Lt lines, i = 1,2,3. We may again assume x\ = x and L2 = L, and by the previous paragraph, we may assume that the flag {x2,Li} belongs to E. Since E is a quadrangle, we must have x2 = xSLSxSh. But then L 3 is incident with both x and xSLSxSL, contradicting the fact that these points are not even collinear. Now let x' be any point not incident with some line L'. We may assume x' — x and L' G E. It follows that x' is incident with at least two lines (the lines through x in E), that L' is incident with at least two points, and that there is a flag {y, M} with x'lMlylL'. The flag {y,M} is unique by the previous paragraph. Hence we have shown that SG,B,N is a GQ (note that by flag transitivity, it cannot be a grid or a dual grid without order). The group G acts as a collineation group, by multiplication at the right, on SQ,B,N- Since every element of the kernel K of that action must in particular fix B, we see that K is a normal subgroup of G contained in B. Conversely, if K' < G, and K' < B, then, since the stabilizer of the flag Bg, g G G, clearly coincides with B9, the group K' stabilizes every flag of SG,B,N, and hence belongs to the kernel K. So K is the biggest normal subgroup of G contained in B, and as such equal to K = p | Bg. geG
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Chapter 11. Moufang Quadrangles with a Translation Point
The group G/K acts faithfully on SG,B,N and the stabilizer of the flag F is B/K. Now we determine the stabilizer of E. Since N stabilizes E and acts transitively on the flags of E, it suffices to determine the elementwise stabilizer S of E, and then NS is the global stabilizer of E. So suppose some b e B stabilizes E. Since E is uniquely determined by the opposite flags F and FSxSLSxSL, this is equivalent with b e B fixing FSxSLS*SL. Hence S = B n B S i S i S i S l . Note that then b automatically belongs to Bw, for every w &W. Hence we can also write S=
p|
BW.
w€W
We have proved the following theorem. Theorem 11.4.3 Let (G, B, N) be a Tits system with Weyl group W. Then the geometry SG,B,N defined above is a generalized quadrangle satisfying the Tits Condition. Setting K= f]B9 geG
andS=
f]
B