On the geometry of the exceptional group G2(q), q even Antonio Cossidente Dipartimento di Matematica Università della Basilicata I-85100 Potenza – Italy cossidente@unibas.it Oliver H. King School of Mathematics & Statistics Newcastle University Newcastle Upon Tyne NE1 7RU, U.K. O.H.King@ncl.ac.uk 1 Proposed Running Head: On the geometry of the exceptional group G2 (q), q even Corresponding Author: Oliver H. King School of Mathematics & Statistics Newcastle University Newcastle Upon Tyne NE1 7RU, U.K. O.H.King@ncl.ac.uk 2 Abstract We study some geometry of the exceptional group G2 (q), q even, in terms of symplectic geometric configurations in the projective space P G(5, q). Using the spin representation of Sp6 (q), we obtain an alternative description of the Split Cayley hexagon H(q) related to G2 (q). We also give another geometric proof of the maximality of G2 (q), q even, in P Sp6 (q). Keywords: exceptional finite groups of Lie type, spin representation, symplectic polarity, generalized hexagon. MSC 2000: 51E20, 20G40. 1 Introduction Let K be a field. Let Q(6, K) be a nonsingular quadric in the projective space P G(6, K) that has on it totally singular planes; for finite K, such a quadric is unique up to equivalence. Let P Ω7 (K) be the projective commutator subgroup of the orthogonal group of Q(6, K). The Cartan–Dickson–Chevalley exceptional group G2 (K) is a subgroup of P Ω7 (K). It occurs as the stabilizer in P Ω7 (K) of a configuration F of points, lines and planes on Q(6, K). The geometry of F was explored first by J. Tits [24] and later, from a somewhat different standpoint, by R.H. Dye [10]. Both approaches involve the geometry associated with the triality of a hyperbolic quadric in P G(7, K). Tits, in particular, obtained a lot of information on the action of G2 (K) by identifying the stabilizer of F with G2 (K) . The group G2 (K) is also contained in the automorphism group of a classical generalized hexagon, called the split Cayley hexagon and denoted by H(K). The points of H(K) are all the points of Q(6, K) and the lines of H(K) are certain lines of Q(6, K). When K is the finite field GF (q), the hexagon is denoted H(q) and has order (q, q), the number of points of Q(6, q) is (q 6 − 1)/(q − 1), and this is also the number of lines of Q(6, q) involved in H(q). For more details, see [25]. When q is even, the orthogonal group P Ω7 (q) is isomorphic to the symplectic group P Sp6 (q), so that G2 (q) may be regarded as a subgroup of P Sp6 (q). The spin representation of Sp6 (q) provides an embedding of P Sp6 (q) in the orthogonal group P Ω+ 8 (q). We shall study G2 (q) as a subgroup of P Sp6 (q) and explore some geometric interrelations between the spin representation of P Sp6 (q) and the geometry of quadrics of P G(5, q) invariant 3 under P Sp6 (q). We shall define special planes and special lines of P G(5, q) that allow us to give an alternative description of the generalized hexagon H(q). This is not a completely new approach but, nevertheless, we think it is of some interest. It is known that G2 (K) is a maximal subgroup of P Ω7 (K). A very nice geometric proof was given by Dye in [10]. Here we will provide an alternative proof when K = GF (q), q even. 2 The symplectic geometry of P G(5, q), q even In this section we shall describe some geometry of the projective space P G(5, q), equipped with a symplectic polarity Λ. We assume from now on that q is even. Let G be the group P Sp6 (q) of Λ (isomorphic to Sp6 (q) since q is even). Although P Sp6 (q) contains a number of subgroups isomorphic to G2 (q), we write G2 (q) for a fixed representative. The number of totally isotropic lines and planes in P G(5, q) is known but, given the nature of some of the calculations given later, it is perhaps more illuminating to derive these numbers. Given a point P of P G(5, q), the totally isotropic lines and planes through P correspond to points and totally isotropic lines in a non-degenerate subspace Σ complementary to P in P ⊥ . These latter each number (q 4 −1)/(q−1), with each point of Σ lying on q + 1 totally isotropic lines. Counting in different ways the numbers of combinations P ∈ ℓ and P ∈ π, where ℓ, π are totally isotropic lines and planes in P G(5, q) respectively, leads directly to there being (q 2 + 1)(q 3 + 1)(q 2 + q + 1) totally isotropic lines and (q + 1)(q 2 + 1)(q 3 + 1) totally isotropic planes. When q is even, an orthogonal polarity is a symplectic one. The group G acts on the set Φ of all quadrics of P G(5, q) inducing the polarity Λ and contains the corresponding orthogonal groups. Let V be the 6–dimensional vector space underlying P G(5, q), and let {e1 , e2 , e3 , e4 , e5 , e6 } be a basis of V so that the basic nonsingular alternating form B has the canonical coordinate form B(x, y) = 3 X (xi y3+i + yi x3+i ). (1) i=1 A quadratic form Q on V has B for its corresponding polar form if and 4 only if B(x, y) = Q(x + y) + Q(x) + Q(y). (2) From (1) it follows that Q is a quadratic form for a quadric in Φ, with associated alternating form B, if and only if Q(x) = x1 x4 + x2 x5 + x3 x6 + 6 X αi x2i , i=1 for some αi ∈ GF (q). Dye shows in [8] that two members Q1 , Q2 of Φ (with corresponding quadratic forms Q1 , Q2 associated with B) are equivalent, i.e., have the same Witt index, if and only if there is a transvection of the group Sp6 (q) of B transforming Q1 into Q2 . The choices for the αi ’s lead to q 6 distinct quadratic forms associated with B. The group G has two orbits on Φ, [6, Theorem 14 ]: an orbit of size q 3 (q 3 + 1)/2 consisting of maximal Witt index quadratic forms and denoted by HG , and an orbit of size q 3 (q 3 − 1)/2 consisting of non–maximal Witt index quadratic forms and denoted by EG . From a projective point of view, members of HG give rise to hyperbolic quadrics Q+ (5, q), and members of EG give rise to elliptic quadrics Q− (5, q). The Arf invariant of Q with respect to the basis {e1 , e2 , e3 , e4 , e5 , e6 } is ∆(Q) = 3 X Q(ei )Q(e3+i ). i=1 Let L = {λ2 + λ : λ ∈ GF (q)}, the image of the homomorphism λ 7→ λ2 + λ of the additive group GF (q)+ of GF (q) whose kernel is GF (2). Thus L is ¯ a subgroup of GF (q)+ isomorphic to GF (q)+ /GF (2)+ . Write ∆(Q) for the class of ∆(Q) modulo L. ¯ Dye shows in [7], [8] that ∆(Q) is independent of the choice of the symplectic base, and two quadratic forms associated with B, say Q1 and Q2 , are ¯ 1 ) = ∆(Q ¯ 2 ). In particular, Q has maximal Witt equivalent if and only if ∆(Q index if and only if ∆(Q) ∈ L. Since q is even, the group G2 (q) contains maximal subgroups isomorphic to SL3 (q) · 2, (c.f., [3], [22]), and there is a single conjugacy class C1 of these subgroups in G2 (q); indeed there is a single conjugacy class of such subgroups in P Sp6 (q). With respect to an appropriate basis,matrices A  A 03 and there from SL3 (q) act as the block diagonal matrix 03 (A−1 )t 5   03 I3 is an involution g0 = normalizing SL3 (q). Geometrically we I3 03 have the full stabilizer in P Sp6 (q) of a pair of uniquely determined, disjoint, totally isotropic planes π1 and π2 (c.f., [16]). In fact there is a unique quadric Q ∈ HG containing π1 and π2 , and the stabilizer in P Sp6 (q) of {π1 , π2 } lies inside the orthogonal group P O6+ (q) of Q where, by [16], it is also a maximal subgroup. Proposition 2.1. The group G2 (q) acts transitively on HG . Proof. It follows from the preceding discussion that to each group in S ∈ C1 there corresponds a unique quadric Q ∈ HG stabilized by S. Since S is a maximal subgroup of G2 (q) and since P O6+ (q) has no subgroups isomorphic to G2 (q), we see that S is the full stabilizer of Q in G2 (q) and that the group P O6+ (q) of Q contains no other member of C1 . The action of G2 (q) on C1 by conjugation is equivalent to an action on HG . The number of subgroups in C1 is |G2 (q)|/|(SL3 (q) · 2| = q 3 (q 3 + 1)/2 = |HG |. Hence G2 (q) acts transitively on HG . Let Q ∈ HG . Maximal totally singular subspaces of Q are planes which fall in two classes L and G (Latin and Greek) that under the Plücker map correspond to planes (the set of all lines in a plane) and line–stars (the set of all lines through a point) of P G(3, q), respectively. Two distinct planes on Q belonging to the same family meet in a point, and planes belonging to different families are either disjoint or meet in a line. Given a point P of Q, its perpendicular space is the join of P to a non-degenerate 3-space Σ. The intersection of Q and Σ is a hyperbolic quadric QΣ containing two families of lines, each of size q + 1. One family corresponds to L and the other to G: two lines from one family are disjoint while two lines from different families meet in a point. The group of QΣ is isomorphic to P SL2 (q) × P SL2 (q) with the factors acting independently on the two families of lines on QΣ (c.f., [12], [23]). Definition 2.2. Suppose that Q ∈ HG and that π1 , π2 are the disjoint planes on Q such that the stabilizer in P Sp6 (q) of π1 , π2 lies in G2 (q). Then π1 , π2 are said to be prescribed planes for Q. A plane π on Q is said to be special for Q if it meets each of π1 , π2 . A plane on Q that is neither prescribed for Q nor special for Q is said to be non-special for Q. 6 Given that π1 , π2 are in different families, if π is special for Q, it meets a πi in a point P and πj (i 6= j) in the line P ⊥ ∩ πj . If π is non-special for Q, then it meets one of π1 , π2 in a point and is disjoint from the other. Each Q ∈ HG has exactly two prescribed planes, these comprising the pair determined by the subgroup in C1 corresponding to Q. Proposition 2.3. Suppose that Q ∈ HG and that π1 , π2 are the prescribed planes for Q. Then the stabilizer, S, of {π1 , π2 } in G2 (q) acts transitively on the set of prescribed planes for Q, on the set of special planes for Q and on the set of non-special planes for Q. Proof. The group S is isomorphic to SL3 (q).2 and contains elements switching π1 and π2 so is transitive on the prescribed planes for Q. Given that S is transitive on the points of π1 ∪ π2 and that the set of special planes for Q is given by {P + (P ⊥ ∩ πj ) : P ∈ πi , j 6= i}, this set forms a single orbit of S. Now let P be a point of π1 and let Σ be a non-degenerate 3-space complementary to P in P ⊥ . Then π1 ∩ Σ and π2 ∩ Σ are lines in the same family in QΣ and the remaining lines of this family are precisely the intersection of QΣ with planes on Q that are non-special for Q. As noted above, the group of QΣ is isomorphic to P SL2 (q) × P SL2 (q) with the factors acting independently on the two families of lines on QΣ . The subgroup structure of P SL2 (q) is well known (c.f., [5], [14] or [18]): the stabilizer of an ordered pair of lines in one family is isomorphic to Cq−1 × P SL2 (q) and acts transitively as the cyclic group of order q − 1 on the remaining lines of this family. Thus the non-special planes for Q that contain P lie in the same orbit under S. Given transitivity on the points of π1 ∪ π2 , it follows that S acts transitively on the non-special planes for Q. We shall show that if a plane π on a quadric Q of HG is special for Q, then it is special for every quadric in HG that contains it. We show first that the properties: prescribed, special and non-special exhibit a certain amount of regularity. Proposition 2.4. Each totally isotropic plane of P G(5, q) lies on exactly q 3 quadrics in HG . Amongst the planes lying on at least one quadric in HG as a prescribed plane, each plane lies on the same number of quadrics as a prescribed plane. Corresponding statements apply to planes lying on at least one quadric in HG as a special plane and to planes lying on at least one quadric in HG as a non-special plane. 7 Proof. Since P Sp6 (q) preserves HG and also, by Witt’s Theorem, acts transitively on totally isotropic planes of P G(5, q), each totally isotropic plane lies on the same number of quadrics in HG . Let us consider the plane π given by x1 = x2 = x3 = 0: it lies on Q ∈ Φ if and only P if the corresponding quadratic form is given by Q(x) = x1 x4 + x2 x5 + x3 x6 + 3i=1 αi x2i , for some αi ∈ GF (q) (and necessarily then Q ∈ HG ). There are q 3 possible choices for the αi ’s so exactly q 3 members of HG containing π. Suppose that π is a plane that is a prescribed plane for k > 0 quadrics in HG say, Q1 , Q2 , . . . , Qk . Then for any g ∈ G2 (q), the quadrics for which πg is a prescribed plane are precisely Q1 g, Q2 g, . . . , Qk g. Given that G2 (q) acts transitively on HG and that the stabilizer in G2 (q) of Q acts transitively on the planes of each type for Q, the number of quadrics for which π is prescribed is independent of the choice of π. The same argument applies to special planes and to non-special planes. We use the following result of Cooperstein to identify the orbits of G2 (q) on totally isotropic planes; later we shall do something similar for lines. Result 2.5 ([3], Lemmas 5.1, 5.2, 5.3). The group G2 (q) has the following properties: (a) It is transitive on points of P G(5, q); (b) It has exactly two orbits on totally isotropic planes of P G(5, q), of lengths (q 6 − 1)/(q − 1) and q 3 (q 3 + 1); (c) It has exactly two orbits on isotropic lines of P G(5, q), of lengths (q 6 − 1)/(q − 1) and q 2 (q 6 − 1)/(q − 1). Proposition 2.6. Suppose that Q ∈ HG and that π is a plane on Q. If π is a special plane for Q, then it is special for every quadric in HG that contains it. If π is prescribed or non-special for Q, then it is prescribed for one quadric in HG that contains it, and it is non-special for the other q 3 − 1 quadrics in HG that contain it. Proof. Let S1 , S2 , S3 denote the sets of totally isotropic planes of P G(5, q) that are respectively prescribed, special and non-special for at least one quadric in HG . Then each Si is an orbit of totally isotropic planes of 8 P G(5, q) under G2 (q) and each totally isotropic plane lies on at least one quadric so lies in at least one Si . The total number of totally isotropic planes is (q 3 + 1)(q 2 + 1)(q + 1). By Result 2.5, G2 (q) has exactly two orbits on the set of such planes. Therefore one type of plane (prescribed, special or non-special) must have the same property for each quadric that it lies on. In each case we can count the number of combinations of a quadric Q ∈ HG and a plane π having the required property for Q, noting that a given quadric in HG has 2 prescribed planes, 2(q 2 + q + 1) special planes (one for each point in a prescribed plane for Q) and thus 2(q 3 − 1) non-special planes (there being 2(q 3 + q 2 + q + 1) planes on a hyperbolic quadric). • If a prescribed plane is always prescribed, then |S1 | = (q 3 + 1)(q 3 /2).2 = q 3 + 1. q3 • If a special plane is always special, then |S2 | = (q 3 + 1)(q 3 /2).2(q 2 + q + 1) = (q 3 + 1)(q 2 + q + 1). q3 • If a non-special plane is always non-special, then |S3 | = (q 3 + 1)(q 3 /2).2(q 3 − 1) = q 6 − 1. q3 Only in the second case do we get a known orbit length. Hence special planes are always special, and form an orbit of length (q 6 − 1)/(q − 1). The other orbit is S1 = S3 and has length q 3 (q 3 + 1), from which we deduce that a prescribed plane is prescribed for one quadric and non-special for q 3 − 1. Following Proposition 2.6, we can define a special plane of P G(5, q): Definition 2.7. A totally isotropic plane is special if it is special for every quadric in HG that contains it, or (equivalently) if it is special for at least one quadric in HG . We now turn to properties of isotropic (necessarily totally isotropic) lines of P G(5, q). 9 Proposition 2.8. Each isotropic line in P G(5, q) lies on exactly q 4 quadrics in Φ. Of these, q 3 (q + 1)/2 are in HG and q 3 (q − 1)/2 are in EG . Proof. Since P Sp6 (q) acts transitively on isotropic lines and preserves each of HG and EG , each isotropic line lies on a fixed number of hyperbolic quadrics and a fixed number of elliptic quadrics. Consider the line ℓ given by x1 = x2 = x3 = x4 = 0. A quadratic form of a quadric Q ∈ Φ contains ℓ if and only if 4 X Q(x) = x1 x4 + x2 x5 + x3 x6 + αi x2i , i=1 4 for some αi ∈ GF (q). There are q choices for the αi ’s so this is the number of quadrics containing ℓ. The Arf invariant of Q is ∆(Q) = α1 α4 , unaffected by choices for α2 and α3 and lying in L for q(q + 1)/2 combinations of α1 and α4 . Hence, of the q 4 quadrics containing ℓ, q 3 (q + 1)/2 are in HG and q 3 (q − 1)/2 are in EG . We identify four types of line on a quadric in HG : Definition 2.9. Suppose that Q ∈ HG . A line ℓ on Q is said to have type A, B, C or D for Q as: A: ℓ lies on a prescribed plane of Q; B: ℓ meets each prescribed plane of Q; C: ℓ meets (but is not on) exactly one prescribed plane of Q; D: ℓ meets neither prescribed plane of Q. Proposition 2.10. Suppose that Q ∈ HG and that S is the stabilizer in P Sp6 (q) of the pair of prescribed planes {π1 , π2 } for Q. Then S acts transitively on the lines of each type for Q, and the numbers of lines in each orbit are as follows: A: 2(q 2 + q + 1); B: (q 2 + q + 1)(q + 1); C: 2(q 2 + q + 1)(q 2 − 1); 10 D: q 2 (q 3 − 1). Proof. We have already observed that S acts transitively on the points of π1 ∪π2 and that a point P ∈ πi determines a line P ⊥ ∩πj on πj (j 6= i). The transitivity on lines of type A and the length of the orbit follow immediately. Further, any line of type B is determined by a point P ∈ π1 and one of q + 1 points in P ⊥ ∩ π2 ; the stabilizer in S of P acts transitively on the points of P ⊥ ∩ π2 . We thus have the required transitivity and orbit length. For a line of type C, the point P of intersection with a prescribed plane can be chosen in 2(q 2 + q + 1) ways. Given a non-degenerate 3-space Σ complementary to P in P ⊥ , the line is determined by a point on a hyperbolic quadric QΣ not lying on either of the lines πi ∩ Σ on QΣ . There are q 2 − 1 such points and so there are 2(q 2 +q +1)(q 2 −1) lines of type C. As discussed in the proof of Proposition 2.3, the stabilizer in P Ω+ 4 (q) of the ordered pair of lines (π1 ∩ Σ, π2 ∩ Σ) has structure Cq−1 × P SL2 (q) and acts transitively on the q 2 − 1 points, from which it follows that lines of type C form a single orbit under S. There are (q 3 + q 2 + q + 1)(q 2 + q + 1) lines in total on Q so there are q 2 (q 3 − 1) lines of type D. Any line of this type may be transformed under S into a line of QΣ relative to some given point of a prescribed line. This time we use the action of Cq−1 × P SL2 (q) on the lines of QΣ : it is transitive on the q − 1 lines disjoint from a given pair. Thus the lines of type D form an orbit under S. Proposition 2.11. Suppose that Q ∈ HG and that ℓ is a line of type B for Q. Then ℓ is of type B for every quadric in HG that contains it. Further, the lines of type B for some quadric in HG form a single orbit under G2 (q) of length (q 6 − 1)/(q − 1). Proof. By Result 2.5, G2 (q) has exactly two orbits on the isotropic lines of P G(5, q). Suppose that O is the orbit of length (q 6 − 1)/(q − 1). Since G2 (q) acts transtively on HG , the number of lines of O that lie on a quadric Q ∈ HG is independent of the choice of Q. Let us denote this number by k and count the number of pairs (Q, ℓ), where Q ∈ HG , ℓ ∈ O and ℓ lies on Q. If we select first the line ℓ, then the number of such pairs is [(q 6 − 1)/(q − 1)][q 3 (q + 1)/2], whereas if we first select the quadric, the number is [q 3 (q 3 + 1)/2]k. Thus k = (q 2 + q + 1)(q + 1). The set of lines of O 11 that lie on quadric Q must be a union of one or more of the orbits of S on the lines on Q. The only way in which this can occur is if the lines in this set are precisely those of type B for Q. It then follows that such lines are of type B for every quadric in HG that contains them. The previous proposition motivates the following definition: Definition 2.12. An isotropic line is special if it is of type B for every quadric in HG that contains it, or (equivalently) if it is of type B for at least one quadric in HG . In summary: we have special planes and special lines (the same number of each) defined in terms of hyperbolic quadrics. Proposition 2.13. Each special plane of P G(5, q) contains q + 1 special lines. Each special line of P G(5, q) lies on q + 1 special planes. Proof. Let π be a special plane of P G(5, q), then π lies on a quadric Q of HG and meets each of the prescribed planes π1 , π2 of Q, one (say πi ) in a point P and the other (πj ) in the line P ⊥ ∩ πj . The special lines on π are precisely the lines through P and they number q + 1. A special line on Q ∈ HG lies on 2 special planes on Q (a line on a hyperbolic quadric lies on exactly two planes, one Latin and one Greek, and each such plane must be special). Let s be the number of special planes containing a given special line ℓ. We can count in two ways the number of pairs (Q, π) with Q ∈ HG and π a special plane on Q containing ℓ, recalling that each special line and each special plane lies on q 3 (q+1)/2 and q 3 quadrics in HG respectively: [q 3 (q + 1)/2].[2] = sq 3 , which tells us that s = q + 1. 3 The spin representation of Sp6(q), q even and its geometry Here we recall the geometric construction of the spin module for the group Sp6 (q), q even, as described in [4] (based on work by A. Brouwer [2] and 12 R. Gow [11] ). Let I be the set of all totally isotropic planes of P G(5, q) with respect to the symplectic polarity Λ. The Grassmannian G0 of planes of P G(5, q) lives in a projective space of dimension 19 (c.f., [13]), and the image of I in G0 spans a projective subspace F of dimension 13. Since q is even, it can be shown that F has a unique maximal subspace N fixed by Sp6 (q). The quotient space M = F/N has projective dimension 7, and corresponds to the spin module for Sp6 (q). We write θ for the Grassman map: I → F and ι for the corresponding mapping I → M. It turns out that g(θ(π)) = θ(g(π)), for all π ∈ I and g ∈ Sp6 (q), that ι is injective and that members of I are mapped bijectively to the points of a quadric Q+ (7, q) of M with the image of a regular spread being an ovoid. Proposition 3.1. Two points of Q+ (7, q) are perpendicular if and only if the corresponding planes of I have nonempty intersection. Proof. P Sp6 (q) acts transitively on ordered pairs of disjoint totally isotropic planes, so any pair of disjoint members of I may be mapped to members of a regular spread and therefore their images under ι are points on an ovoid that are thus non-perpendicular. Given π ∈ I, the number of planes in I meeting π in exactly a line is q(q 2 + q + 1), while the number of planes in I meeting π in exactly a point is q 3 (q 2 + q + 1). Thus the number of planes not disjoint from π is 1+(q 3 +q)(q 2 +q +1) (including π itself). This is precisely the number of points in Q+ (7, q) perpendicular to ι(π). Hence two points of Q+ (7, q) are perpendicular if and only if the corresponding planes of I have nonempty intersection. We note that Q+ (7, q) has exactly (q + 1)(q 2 + 1)(q 3 + 1) points (c.f., [13, 22.5.1]). Further G has exactly two orbits on points of M: the set of points on Q+ (7, q) and its complement. The following result shows how G2 (q) arises as a subgroup of P Sp6 (q). Result 3.2 ([25], [19], [21]). If X is a point of M not on Q+ (7, q), then the stabilizer of X in P Sp6 (q) is isomorphic to G2 (q) and there is a single conjugacy class of subgroups G2 (q) in P Sp6 (q). We now describe the lines of Q+ (7, q) in terms of members of I. We need the following definition: 13 Definition 3.3. Let Q ∈ HG . A plane–star on Q is a set of q + 1 planes on Q in the same family (i.e., L or G) passing through a point P ∈ Q. A plane–star of P G(5, q) is a plane–star on some Q ∈ HG . Theorem 3.4. Let L be a line of Q+ (7, q). Then, in the spin representation, either (a) L arises from a pencil of planes of I through a given totally isotropic line, or (b) L arises from a plane–star on Q for some Q ∈ HG . Furthermore, P Sp6 (q) acts transitively on the sets of lines on Q+ (7, q) of each type. Proof. By Proposition 3.1, two points of Q+ (7, q) are collinear (perpendicular with respect to the orthogonal polarity induced by Q+ (7, q)) if and only if the corresponding planes of I meet nontrivially. The image in G0 of any pencil of planes through a line of P G(5, q) is a line, so θ maps a pencil of planes of I through a given totally isotropic line to a line of F whose projection onto M must be a line of Q+ (7, q). There are exactly (q 2 + 1)(q 6 − 1)/(q − 1) totally isotropic lines of P G(5, q) and, by Witt’s theorem, they from a single orbit under P Sp6 (q). Hence the corresponding plane pencils give rise to an orbit of lines on Q+ (7, q) under P Sp6 (q). Now consider the plane–stars of P G(5, q). We begin by showing that the planes in a plane–star correspond to the points of a line on Q+ (7, q). Let L be the set of points on Q+ (7, q) corresponding to a given plane–star. By Proposition 3.1, the points of L are pairwise perpendicular. They span one of: a line on Q+ (7, q), a plane on Q+ (7, q), or a solid on Q+ (7, q). The points perpendicular to a solid on Q+ (7, q) are just the points of the solid, numbering q 3 + q 2 + q + 1. The points of Q+ (7, q) perpendicular to a plane π on Q+ (7, q) lie in one of the two solids on Q+ (7, q) that contain π: each has q 3 points in addition to those on π and thus q 2 + q + 1 + 2q 3 in total. Hence if there are more than q 2 + q + 1 + 2q 3 points on Q+ (7, q) perpendicular to each point of L, then the points of L must form a line on Q+ (7, q). Now take a plane–star P through P in P G(5, q) for a quadric Q ∈ HG and let π2 , π3 be two planes in P. Now a totally isotropic plane of P G(5, q) meeting each plane in the P is either a plane through P : there are q 3 +q 2 +q+1 of these (one for each totally isotropic line in a complement Σ to P in P ⊥ ), or it is not through P . In the latter case it meets π2 in a point P2 6= P (there 14 are q 2 + q possibilities) and π3 ∩ P2⊥ in a point P3 6= P (here there are q possibilities). Choose a complementary 3-space Σ to P in P ⊥ containing P2 and P3 . The line P2 P3 is a line on Q meeting π2 ∩ Σ and π3 ∩ Σ so it meets π ∩ Σ in a point for each π ∈ P. For each of the q points P4 6= P ∈ Σ⊥ , the plane P2 P3 P4 is totally isotropic (not necessarily totally singular) and meets each plane of P. There are thus q 2 (q 2 +q) totally isotropic planes of P G(5, q) meeting each of the planes in P but not passing through P . In total we have q 4 + 2q 3 + q 2 + q + 1 planes meeting each of the planes in P. These correspond to q 4 + 2q 3 + q 2 + q + 1 points on Q+ (7, q) each perpendicular to each point of L. It follows that the q + 1 points of L lie on a line. As we have seen, there are q 3 (q 3 + 1)/2 members of HG , and P Sp6 (q) acts transitively on them. Let Q ∈ HG . Through each point P of Q there pass exactly q + 1 planes of L and q + 1 planes of G, giving rise to two distinct plane–stars. There are (q 2 + 1)(q 2 + q + 1) points on Q and therefore 2(q 2 + 1)(q 2 + q + 1) plane–stars of Q. The stabilizer of Q in P Sp6 (q) is a group P O6+ (q) that acts transitively on the points of Q; the stabilizer in P O6+ (q) of a point of Q contains elements of P O6+ (q)\P SO6+ (q), i.e., elements that switch Latin and Greek planes. Hence P O6+ (q) acts transitively on the plane–stars of Q. We show that each plane–star of Q lies on exactly q members of HG . By the transtivity we may consider the quadric given by Q(x) = x1 x4 +x2 x5 +x3 x6 and the plane–star through the point (1, 0, 0, 0, 0, 0) containing the planes x4 = x5 = x6 = 0 and x2 = x3 = x4 = 0. The quadrics in HG containing this plane–star are precisely those corresponding to the quadratic forms x1 x4 + x2 x5 + x3 x6 + α4 x24 with α4 ∈ GF (q) and are thus q in number. Let n be the number of distinct plane–stars and consider the number of ways of selecting a quadric in HG and a plane–star on that quadric. If we count first the number of quadrics, we reach a total of [q 3 (q 3 + 1)/2][2(q 2 + 1)(q 2 + q + 1)], whilst if we count first the number of plane–stars we get nq. Thus n = [q 3 (q 3 + 1)/2][2(q 2 + 1)(q 2 + q + 1)]/q = q 2 (q 2 + 1)(q 3 + 1)(q 2 + q + 1). Hence the lines on Q+ (7, q) arising from plane–stars of P G(5, q) form an orbit under P Sp6 (q). Summing the lengths of the orbits arising from plane–pencils and plane– stars, we have (q 2 + 1)2 (q 3 + 1)(q 2 + q + 1) lines of Q+ (7, q), but this is exactly the total number of lines of Q+ (7, q) (c.f., [13, 22.5.1]). Thus P Sp6 (q) has 15 exactly two orbits of lines on Q+ (7, q). 4 A representation of H(q) In this section we will represent the generalized hexagon H(q), q even, of order (q, q), in terms of the geometry of quadrics of P G(5, q) invariant under G = P Sp6 (q). Firstly, we recall the classical construction of H(q). Consider the quadric Q(6, q) in P G(6, q) given by the equation X0 X4 + X1 X5 + X2 X6 = X32 , where X0 , X1 , . . . , X6 are projective homogeneous coordinates in P G(6, q). The points of H(q) are all points of Q(6, q). The lines of H(q) are the lines of Q(6, q) whose Grassmannian coordinates satisfy the equations p01 = p36 , p12 = p34 , p20 = p35 , p03 = p56 , p13 = p64 and p23 = p45 . There exists another representation of H(q), arising from the natural representation projecting from the nucleus N of Q(6, q) onto a hyperplane on P G(6, q) (perfect symplectic hexagon) [25]. We can consider H(q) ⊂ Q(6, q) as a nondegenerate section of the quadric + Q (7, q) on which G acts in its spin representation. This allows us to give a new interpretation of H(q) in P G(5, q) arising from quadrics in HG . Theorem 4.1. Let P be the set of special planes in P G(5, q), let L be the set of special lines and let incidence be inclusion. Then P, L form a generalized hexagon, isomorphic to H(q) and with G2 (q) as an automorphism group. Proof. By Proposition 2.13, each member of P is incident with q + 1 members of L and vice-versa. The crux of the argument we present is that G2 (q) simultaneously lies in two subgroups of P Ω+ 8 (q) isomorphic to P Sp6 (q), one a spin representation (where we use G for P Sp6 (q)) and the other a natural representation in which P Sp6 (q) is isomorphic to P SO7 (q) which is in turn the stabilizer of a point of X ∈ P G(7, q) \ Q+ (7, q) (here we use P SO7 (q)). As noted in Result 3.2, the group G2 (q) then arises as the stabilizer in G of X. In the standard construction, the points of H(q) are all the points of Q+ (7, q) ∩ X ⊥ . We show that these points are the images under ι of the special planes of P G(5, q). The lines of H(q) are (q 6 − 1)/(q − 1) of the lines of Q+ (7, q) ∩ X ⊥ . We show that these lines arise from plane-pencils on 16 special lines of P G(5, q). Note that G2 (q) acts transitively on both points and lines of H(q). We start with the spin representation. Totally isotropic planes in P G(5, q) map bijectively to points of Q+ (7, q), with a corresponding irreducible representation of G. By Result 2.5, G2 (q) has two orbits on totally istropic planes in P G(5, q), so it has two orbits on the points on Q+ (7, q): these can only be those points of Q+ (7, q) in X ⊥ and those not. The number of points in Q+ (7, q)∩X ⊥ is (q 6 −1)/(q −1) so these must be the images of special planes of P G(5, q). We conclude that ι maps special planes of P G(5, q) bijectively to points in Q+ (7, q) ∩ X ⊥ . By Theorem 3.4, G has two orbits on the lines on Q+ (7, q), those arising from plane-pencils and those from plane-stars: we refer to these as planepencil lines and plane-star lines respectively. As P Sp6 (q) is irreducible on P G(7, q), neither of the orbits can consist solely of lines in X ⊥ . Consider a quadric Q ∈ HG of P G(5, q) with prescribed planes π1 , π2 , and a special plane π on Q meeting π1 and π2 in a point P and the line ℓ = P ⊥ ∩ π2 respectively . Let R be a point of ℓ, let m = P R and let π ′ be the second plane (necessarily a special plane) on Q passing through m. Then ι(π) and ι(π ′ ) both lie in Q+ (7, q)∩X ⊥ and the set of all totally istropic planes through m form a plane-pencil that can only map to a plane-pencil line joining ι(π) and ι(π), and thus lying in Q+ (7, q) ∩ X ⊥ . Since G2 (q) acts transtively on special lines of P G(5, q), it also acts transitively on planepencils on special lines and therefore has an orbit of length (q 6 − 1)/(q − 1) on plane-pencil lines in Q+ (7, q) ∩ X ⊥ . Since G2 (q) has only one other orbit on isotropic lines of P G(5, q) it must have one other orbit on plane-pencil lines on Q+ (7, q), of length q 2 (q 6 − 1)/(q − 1). As the plane-pencil lines do not all lie in Q+ (7, q) ∩ X ⊥ , no plane-pencil line in the larger orbit of G2 (q) lies in Q+ (7, q) ∩ X ⊥ . Now turn to the P SO7 (q): it acts transitively on the lines in Q+ (7, q) ∩ X ⊥ . In this context, G2 (q) has two orbits on the lines of Q+ (7, q) ∩ X ⊥ . One orbit we already know about, the other (of length q 2 (q 6 − 1)/(q − 1)) must consist of plane-star lines. We conclude that the lines of H(q), being (q 6 −1)/(q−1) lines of Q+ (7, q)∩ X ⊥ in a single orbit under G2 (q), must be precisely the plane-pencil lines arising from special lines of P G(5, q). Now a plane π ∈ P and a line ℓ ∈ L are incident if and only if ι(π) is incident with ι(ℓ), the plane-pencil line of X ⊥ corresponding to the pencil of planes through ℓ. Hence, as an incidence structure, (P, L) is isomorphic to 17 H(q) and admits G2 (q) as an automorphism group. Remark 4.2. It should be noted that in the natural representation of H(q), the points which lie on a pair of prescribed planes π1 and π2 , are exactly the points of a thin subhexagon H(1, q), i.e., the double of the projective plane P G(2, q). The lines of this subhexagon are the special lines we defined above. The special planes are exactly the hexagon planes, i.e. the planes of Q(6, q) containing hexagon lines. 5 G2(q), q even, is maximal in P Sp6(q): another geometric proof In the study of classical groups over a finite field, Aschbacher’s theorem plays a major part (c.f., [1], [20]). Any subgroup of a classical group G either lies inside a maximal subgroup belonging to one of eight “geometric” classes of subgroups of G or has the form H = NG (X), for some quasisimple subgroup X of G satisfying some extra conditions, and in that case we say that it belongs to the class S. Given such a group H not lying in one of the eight classes of C, the main problem is to determine whether or not H is maximal in G. Here we are interested in the maximality of the group G2 (q), q > 2 even, in the class S of the symplectic group G = P Sp6 (q). A quick geometric proof of the maximality of G2 (K), K any field, was given by Dye in [10]. When K is finite, the proof relies on the classification of irreducible groups generated by long root elements [15]. Here we present a very short alternative proof for even q based on the geometry of quadrics of P G(5, q). Theorem 5.1. If q is even, then G2 (q) is a maximal subgroup of P Sp6 (q). Proof. Suppose G2 (q) < F ≤ P Sp6 (q) and let Q ∈ HG . By proposition 2.1, G2 (q) acts transitively on HG and so the same applies to F . Therefore S < H ≤ P O6+ (q) where H = StabF Q. It is proved in [17] that S is maximal in P O6+ (q), so H = P O6+ (q) = StabG Q and hence F = P Sp6 (q). 18 6 Lines on elliptic quadrics in EG This section is something of a digression. We view the special lines of P G(5, q) from a different perspective, namely from the point of view of elliptic quadrics in EG , and observe the way in which certain line–spreads of H(q) arise. It is well known (c.f., [25]) that the generalized hexagon H(q) in its standard representation has a line–spread arising from elliptic quadrics of P G(5, q). We shall realise this line–spread in the context of the representation of H(q) given in Section 4. In order to avoid confusion, we note that while a line–spread of an elliptic quadric is a partition of the point set of the quadric into lines, the same is not the case for line–spreads of H(q) (we shall not go into details here, but note that they may be found in [25]). Consider the class of maximal subgroups of G2 (q) isomorphic to SU3 (q 2 )·2 (c.f., [3], [22]), lying in a single conjugacy class C2 ; indeed there is a single conjugacy class in P Sp6 (q). Given such a subgroup T of P Sp6 (q), it is always contained in the orthogonal group P O6− (q) of a quadric Q ∈ EG , i.e., T stabilizes Q. Proposition 6.1. The group G2 (q) acts transitively on EG . Proof. Let T ∈ C2 and let Q be a quadric in EG stabilized by T . Then T must be the intersection of the orthogonal group of Q with G2 (q) and hence T is the full stabilizer of Q in G2 (q). Thus the groups in C2 correspond to quadrics in EG . The number of subgroups in C2 is given by |G2 (q)|/|SU3 (q 2 ) · 2| = q 3 (q 3 − 1)/2. Hence G2 (q) acts transitively on EG . Let Q ∈ EG . The configuration of Q stabilized by the group T isomorphic to SU3 (q 2 ) · 2 is easily described. It is a so–called line–spread of Q, i.e., a partition of the pointset of Q into q 3 + 1 lines; this particular line–spread is known as an Hermitian spread of Q. The particular line–spread in Q is determined by T ; we call it the prescribed line–spread for Q. This embedding was studied by Dye in [9]. He showed that the points of P G(2, q 2 ) correspond to a line–spread of P G(5, q) with the points of a Hermitian curve in P G(2, q 2 ) corresponding to the line–spread of Q. Proposition 6.2. The lines of a prescribed line–spread of Q ∈ EG are special lines of P G(5, q) and they lie on the prescribed line–spread of each quadric in EG that contains them. 19 Proof. Let Q ∈ EG and let S be the precribed line–spread of Q. It is clear from Dye’s work that the subgroup T isomorphic to SU3 (q 2 ) · 2 stabilizing S in G2 (q) is doubly transitive on the lines in S, and that it is transitive on the lines on Q that do not lie in S. Hence T has two orbits on the lines on Q. By Result 2.5, G2 (q) has two orbits on isotropic lines of P G(5, q). Therefore an isotropic line is either on a prescribed line–spread for all quadrics in EG containing it or it is off the prescribed line–spread for all quadrics. The number of lines on a line–spread of Q ∈ EG is q 3 + 1. There are 3 3 q (q − 1)/2 quadrics in EG and, by Proposition 2.8, each isotropic line of P G(5, q) lies on q 3 (q−1)/2 quadrics in EG . Hence the lines lying in prescribed line-spreads form an orbit of length (q 6 − 1)/(q − 1). By Proposition 2.11, the lines in prescribed line–spreads are precisely the special lines of P G(5, q) (as defined in Definition 2.12). Corollary 6.3. Let Q ∈ EG and let S be the prescribed (Hermitian) line– spread of Q with group T isomorphic to SU3 (q 2 ) · 2. Then the plane-pencils on lines of S form a line–spread of H(q). References [1] M. Aschbacher, On the maximal subgroups of the finite classical groups, Invent. Math. 76 (1984), 469-514. [2] A.E. 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