Rank 3 Permutation Characters and Maximal Subgroups-PhD Thesis by Hung Phi Tong-Viet | Papers by Hung Phi

Ph.D Thesis

Rank 3 permutation characters, Guralnick-Magaard problem, Rank 3 permutation modules, and Maximal subgroups of classical groups

rank 3 permutation Characters and maximal subgroups by Tong Viet Phi Hung A thesis submitted to The University of Birmingham for the degree of DOCTOR OF PHILOSOPHY School of Mathematics The University of Birmingham June 2009 Abstract Let G be a transitive permutation group acting on a ﬁnite set E. Let P be a stabilizer in G of a point in E. We say G is primitive rank 3 on E if P is maximal in G and P has exactly three orbits on E. For any subgroup H of G, we denote by 1G the permutation character H (or permutation module over C) of G on the cosets G/H. Let H and K be subgroups of G. We say 1G H 1G if 1G − 1G is either 0 or a character of G. Also a ﬁnite group G H K K is called nearly simple primitive rank 3 if there exists a quasi-simple group L such that L/Z(L) G/Z(L) Aut(L/Z(L)) and G acts as a primitive rank 3 permutation group on the set of cosets of a subgroup of L. In this thesis we classify all maximal subgroups M of a class of nearly simple primitive rank 3 groups G acting on E such that 1G P 1G , M where P is a stabilizer of a point in E. This result has an application to the study of minimal genus of algebraic curves which admit group actions. Acknowledgements First of all, I would like to thank my thesis supervisor, Prof. Kay Magaard, for his direction, his guidance and his patience. Without him, this work cannot be done. I also wish to thank all the members of the pure mathematics group at University of Birmingham, especially Prof. Christopher Parker for his teaching and support, Prof. Robert Curtis , Prof. Sergey Shpectorov, Dr. Corneliu Hoﬀman and Dr. Paul Flavell for helping me many times with my research. I am very grateful for the ﬁnancial support from the School of Mathematics, University of Birmingham during the pursuance of my Ph.D. I also thank Wayne State University for the ﬁnancial support via the Graduate Teaching Assistantship from 2005-2007. I would like to thank my examiners, Dr. Ralf Gramlich and Prof. Peter Fleischmann. Their valuable feedback helped me to improve my thesis. I am grateful to Prof. David Gluck and Prof. Daniel Frohardt for their teaching and their help when I was in Wayne State University. Special thanks should be given to my best friend, Dr. Le Thien Tung, who constantly helps me not only in mathematics but also in many aspects of my life. I also wish to thank my family and my wife who support me spiritually throughout my life. Finally best wishes and love to my new born daughter. Contents 1 Introduction 2 Preliminaries 2.1 Finite classical groups . . . . . . . . . . . . . 2.2 Deﬁnitions and Structures of classes C and S . 2.3 Properties of ﬁnite simple groups . . . . . . . 2.4 Representations of Symmetric groups . . . . . 2.5 Representations of Finite groups of Lie type in 2.6 Representations of Finite groups of Lie type in 1 8 8 17 20 26 35 35 43 43 47 48 49 49 51 70 95 95 97 133 162 165 171 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cross characteristic . . deﬁning characteristic . . . . . . . . . . . . . . . . . . 3 Rank 3 permutation characters and maximal subgroups 3.1 Higman rank 3 parameters and the equation . . . . . . . . 3.2 An example . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Main Hypothesis and Notations . . . . . . . . . . . . . . . 3.4 Ω2m+1 (3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Parameters for Ω2m+1 (3) . . . . . . . . . . . . . . . 3.4.2 Permutation characters of maximal subgroups in C 3.4.3 Permutation characters of maximal subgroups in S 3.5 Ω± (3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2m 3.5.1 Parameters for Ω± (3) . . . . . . . . . . . . . . . . 2m 3.5.2 Permutation characters of maximal subgroups in C 3.5.3 Permutation Characters of Maximal subgroups in S A Classiﬁcation of nearly simple primitive rank 3 groups B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nearly simple groups of type Ωε (2), SUm (2) and sporadic 2m List of References Chapter 1 Introduction Let G be a transitive permutation group acting on a ﬁnite set E. Let P be a stabilizer of a point in G. We say that G is primitive on E if and only if P is maximal in G. We deﬁne the rank of G on E to be the number of P -orbits on E. For any subgroup H of G, we denote by 1G the permutation character of G on the cosets G/H. We also use the H same notation 1G for the permutation module. Let H, K be subgroups of G. Consider the H permutation characters 1G and 1G , we say 1G H K H G. In terms of permutation modules, 1G H 1G if 1G − 1G is zero or a character of H K K 1G if 1G is isomorphic to a submodule of 1G . K H K A ﬁnite group L is said to be quasi-simple if L is perfect and L/Z(L) is simple. A ﬁnite group G is called nearly simple of type L if L G and L/Z(L) G/Z(L) Aut(L/Z(L)) for some quasi-simple group L. Moreover a ﬁnite group G is called almost simple of type L if L G Aut(L) for some ﬁnite simple group L. It follows from deﬁnitions that if G is 3 nearly simple of type L then G/Z(L) is almost simple of type L/Z(L). Assume that m is an integer. Let L be one of the following quasi-simple groups Ω2m+1 (3), Ωε (3), Ωε (2) 2m 2m or SUm (2). Let G be a nearly simple group of type L such that G acts on the L-orbit E(V ) of non-singular points in the natural module V for L. Then G is a primitive rank 3 group on E(V ) with socle L. (Theorem A.1 in Appendix A). In this situation, we say that G is a nearly simple primitive rank 3 group of type L. Now ﬁx groups L, G as above 1 and also ﬁx a stabilizer P in G of a non-singular point x in V. In this work, we classify all maximal subgroups M of G such that 1G P 1G . M 3, Theorem 1.1 Let L be one of the following groups Ω2m+1 (3), or Ωε (3), with m 2m and G be a nearly simple primitive rank 3 group of type L. Let P be the stabilizer of a non-singular point in V. Let M be any maximal subgroup of G. Then 1G P pairs (L, M ) appear in Tables 1.1-1.3. Remark of Theorem 1.1. For all pairs (L, M ) in Table 1.1 and 1.2, 1G is not contained P in 1G . The pairs in Table 1.3 are the cases that we have not determined whether or not M there is containment. A similar result for nearly simple primitive rank 3 group of type Ωε (2), SUm (2) and almost simple groups of sporadic type can be found in Appendix B. 2m The main motivation for this work comes from algebraic curves which admit group actions. From Riemann’s Existence Theorem, we know that for every ﬁnite group there are inﬁnitely many Riemann surfaces with automorphism group G. We would like to identify G-curves of smallest possible genus. Now Theorem 1.1 and Corollary 8.2 in [12] yield the following corollary. Corollary 1.2 Assume the assumption and notations of Theorem 1.1. Let X be a compact Riemann surface with Aut(X) ∼ G, and inertia groups g1 , · · · , gr over X/G. = Then g(X/P ) Tables 1.1-1.3. A triple S = (P, B, I), where P, B, I are sets with P ∩ B = ∅ and I ⊆ P × B is called an incidence structure. The elements of P, B and I are called points, blocks and ﬂags, respectively. If (p, b) ∈ I then we say p and b are incident and write pIb. If there exists a ﬁnite group G of incidence preserving permutations of P and B which is transitive on pairs (p, b) ∈ P × B with pIb, then S = (P, B, I) is said to have a ﬂag transitive automorphism groups. Such an incidence structure can arise from subgroups of G as follows. Let P, M 2 g(X/M ) for any maximal subgroup M of G which does not appear in 1G unless the M be subgroups of G and let P, B be the sets of right cosets of P, M, respectively. We deﬁne an incidence structure I by P x I M y if and only if P x ∩ M y = ∅. We now ﬁx orderings of P and B. The group G deﬁnes permutation representations π1 and π2 on P and B, respectively by right multiplications. An incidence matrix of S denoted by M = M(S) is a matrix whose rows and columns are indexed by P and B, respectively with entries 1 on elements of I and 0 elsewhere. As G preserves the incidence structure, we have for any g ∈ G, π1 (g)Mπ2 (g)t = M. Thus if M has maximal rank then π1 or equivalently 1G P π2 1G as π1 = 1G and π2 = 1G (Lemma 2.3 in [33]). Now let X be a M P M e f n − e and denote by Xe , ﬁnite set of size n, and let e, f be integers such that 1 and Xf the collections of subsets of X of size e and f, respectively. Let Sef (X) be the incidence structure (Xe , Xf , I) where pIb if p ⊆ b for any (p, b) ∈ Xe ×Xf . Let V be an Fq vector space of dimension n with q a prime power. The collection of e-subspaces of V is denoted by Ve . Similarly one can form the incidence structure Sef V = (Ve , Vf , I). These two structures share many properties. For example, Kantor [27] showed that whenever 1 e f n−e, the incidence matrices of Sef (X) and Sef (V ) have maximal rank. This can be interpreted as the relation between a group with BN -pairs and its Weyl groups. In [33], Lehrer generalized this observation to all classical groups and he managed to show that if G is any ﬁnite classical group then 1Ge P 1Gf , whenever 1 P e<f n − e except + when G = O2n (q) and e = 1, f = n − 1, where Pe , Pf denote the parabolic subgroups which are the stabilizers of a totally singular or isotropic subspaces of dimension e, f, respectively and n is the dimension of any maximal totally singular (isotropic) subspaces of G. (Theorem 5.1[33]). These results in fact follow from the statements about its Weyl groups and the passage between these two structures are provided by generic algebras introduced by Tits (Theorem 3.2[33]). The permutation character containment can be determined by the rank of suitable incidence matrix as above. Now if G is an almost simple group which is doubly transitive 3 on E with point stabilizer P, then either 1G P 1G or G = P M for any maximal subgroup M M of G. As the maximal factorization of almost simple groups was classiﬁed completely by M. Liebeck, C. Praeger and J. Saxl in [34], we can tell exactly for which M we have the containment 1G P 1G . In case of rank 3, M. Aschbacher, R. Guralnick and K. Magaard M ([2]) have a criterion in terms of the Higman rank 3 parameters. In that paper, they consider the case when G is a nearly simple classical group acting on the set of singular points on its natural module. Also a partial result of this case has been dealt with by D. Frohardt, the second and the third authors above in [12]. In the case when G is a nearly simple primitive rank 3 group of sporadic type, the containment of the permutation characters of G is completely determined since all the permutation characters of maximal subgroups of G are stored in [13], except HS.2, F i22 .2 and F i24 .2. We now describe our strategy. Let L be a ﬁnite simple classical group of degree d 2, deﬁned over a ﬁnite ﬁeld F = Fq , q a prime power, and let V be the natural module for L. Assume that G is an almost simple group with simple socle L. We have a powerful theorem on the subgroup structure of G by M. Aschbacher. The theorem says that if M is a subgroup of G then M is either belongs to a collection C(G) of geometric subgroup of G or M ∈ S(G), that is, M is an almost simple group and the full covering group of the socle of M acts absolutely irreducible on the natural module V for G and cannot be realized over any proper subﬁeld. Thus if M is a maximal subgroup of G then either M ∈ C(G) or M ∈ S(G). The subgroup structure and the maximality among members of C(G) has been determined by P. Kleidman and M. Liebeck in [29] when the degree is at least 13. For this case, using the geometrical properties of the groups, we can solve the problem completely. When M is not a geometric subgroup, that is, M ∈ S(G), the problem is much more complicated as we still do not know which members of S(G) are maximal. Now assume that M ∈ S(G). Denote by S the socle of M. So S is a non-abelian ﬁnite simple group. According to the Classiﬁcation of Finite Simple Groups, S is an alternating 4 group of degree at least 5, a ﬁnite group of Lie type or one of the 26 sporadic groups. By way of contradiction, we assume that 1G P 1G . From this assumption, we will get an M upper bound for the dimension of V in terms of the size of the automorphism group of S. From the deﬁnition of members in S(G), the full covering group S of S acts absolutely irreducible on V. Now using the information on the lower bound for the dimension of the absolutely irreducible representations of ﬁnite simple groups, we will get a ﬁnite list of cases that we can handle either by constructing the representations or by computer program GAP [13]. As in the almost simple doubly transitive case, we can get a list of maximal subgroups M such that 1G P 1G . In Table 1.1, we list all the cases when M ∈ C(G), Table 1.2 M contains all cases when M ∈ S(C), and in the last table, we list the cases that we have not determined whether or not there is containment. Notice that we only have a ﬁnite number of exceptions in Table 1.3. Also there is a ﬁnite number of cases in Table 1.2. Chapter 2 is the preliminaries, in this chapter, we will ﬁx some notations and give some deﬁnitions as well as some properties of ﬁnite simple groups and also some information on representations of these groups. The proof of Theorem 1.1 is carried out in Chapter 3. We will deal with each type of quasi-simple groups separately. For the notations in the tables of Theorem 1.1, the columns ‘orbits’ give the number of orbits of M on E(V ) and this is also the number of double cosets of group G on P and M. The ﬁrst columns are the type of the nearly simple group G. The last columns ‘Ref’ give the references for the result. For example 3.18 means that the case is dealt with in Proposition 3.18. Other notations will be explained in Chapter 3. 5 L Ω2m+1 (3) type of M ε O1 (3) ⊥ O2m (3) Pα O1 (3) Sn Oα (33 ) O1 (3) ⊥ O2m−1 (3) Pα O1 (3) Sn Om (3) S2 GLm (3).2 + O2a (33 ) + Om (32 ) − O2a (33 ) − Om (32 ) Om (32 ) GUm (3) Sp2 (3) ⊗ Spm (3) Table 1.1: M ∈ C conditions 1 α m (n, ξ, r) = (5, ±, t) (7, +, t) Remarks orbits 2 n = 2m + 1 2m + 1 = 3α 2 2 3 2 Ref 3.10 3.13 3.17 3.20 3.37 3.38 3.39 3.41 3.42 3.45 3.46 3.47 3.48 3.53 Ωε (3) 2m 1 α m (ε, r) = (−, t), (ε, r, ξ) = (−, t, ) (ε, r) = (−, t), r=s r=t G=I G = I if ε = + ε=+ n=6 n = 10 m odd m = 3a m is even m = 3a m is even m odd m even 2 3 3 1 3 2 3 2 2 1 1 Table 1.2: M ∈ S socle of M modules Remarks A9 λ = (8, 1) P Sp6 (2) P Sp6 (3) λ2 G2 (3) λ1 , λ2 F4 (3) λ4 − Ωn−1−ε3 (n) (3) An , n = 5, 6, 7 λ = (n − 1, 1) Ω+ (3) A12 ξ= 10 Ω− (3) A18 ξ = ,r = t 16 Ω− (3) L3 (4) 6 + Ω8 (3) Ω+ (2) 8 Ω− (3) Ω5 (3) 2λ1 r=t 10 + Ω8 (3) Ω7 (3) Spin module Ω+ (3) Ω9 (3) Spin module 16 + Ω24 (3) 2.Co1 Leech lattice L Ω7 (3) Ω7 (3) Ω13 (3) Ω7 (3) Ω25 (3) orbits 2 2 2 1 2 2 2 4 1 2 5 1 1 2 Ref 3.26 3.27 3.30 3.32 3.57 3.60 3.65 3.69 6 Table 1.3: Exceptions L socle of M modules Remarks Ω41 (3) S8 (3) λ1 Ω77 (3) E6 (3) adjoint module Ω133 (3) E7 (3) adjoint module Ω+ (3) F4 (2) 52 Ωε (3) P Sp2 (3) λ −1 = 5, 6, 7 2m − i Ω36 (3) P Sp6 (3) (1 + 3 )λ 1 i 2 Ωε (3) F4 (3) adjoint module 52 Ref 3.30 3.32 3.60 3.66 3.68 7 Chapter 2 Preliminaries 2.1 Finite classical groups We adopt the constructions and notations of [29]. We denote by V a vector space of dimension n over the ﬁeld F, where F is either a ﬁnite ﬁeld or an algebraically closed ﬁeld of characteristic p. The group of all non-singular F-linear transformations of V is denoted by GL(V, F), the general linear group of V over F. The special linear group of V over F, SL(V, F), is the group of all elements of GL(V, F) with determinant 1. Fix a basis β = {v1 , v2 , · · · , vn } of V. For g ∈ GL(V, F), gβ denotes the n × n matrix which satisﬁes vi gβ = Σn (gβ )ij vj . For λi ∈ F∗ , i = 1 . . . n, we denote by diagβ (λ1 , . . . , λn ) j=1 the diagonal linear transformation which satisﬁes vi diagβ (λ1 , . . . , λn ) = λi vi . The element diagβ (λ, . . . , λ), with λ ∈ F∗ , is called a scalar linear transformation, or a scalar. The center of GL(V, F) is the group of all non-zero scalars, which is isomorphic to F∗ . We will write P GL(V, F) for the projective general linear group GL(V, F)/F∗ . For any subgroup of GL(V, F), we write P X for the corresponding projective group X/X ∩F∗ . We also use the bar convention to denote reduction modulo scalars, for example, GL(V, F) = P GL(V, F). A map g from V to itself is called an F-semilinear transformation of V if there is a ﬁeld 8 automorphism σ(g) ∈ Aut(F) such that for all v, w ∈ V and λ ∈ F, (v + w)g = vg + wg and (λv)g = λσ(g) (vg) (2.1) An F-semilinear transformation g is non-singular if {v ∈ V |vg = 0} = {0}. We now deﬁne Γ L(V, F) to be the set of all non-singular F-semilinear transformations of V. It is a group, called the general semilinear group of V over F. The map σ from Γ L(V, F) to Aut(F) is a surjective homomorphism with kernel GL(V, F). As F∗ Γ L(V, F), we can factor out the scalars to get the projective general semilinear group P Γ L(V, F). With the basis β as above, each element g ∈ Γ L(V, F) is determined by its action on β together with σ(g). If α ∈ Aut(F) then we deﬁne φβ (α) as the unique element of Γ L(V, F) which lies in σ −1 (α) and ﬁxes each vi . Then n n ( i=1 λi vi )φβ (α) = i=1 λα vi i (2.2) A left linear form on V is a map f : V × V → F such that for each v ∈ V, the map V → F given by x → f (x, v) is a linear map. There is an analogous deﬁnition for right linear form. A bilinear form is a map which is both a left linear and a right linear form. For any map Q : V → F, deﬁne fQ : V × V → F by fQ (v, w) = Q(v + w) − Q(v) − Q(w). The map Q is called a quadratic form if Q(λv) = λ2 Q(v) for all v ∈ V and λ ∈ F, and fQ is a bilinear form. When Q is a quadratic form, fQ is called the associated bilinear form. If f is a map form V × V to F and β = {v1 , v2 , . . . , vn } is any basis for V, then we deﬁne fβ as the matrix satisfying (fβ )ij = f (vi , vj ). For v ∈ V, we call f (v, v) the f − norm of v. We write (v, w) instead of f (v, w). Let κ be either a left linear form or a quadratic form. Then κ is a map from V k to F, where k = 1 or 2. We have κ(λv1 , . . . , vk ) = λ3−k κ(v1 , . . . , vk ). Assume that (V, F, κ) and (V , F, κ ) are two spaces of dimension n over F, where κ and κ are either both left linear of both quadratic forms. An isometry is an invertible element 9 g ∈ HomF (V, V ) which satisﬁes κ (vg) = κ(v) for all v ∈ V k . If such an isometry exists, then (V, F, κ) and (V , F, κ ) are said to be isometric and we write (V, F, κ) ∼ (V , F, κ ). = An invertible element g ∈ HomF (V, V ) is a similarity if there exists λ ∈ F∗ such that κ (vg) = λκ(v) for all v ∈ V k . If such a similarity exists, then (V, F, κ) and (V , F, κ ) are said to be similar. In the case (V, F, κ) = (V , F, κ ), the isometries are called κisometries and the set of all κ-isometries forms a subgroup of GL(V, F). This subgroup is called κ-isometry group and is denoted by I(V, F, κ). An element of I(V, F, κ) ∩ SL(V, F) is a special κ-isometry. The special κ-isometry group is denoted by S(V, F, κ). The set of all κ-similarities forms the similarity group, denoted by Λ(V, F, κ). (This is ∆(V, F, κ) in [29].) Finally an element g ∈ Γ L(V, F) is called a κ-semisimilarity if there exist λ ∈ F∗ and α ∈ Aut(F) such that κ(vg) = λκ(v)α for all v ∈ V k . (2.3) The set of all κ-semisimilarities forms a group, denoted by Ξ(V, F, κ). (This group is denoted by Γ (V, F, κ) in [29].) Suppose that κ is surjective and g ∈ Ξ(V, F, κ) satisfying (2.3). Then λ in (2.3) is uniquely determined by g and the map τ : Ξ(V, F, κ) −→ F∗ , τ (g) = λ is well-deﬁned. The restriction of τ to Λ(V, F, κ) is a homomorphism to F∗ with kernel I(V, F, κ). The ﬁeld automorphism α in (2.3) is exactly σ(g) given in (2.1). The restriction of σ to Ξ(V, F, κ) is a homomorphism to Aut(F) with kernel Λ(V, F, κ). Assume that f is a map from V × V to F. Then F is called non-degenerate if for each v ∈ V \ 0, the maps from V → F given by x → (x, v) and x → (v, x) are non-zero. A quadratic form Q is called non-degenerate if its associated bilinear form fQ is non-degenerate. The map f is symmetric if f (v, w) = f (w, v) for all v, w ∈ V. The map f is skew - symmetric if f (v, w) = −f (w, v) for all v, w ∈ V. The map f is symplectic if f is skew-symmetric, bilinear, and f (v, v) = 0 for all v ∈ V. Finally, f is said to be unitary if F has an involutory ﬁeld automorphism α, f is left linear and 10 f (v, w) = f (w, v)α for all v, w ∈ V. Let F be a ﬁnite ﬁeld of characteristic p, and let κ be a left linear form f or a quadratic form Q appearing in one of the four cases below: case L : κ is identically 0; case S : κ = f , a non-degenerate symplectic form; case O : κ = Q, a non-degenerate quadratic form; case U : κ = f , a non-degenerate unitary form. We deﬁne    Ξι A=  Ξ  in case L with n otherwise 3    certain subgroup of index 2 in S in case O Ω=  S  otherwise where ι is an inverse-transpose automorphism of GL(V, F). Hence we get a chain of groups Ω S I Λ Ξ A. (2.4) This chain is A-invariant and F∗ A, so we can deﬁne projective groups for any subgroups of A. We use the notation¨to denote reduction modulo ΩF∗ in A. We get a chain ¨ 1=Ω ¨ S ¨ I ¨ Λ ¨ Ξ ¨ A. (2.5) A ﬁnite classical group is any group G satisfying Ω G A or Ω G A in one of the four cases L, S, O or U. If G is such a classical group, then we call G a linear group, symplectic group, orthogonal group or unitary group, respectively. The convention for the 11 ﬁeld F is    1 f F = Fqu , where q = p and u =  2  in cases L, S and O in case U. We also denote by α the automorphism of F given by λα = λq for λ ∈ F. We write    Ln (q) if the sign is + ± ± Ln (q) = P SLn (q) =  U (q) if the sign is −.  n Assume F, κ, f , Q, σ and τ as in above discussion, and let X ∈ {Ω, S, I, Λ, Ξ, A}. We shall call (V, F, κ) or brieﬂy, (V, κ), a classical geometry. If case L, S, O or U holds then we call (V, F, κ) a linear, symplectic, orthogonal or unitary geometry, respectively. If W is a subspace of V, and κW is the restriction of κ to W ( W × W ) such that κ is either non-degenerate or identically zero, then (W, F, κW ) is also a classical geometry and we call it a sub-geometry of (V, F, κ). W is said to be non-degenerate if κW is non-degenerate and W is totally singular if κW = 0. We also deﬁne W to be totally isotropic if fW = 0. Let v ∈ V \ 0. We say that v is singular or isotropic if f (v, v) = 0 in cases L, S, and U. In case O, we say that v is isotropic if f (v, v) = 0, and singular if Q(v) = 0. Vectors which are not singular are called non-singular and those which are not isotropic will be called non-isotropic. We now state the Witt’s Lemma concerning the classical geometries. Proposition 2.1 (Witt’s Lemma, Proposition 2.1.6[29]). Assume that (V1 , κ1 ), (V2 , κ2 ) are two isometric classical geometries and that Wi is a subspace of Vi for i = 1, 2. Further assume that there is an isometry g from (W1 , κ1 ) to (W2 , κ2 ). Then g extends to an isometry from (V1 , κ1 ) to (V2 , κ2 ). Corollary 2.2 (Corollary 2.1.7[29]). All maximal totally singular subspaces of (V, κ) have the same dimension. And if κ is non-degenerate, this dimension is at most n . 2 Let (V, F, κ) be a classical geometry with dimF (V ) = n and G be an irreducible 12 subgroup of GL(V, F). For any ﬁeld extension E of F, we form the tensor product V ⊗ E, an n-dimensional space over E. Let G act naturally on V ⊗ E via (v ⊗ λ)g = vg ⊗ λ, (v ∈ V, g ∈ G, λ ∈ E), and we may regard G GL(V ⊗ E). We say that G is absolutely irreducible in GL(V, F) if G remains irreducible in GL(V ⊗ E, E) for all extension ﬁeld E of F. A subgroup of P GL(V, F) is said to be absolutely irreducible if its pre-image in GL(V, F) is absolutely irreducible. For λ ∈ F, deﬁne    {v ∈ V \ 0 | f (v, v) = λ} in cases L, S, U; Vλ =  {v ∈ V \ 0 | Q(v) = λ}  in cases O. (2.6) The following lemma describes the orbits of Ω on V. Write Ω for Ω(V, F, κ). Lemma 2.3 ([29], Lemma 2.10.5). Assume that n 2. (i) In cases L and S, the group Ω is transitive on V \ 0. (ii) Suppose that either case U holds with n transitive on Vλ for any λ. (iii) SU2 (q) has q + 1 orbits on V0 (each of size q 2 − 1), and is transitive on Vλ for λ = 0. (iv) Ω3 (q) has two orbits on V0 (each of size 1 (q 2 − 1)), and is transitive on Vλ for λ = 0. 2 (v) Ω+ (q) has 2(2, q − 1) orbits on V0 , and Ω± (q) has (2, q − 1) orbits on Vλ for λ = 0. 2 2 Let (V, F, Q) be a classical orthogonal geometry, where F := Fq , and Q is a quadratic form on V. Let f = fQ , the associated bilinear form. Then f is a symmetric bilinear form on V. Write (v, w) for f (v, w). Then for any v ∈ V, we have 2Q(v) = (v, v). Suppose that q is odd then Q(v) = 1 (v, v). Sometimes, it will be convenient to work with the standard 2 basis for L, deﬁned as follows: Deﬁnition 2.4 ([29], Proposition 2.5.3). The space (V, F, Q) has a basis of one of the following forms: 13 3, or case O holds with n 4. Then Ω is (i) {e1 , . . . , em , f1 , . . . , fm }, n = 2m, where Q(ei ) = Q(fi ) = 0 and (ei , fj ) = δij , ∀i, j; (ii) {e1 , . . . , em−1 , f1 , . . . , fm−1 , x, y}, n = 2m, where Q(ei ) = Q(fi ) = 0, (ei , fj ) = δij and (ei , x) = (ei , y) = (fi , x) = (fi , y) = 0, for all i, j, and {x, y} satisfying the following condition (Q(x), Q(y), (x, y)) = (1, ζ, 1), where t2 + t + ζ is irreducible over F. (iii) {e1 , . . . , em , f1 , . . . , fm , x}, n = 2m + 1, where Q(ei ) = Q(fi ) = 0, (ei , fj ) = δij and (ei , x) = (fi , x) = 0, for all i, j, and x is non-singular. Deﬁne    ◦    sgn(Q) = +      − if n is odd if n is even and (V, Q) has a basis of type 2.4(i) if n is even and (V, Q) has a basis of type 2.4(ii) (2.7) We divide case O into cases O◦ , O+ and O− , accordingly. Proposition 2.5 ([29], Proposition 2.5.4). For each n, there are precisely two isometry classes of orthogonal geometries in dimension n. (i) If n is even, then the two isometry types are distinguished by the dimension of their maximal totally singular subspaces. Indeed, the maximal totally singular subspaces have dimension n 2 or n 2 − 1 according as sgn(Q) = + or sgn(Q) = −. (ii) If n is odd, then the two isometry types are distinguished by the value of Q(x) modulo (F∗ )2 , where x is given in Deﬁnition 2.4(iii). The two geometries are similar, and all 1 maximal totally singular subspaces have dimension 2 (n − 1). Let β be a basis of V, the discriminant D(Q) of Q, is D(Q) ≡ det(fβ )(mod(F∗ )2 ) ∈ F∗ /(F∗ )2 . We write D(Q) = or , according as D(Q) is a square or a non-square. When q is odd and n is even, the next result determines sgn(Q). 14 Proposition 2.6 ([29], Proposition 2.5.10, 2.5.12, 2.5.13). Assume that q is odd and n = 2m is even. We have: (i) If sgn(Q) = +, then D(Q) = (ii) If sgn(Q) = −, then D(Q) = (iii) D(Q) = 1 if and only if 2 m(q − 1) is even; if and only if 1 m(q − 1) is odd; 2 1 if and only if sgn(Q) = (−) 2 (q−1)m . (iv) V has a basis β such that fβ is either In or diag(λ, 1, · · · , 1), according as D(Q) = or D(Q) = , where λ is a generator of F∗ . Proposition 2.7 ([29], Propposition 2.5.11). Assume that V = V1 ⊥ · · · ⊥ Vt , where each Vi is a non-degenerate subspace of V. (i) D(Q) = t i=1 D(QVi ); t i=1 (ii) If dim(Vi ) is even for all i, then sgn(Q) = sgn(QVi ). Proposition 2.8 ([29], Proposition 2.6.1). Let (V, F, Q) be a classical orthogonal geometry in odd dimension n = 2m + 1. Then there exists a basis β of (V, F, Q) such that fβ = λIn , where D ≡ λ(mod (F∗ )2 ). Let U be a non-degenerate subspace of V. Write D(U ) = D(QU ) and sgn(U ) = sgn(QU ). Let (V, F, Q) be a classical orthogonal geometry of type ε ∈ {◦, ±} with q odd and dimV = n. For any non-zero vector x in V, a one-space with representative x will be called a point in V and denoted by x . We now deﬁne a type function ρ = ρV on V \ {0} as follows: If x is a singular vector in V, that is, x ∈ V \ {0} and Q(x) = 0, then ρ(x) = 0. If n = dimV = 2m is even, then ρ(x) = Q(x). If n = 2m + 1 is odd, then ρ(x) = sgn(x⊥ ). Assume that dimV is odd. Let x be a non-singular vector in V. We say that x is a plus vector if ρ(x) = +; and x is a minus vector if ρ(x) = −. We also say point x is of plus or minus type according to whether its representative x is a plus or minus vector. Let x ∈ V be a non-singular vector with ρ(x) = ξ. Deﬁne Eε (V ) to be the set of all non-singular ξ points of type ξ in V. We can omit either the index ξ or the vector space V, when they 15 are understood. From deﬁnition, we have Eε (V ) = { v ⊆ V | Q(v) = 0 and ρ(v) = ξ}. ξ Recall that Vγ = {v ∈ V \ 0 | Q(v) = γ}, for any γ ∈ F. Remark 2.9 If dimV is even then Eε (V ) = { v | v ∈ Vξ }. However, if dimV is odd, ξ they may be not equal. Lemma 2.10 Assume that dimV is odd. Let x, y be two non-singular vectors V. Then two vectors x, y (or two non-singular points x , y ) have the same type if and only if Q(x) ≡ Q(y) (mod (F∗ )2 ). Thus if x is a non-singular vector in V with Q(x) = γ = 0 and ρ(x) = sgn(x⊥ ) = ξ. Then Eε (V ) = ξ Eξ (V ) = { v | v ∈ Vγ }. Proof. Assume ﬁrst that Q(x) ≡ Q(y) (mod (F∗ )2 ). According to Proposition 2.5(ii), x and y are isometric. By Witt’s Lemma, this isometry extends to an isometry g of V such that x g = y . As x , y are non-degenerate, x⊥ g = y ⊥ . It follows that x⊥ and y ⊥ are isometric, and hence sgn(x⊥ ) = sgn(y ⊥ ), so that ρ(x) = ρ(y). Now, assume that x, y have the same type. By Witt’s Lemma and Proposition 2.5(i), there exists an isometry between x⊥ and y ⊥ . This isometry can extend to an isometry g of V such that (x⊥ )g = y ⊥ . As (x⊥ )⊥ = x , and (y ⊥ )⊥ = y , ( x )g = y . Thus xg = µy for some µ ∈ F∗ . Therefore Q(x) = Q(xg) = Q(µy) = µ2 Q(y). If F = F3 , then F∗2 = {1}. The result follows. When n is odd, for any non-zero vector x in V, we denote by S(n, x) the number of all vectors v ∈ V with ρ(v) = ρ(x) = ξ ∈ {0, ±}, and Q(v) = Q(x). When n is even, for any γ ∈ F, set S ε (n, γ) = |Vγ |, the number of solutions to Q(v) = γ in V. Lemma 2.11 Assume q is odd, x is a non-zero vector in V with ρ(x) = ξ ∈ {0, ±}, and γ ∈ F. 16 γ∈ξF∗2 { v |v ∈ Vγ }. In particular, if q = 3 then 1. S ε (2k, γ) =   2k−1  q + ε(q k − q k−1 ) if γ = 0,  q 2k−1 − εq k−1  if γ = 0; 2k k 2. S(2k + 1, x) = q + ξq . Proof. (1) follows from Proposition 9.10 in [14]. For (2), let γ = Q(x). Firstly, assume that x is a non-singular vector. Then V = x ⊥ x⊥ , sgnx⊥ = ρ(x) = ξ and dimx⊥ = 2k. According to Lemma 2.10, for any v ∈ V, if Q(v) = Q(x) then ρ(v) = ρ(x). Thus S(2k + 1, x) = |Vγ |. Now, for any v ∈ V with Q(v) = γ, write v = ϕx + v0 , where ϕ ∈ F and v0 ∈ x⊥ . Then Q(v0 ) = Q(v) − ϕ2 Q(x) = γ(1 − ϕ2 ). If ϕ = ±1, then Q(v0 ) = 0, hence by (1), there are 2S ξ (2k, 0) = 2(q 2k−1 + ξ(q k − q k−1 )) such v. If ϕ = ±1, then Q(v0 ) = γ(1 − ϕ2 ) = 0, hence by (1), again, there are (q − 2)S ξ (2k, γ(1 − ϕ2 )) = (q − 2)(q 2k−1 − ξq k−1 )) such v. Thus, S(2k + 1, x) = 2S ξ (2k, 0) + (q − 2)S ξ (2k, γ(1 − ϕ2 )) = q 2k + ξq k . Finally, assume that x is a singular vector. Then S(2k + 1, x) = |V0 |. Observe that V always contains a non-singular vector y. Let η = ρ(y) and µ = Q(y). Arguing as previous case, we have V = y ⊥ y ⊥ , sgn(y ⊥ ) = η, Q(y) = µ ∈ F∗ and dimy ⊥ = 2k. For any v ∈ V with Q(v) = 0, write v = ϕy + v0 , where ϕ ∈ F and v0 ∈ y ⊥ . Then Q(v0 ) = Q(v) − ϕ2 Q(y) = −µϕ2 . If ϕ = 0, then Q(v0 ) = 0, hence by (1), there are S η (2k, 0) = q 2k−1 + η(q k − q k−1 ) such v. If ϕ = 0, then Q(v0 ) = −µϕ2 = 0, hence by (1), again, there are (q − 1)S η (2k, −µϕ2 ) = (q − 1)(q 2k−1 − ηq k−1 )) such v. Thus, S(2k + 1, x) = S η (2k, 0) + (q − 1)S η (2k, −µϕ2 ) = q 2k . 2.2 Deﬁnitions and Structures of classes C and S Let G0 be one of the ﬁnite classical groups P SLn (q), P SUn (q), P Spn (q)( n even ), P Ω± (q)( n even), Ωn (q)( nq odd). n (2.8) 17 Ci C1 C2 C3 C4 C5 C6 C7 Table 2.1: Rough description of classes Ci rough description rough structure in GLn (q) stabilizers of totally maximal parabolic singular or non-singular subspaces stabilizers of decompositions V = t Vi , dim(Vi )=a GLa (q) St , n = at i=1 stabilizers of extension ﬁelds GLa (q b ) · b, n = ab, b prime of Fq of prime index b stabilizers of tensor product GLa (q) ◦ GLb (q), n = ab decompositions V = V1 V2 b stabilizers of subﬁelds of GLn (q0 ), q = q0 , b prime Fq of prime index b normalizers of sympletic-type r-groups (r prime) in absolutely (Zq−1 ◦ r1+2a ) · Sp2a (r), n = ra irreducible representations stabilizers of decopositions t V = C8 t i=1 Vi , dim(Vi )=a classical subgroups (GLa (q) ◦ · · · ◦ GLa (q)) · St Spn (q), n even ε On (q), q odd √ GUn ( q), q a square Let V be the natural n-dimensional vector space over the ﬁnite ﬁeld F associated with G0 , and let Ξ be the full semilinear classical group corresponding to G0 . M. Aschbacher deﬁned eight collections Ci (Ξ), (i = 1, . . . , 8) of natural subgroups of Ξ. The rough description of the subgroups H ∩ GLn (q), where H ∈ Ci (Ξ) and G0 = P SLn (q) is given in Table 2.1. For each group X satisfying either Ω X A or Ω X A, the collection C(X) is a union of families Ci (X), where i = 1, · · · , 8, we also deﬁne Ci = ∪Ci (X) and C = ∪C(X). In each collection Ci , there are many sub-collections called type. If T is a type in Ci , and H belongs to T, then we say that H is of type T. Moreover, for any group HΞ ∈ C(Ξ), there are corresponding group HX = HΞ ∩ X ∈ CX , for any X as above. If Y satisﬁes the same condition as X, then HX is called X-associate of HY . Now we deﬁne the class S of subgroups of G, where G0 G Aut(G0 ) and G0 as in (2.8). Deﬁnition 2.12 The subgroup H of G lies in S if and only if the following hold. 18 (a) The socle S of H is a non-abelian simple group (i.e., H is almost simple). (b) If L is the full covering group of S, and if ρ : L → GL(V ) is a representation of L such that ρ(L) = S, then ρ is absolutely irreducible. (c) ρ(L) cannot be realized over a proper subﬁeld of F. (d) If ρ(L) ﬁxes a non-degenerate quadratic form on V, then G0 = P Ωε (q). n (e) If ρ(L) ﬁxes a non-degenerate symplectic form on V, but no non-degenerate quadratic form, then G0 = P Spn (q). (f ) If ρ(L) ﬁxes a non-degenerate unitary form on V, then G0 = P SUn (q). (g) If ρ(L) does not satisfy the conditions in (d), (e) or (f ), then G0 = P SLn (q). We now state the subgroup structure theorem due to Aschbacher. Theorem 2.13 Let G be a group such that G0 G Ξ, with G0 and Ξ as in (2.8) above, and let H be a subgroup of G not containing G0 . Then either H is contained in a member of C(G) or H ∈ S. Kleidman and Liebeck have determined the maximality of subgroups of C(G) when n Here is the main theorem of [29]. Theorem 2.14 Let G0 be as in (2.8) and let G0 G Aut(G0 ). Then 13. (A) the group-theoretic structure of each H ∈ C(G) is known; (B) the conjugacy amongst the members of C(G) is known; (C) for H ∈ C(G) all overgroups of H which lie in C(G) S are known (for n 13). In order to determine the structure and conjugacy amongst members of C, we need the following deﬁnitions. For any group X, Y satisfying Y Ω X A, deﬁne [Y ]X as the set of X-conjugates of Y. Write [Y ] = [Y ]Ω and deﬁne [Y ]X to be the set of Ω-classes contained in the X-class [Y ]X . Thus [Y ]X = {[Y x ] | x ∈ X}. Deﬁne e = |[HΩ ]A |, (this is parameter c in [29]) and write [HΩ ]A = {[H1 ], · · · , [He ]}. Also deﬁne Xi = X[Hi ] = 19 Table 2.2: The simple classical groups Lie notation L Lie rank d |Out(L)| |L| n 1 n(n−1)/2 i Ln (q) An−1 (q) (n, q − 1) 2df, n 3 q i=2 (q − 1) d n−1 df, n = 2 n 1 n(n−1)/2 i i 2 Un (q) (n, q + 1) 2df, n 3 An−1 (q) q i=2 (q − (−1) ) d [n] df, n = 2 2 n 1 n2 2i P Sp2n (q) Cn (q) (2, q − 1) df, n 3 q i=1 (q − 1) d n 2f, n = 2 n 1 n2 2i Ω2n+1 (q) Bn (q) 2 2f q i=1 (q − 1) d q odd n + 1 P Ω2n (q) Dn (q) (4, q n − 1) 2df, n = 4 d q n(n−1) (q n − 1) n−1 (q 2i − 1) i=1 n 3 n 6df, n = 4 1 n(n−1) n 2 P Ω− (q) (4, q n + 1) 2df Dn (q) q (q + 1) n−1 (q 2i − 1) 2n i=1 d n 2 n−1 ¨ NX (Hi )Ω. Then Xi = NX (Hi )Ω/Ω. For each i = 1, · · · , e, write HG,i for G-associate of ¨ ¨ Hi . The action of A on [HΩ ]A induces a homomorphism π = πHΩ from A to the symmetric group Se . 2.3 Properties of ﬁnite simple groups 5, a ﬁnite simple group of Lie type, or one of By the theorem of classiﬁcation of ﬁnite simple groups, every non-abelian ﬁnite simple group is either an alternating group An , n the 26 sporadic groups. The ﬁnite simple groups of Lie type is divided into the classical groups and the exceptional groups. The orders of alternating group An is 1 n!. The orders 2 of remaining simple groups are given in Tables 2.2, 2.3 and 2.4. Theorem 2.15 The excetional groups in Table 2.3 are simple except for 2 B2 (2) ∼ Z5 : Z4 = G2 (2) ∼ Aut(U3 (3)) ∼ U3 (3).2 = = 2 2 G2 (3) ∼ Aut(L2 (8)) ∼ L2 (8).3 = = F4 (2) ∼ 2 F4 (2) .2. = 20 L G2 (q) F4 (q) E6 (q) E7 (q) E8 (q) 2 B2 (q), q = 22m+1 2 G2 (q), q = 32m+1 2 F4 (q), q = 22m+1 3 D4 (q) 2 E6 (q) Table 2.3: The simple exceptional groups d |Out(L)| |L| 6 2 2 1 f if p = 3 q (q − 1)(q 6 − 1) 2f if p = 3 4 1 (2, p)f q 24 (q 2 − 1)(q 6 − 1)(q 8 − 1)(q 12 − 1) 1 36 i 6 (3, q − 1) 2df q i∈{2,5,6,8,9,12} (q − 1) d 1 63 i 7 (2, q − 1) df q i∈{2,6,8,10,12,14,18} (q − 1) d 8 1 f q 120 i∈{2,8,12,14,18,20,24,30} (q i − 1) 1 1 f q 2 (q 2 + 1)(q − 1) 1 1 f q 3 (q 3 + 1)(q − 1) 2 1 f q 12 (q 6 + 1)(q 4 − 1)(q 3 + 1)(q − 1) 2 1 3f q 12 (q 8 + q 4 + 1)(q 6 − 1)(q 2 − 1) 1 36 i i 4 (3, q + 1) 2df q i∈{2,5,6,8,9,12} (q − (−1) ) d The group 2 F4 (2) is simple and is called the Tits group. We deﬁne Lie(p) to be the set of simple groups of Lie type over ﬁelds of characteristic p. We include P Sp4 (2) and 2 F4 (2) in Lie(2). We also deﬁne Lie(p ) = ∪r=p Lie(r). A member of Lie(p) is said to be twisted if it lies in one of the families 2 A ,2 B2 ,2 D ,3 D4 ,2 E6 ,2 F4 or 2 G2 . Otherwise, L is untwisted. Theorem 2.16 The isomorphisms among the groups in Tables 2.2, 2.3 and 2.4 and the alternating groups are precisely those given in Theorem 2.15 together with the following isomorphisms: (i) SL2 (q) ∼ Sp2 (q) ∼ SU2 (q). = = (ii) For q odd, L2 (q) ∼ Ω3 (q). = ± (iii) O2 (q) ∼ D2(q = ± 1) , SO2 (q) ∼ Zq 1 .(2, q) and Ω± (q) ∼ Zq = = 2 1/(2,q) . (iv) Ω+ (q) ∼ SL2 (q) ◦ SL2 (q). = 4 (v) Ω− (q) ∼ L2 (q 2 ). = 4 (vi) For q odd, P Sp4 (q) ∼ Ω5 (q). = (vii) P Ω± (q) ∼ L± (q). = 4 6 (viii) L2 (2) ∼ S3 . = 21 Table 2.4: The simple sporadic groups L d |Out(L)| |L| 4 2 M111 1 1 2 .3 .5.11 M12 2 2 26 .33 .5.11 M22 12 2 27 .32 .5.7.11 M23 1 1 27 .32 .5.7.11.23 M24 1 1 210 .33 .5.7.11.23 J1 1 1 23 .3.5.7.11.19 J2 2 2 27 .33 .52 .7 7 5 J3 3 2 2 .3 .5.17.19 21 3 J4 1 1 2 .3 .5.7.113 .23.29.31.37.43 HS 2 2 29 .32 .53 .7.11 13 7 2 Suz 6 2 2 .3 .5 .7.11.13 M cL 3 2 27 .36 .53 .7.11 Ru 2 1 214 .33 .53 .7.13.29 He = F7 1 2 210 .33 .52 .73 .17 8 7 6 Ly 1 1 2 .3 .5 .7.11.31.37.67 ON 3 2 29 .34 .5.73 .11.19.31 Co1 2 1 221 .39 .54 .72 .11.13, 23 Co2 1 1 218 .36 .53 .7.11.23 Co3 1 1 217 .37 .53 .7.11.23 F i22 6 2 217 .39 .52 .7.11.13 F i23 1 1 218 .313 .52 .7.11.13.17.23 21 16 2 3 F i24 3 2 2 .3 .5 .7 .11.13.17.23.29 HN = F5 1 2 214 .36 .56 .7.11.19 T h = F3 1 1 215 .310 .53 .72 .13.19.31 41 13 6 2 BM = F2 2 1 2 .3 .5 .7 .11.13.17.19.23.31.47 M = F1 1 1 246 .320 .59 .76 .112 × 133 .17.19.23.29.31.41.47.59.71 22 (ix) L2 (3) ∼ A4 . = (x) L2 (4) ∼ L2 (5) ∼ A5 . = = (xi) L2 (7) ∼ L3 (2). = (xii) L2 (9) ∼ A6 . = (xiii) L4 (2) ∼ A8 . = (xiv) U3 (2) ∼ 32 .Q8 . = (xv) U4 (2) ∼ P Sp4 (3). = (xvi) Sp4 (2) ∼ S6 = Let L be a simple group. Denote by Out(L) the outer automorphism group of L, which is Aut(L)/L. The orders of Out(L) are given in Tables 2.2, 2.3 and 2.4 for simple groups of Lie type and sporadic groups. The orders of Out(An ), n theorem. Theorem 2.17 (Theorem 5.1.3[29].) If n |Out(An )| = 2. However |Out(A6 )| = 4. Lemma 2.18 (Lemma 2.1[47]). Let 2 Then (q a1 + ε1 ) . . . (q a + ε ) 1 < < 2. 2 q a1 +···+a Using the above lemma, we can easily verify the following: Lemma 2.19 Assume that L is a ﬁnite simple group of Lie type. Then |Aut(L)| where f (L) is given in Table 2.5. Primitive prime divisors Let q and n be integers with q 2 and n 3. A prime s is a primitive prime divisor of f (L), a1 < · · · < a be integers, ε1 , . . . , ε ∈ {±1}. 5 and n = 6, then Aut(An ) = Sn and hence 5 are given in the following q n − 1, denoted by qn , if s|q n − 1 but s does not divide q i − 1 for i < n. The Zsigmondy’s theorem (Zsigmondy [48]) asserted the existence of primitive prime divisor. 23 Table 2.5: Upper bounds for order of Aut(L). L f (L) L f (L) n2 Ln (q), n 3 q E8 (q) q 249 2 +n+1 P Sp2n (q) q 2n F4 (q) q 53 2 2 Un (q), n 3 qn E6 (q) q 79 + 2n2 −n+1 P Ω2n (q), n = 4 q G2 (q) q 15 3 P Ω+ (q) 3q 29 D4 (q) 3q 29 8 2 −2n+3 2 P Ω− (q) q 2n F4 (q) q 27 2n 2 +n+1 Ω2n+1 (q) q 2n Sz(q) q6 79 2 E6 (q) q G2 (q) q8 2 −2n+2 E7 (q) q 134 P Ω− (q) 2q 2n 2n Theorem 2.20 There exists a primitive prime divisor of q n − 1, provided q and (q, n) = (2, 6). When n is odd and (q, n) = (2, 3), then there is a primitive prime divisor q2n of q 2n − 1, and we write q−n = q2n . Representations of simple groups Let G be a ﬁnite group and P, M be subgroups of G. Denote by M \G/P a set of representatives for double cosets of G on P and M. Let E = G/P, the right cosets of G on P. Then M acts on E by right multiplication. Lemma 2.21 Let M, P be subgroups of a ﬁnite group G. Then (i) M has |M \G/P | orbits on G/P, where M acts on G/P by right multiplication. (ii) (1G , 1G ) = |M \G/P |. M P Proof. (i) Suppose that M \G/P = {x1 , . . . , xk }. Then P xi M ∩ P xj M = ∅ if i = j, i, j = 1, . . . , k. Clearly, P xi M are distinct orbits of M on G/P. As G = ∪k P xi M, {P xi M }k i=1 i=1 is a complete set of orbits of M on G/P. 2, n 3 24 (ii) By Mackey’s formula ([23]), (1G , 1G ) = M P t∈M \G/P (1t t ∩P , 1M t ∩P ) = M t∈M \G/P 1 = |M \G/P |. This proves the lemma. Let G be a group and p a prime. A p-modular representation of G is a homomorphism form G to GLn (F) for some n and some ﬁeld F of characteristic p. A projective p-modular representation of G is a homomorphism from G to P GLn (F). If G has a faithful projective p modular representation of degree n, then G embeds in P GLn (F), written G We deﬁne P GLn (F). RF (G) = min{n | G Rp (G) P GLn (F)}, (2.9) = min{RF (G) | F a ﬁeld of characteristic p}, Rp (G) = min{Rs (G) | s prime, s = p}, R(G) = min{Rp (G) | all primes p}. P GLn (Fp )}. If Fp is the algebraic closure of Fp , then Rp (G) = min{n | G For irreducible projective representations, we have similar deﬁnitions. i RF (G) = min{n | G embeds irreducibly in P GLn (F)} i Rp (G) i = min{RF (G) | F a ﬁeld of characteristic p} = min{n | G embeds irreducibly in P GLn (Fp )} i i Rp (G) = min{Rs (G) | s prime, s = p}, (2.10) Ri (G) i = min{Rp (G) | all primes p}. Embedding of ﬁnite simple groups We ﬁrst deﬁne the Frobenius-Schur indicators of the irreducible Brauer characters. Let G be a ﬁnite group, F be a ﬁeld and M be a ﬁnitely generated, absolutely irreducible 25 FG-module which aﬀords the Brauer character χ. If M is self-dual, then M carries a non-degenerate G-invariant bilinear form, which is symmetric or alternating and unique up to scalar multiplication. If the characteristic of F is 2, and M is not the trivial module, then the invariant form is alternating. If the characteristic of F is odd and M carries a non-degenerate symmetric bilinear form, this form is the associated bilinear form of a quadratic form on M , which is non-degenerate and G−invariant. If M is not self-dual, it does not carry any non-trivial G−invariant bilinear form. The Frobenius-Schur indicator of M, written ind(M ) or ind(χ), is an integer of the set {−1, 0, +1}. It is deﬁned to be 0 if and only if M is not self-dual. The Frobenius-Schur indicator of M is set to be +1 if and only if M carries non-degenerate G−invariant quadratic form. Thus the indicator of M is -1 if and only if M is self-dual and carries a non-degenerate G−invariant bilinear alternating form, but no G−invariant quadratic form. If ind(χ) = +1, then G embeds in an orthogonal groups and the value of χ determine the smallest ﬁeld F such that G ε embeds in GOn (F) for some ε ∈ {+, −}, where n = dim(M ). If characteristic of F is 2 and n is even, then G embeds in Spn (F). If ind(χ) = −1, then G embeds into Spn (F), where the ﬁeld F is determined by the value of χ. If ind(χ) = 0, then G may or may not embed in a suitable unitary groups. 2.4 Representations of Symmetric groups We collect some information on representations of symmetric and alternating groups in characteristic p. We are interested in the lower bounds for absolutely irreducible p-modular representations of these groups. We begin by some basic deﬁnitions. Deﬁnition 2.22 (i) λ = (λ1 , . . . , λk ) is a partition of n, written λ 1, . . . , k are integers, with λ1 ... λk > 0 and k i=1 n, provided λi , i = λi = n. We often gather the equal h i=1 parts together and write λ = (λa1 , · · · , λah ), where λ1 > . . . λh > 0 and 1 h ai λi = n. ai p−1 (ii) For any prime p, a partition λ = (λa1 , λa2 , · · · , λah ) is called p-regular if 1 1 2 h 26 for all i = 1, . . . , h. (iii) Let λ = (λa1 , λa2 , · · · , λah ) be a p-regular partition. Then λ is called a JS- partition 2 1 h if λi − λi+1 + ai + ai+1 ≡ 0 mod p. Next we construct the fully deleted module for alternating groups. This is the smallest faithful irreducible representation of An when n is at least 9. Let n 5, let p be a prime, n and let Sn act on the permutation module Fp by permuting the coordinates naturally. n Deﬁne submodules U, W of Fp by n U = {(a1 , a2 , . . . , an )| i=1 ai = 0}, W = {(a, a, . . . , a)|a ∈ Fp }. Let V = U/(U ∩ W ), and    εp (n) = 0 if p does not divide n  1 if p|n.  Then dim(V ) = n − 1 − εp (n). V is said to be the fully deleted permutation module for An over Fp . Deﬁne the natural symmetric bilinear form f : U × U → Fp given by f ((a1 , . . . , an ), (b1 , . . . , bn )) = n i=1 ai bi . An preserves this form and f induces a symmetric bilinear form on the fully deleted permutation module V, and we obtain the embedding An Ω(V, Fp , f ). 10 and V be a non-trivial irreducible An n. Then V is isomorphic to the fully Proposition 2.23 ([25], Theorem 6). Let n module in characteristic p. Suppose that dim(V ) deleted module for An . Let λ = (λ1 , λ2 , · · · , λh ) be a partition of n, where λ1 ··· λh and λ1 + · · · + λh = n. The subgroup Sλ = S{1,2,··· ,λ1 } × S{λ1 +1,··· ,λ1 +λ2 } × S{λ1 +···+λh−1 +1,··· ,λ1 +···+λh } is called a Young subgroup of Sn associated to λ. Denote by M λ the permutation module 1Sn . Let Sλ 27 λ = (λ1 , λ2 , . . . , λh ) n be a partition of n. The Young diagram associated to λ, written i, 1 j λi }, which is consisting of n nodes. The [λ], is the subset (i, j) ∈ {Z × Z | 1 λ-tableaux is one of the n! arrays of integers obtained by replacing each node in [λ] by one of the integers 1, 2, · · · , n, without repetition. The symmetric group Sn acts naturally on the set of λ-tableaux. Let t be a tableaux. The row-stabilizer, Rt , is a subgroup of Sn which ﬁxes the rows of t. The column stabilizer, Ct , of t is deﬁned similarly. Using the row stabilizer, we can deﬁne an equivalence relation on the set of all λ-tableaux by saying that t1 ∼ t2 if and only if t1 π = t2 for some π ∈ Rt1 . The tabloid {t} containing t is the equivalence class of t under this equivalence relation. The group Sn acts on tabloids as follows: {t}π = {tπ} for any tabloid t and π ∈ Sn . Let F be any ﬁeld. Then M λ becomes an FSn -module via the action of Sn on λ-tabloids. Let t be a λ-tableau. The element κt = π∈Ct (sgnπ)π ∈ FSn is called the signed column sum. The polytabloid, et , associated with the tabloid t is et = {t}κt . The Specht module S λ for the partition λ is the submodule of M λ generated by polytabloids. We next ﬁnd a basis for Specht module. Let t be a λ-tableaux. Then t is a standard tableaux if the numbers increase along the rows and down the columns of t. We say {t} is a standard tabloid if there is a standard tableaux in the equivalence class {t}. Also et is a standard polytabloid if t is standard. Theorem 2.24 ( [24], 8.4) {et | t is a standard λ-tableaux} is a basis for S λ . The dimension of Specht modules can be given by the Hook formula. We need some deﬁnitions before we can give the formula. Let [λ] be a Young diagram. The conjugate diagram [λ ] is obtained by interchanging the rows and columns in [λ]. If λ = (λ1 , λ2 , · · · ) then λ = (λ1 , λ2 , · · · ), where λi is the length of ith column in [λ]. The (i, j) − hook of [λ] consists of the (i, j)-node along with the λi − j nodes to the right of it and the λj − i nodes below it. The length of (i, j)-hook is hij = λi + λj + 1 − i − j. The hook graph of [λ] is obtained from the Young diagram [λ] by replacing the (i, j)-node by the numbers hij . 28 Theorem 2.25 ( [24], 20.1). The dimension of the Specht module S λ is given by n! . (hook lengths in [λ]) dimS λ = We can deﬁne a non-degenerate symmetric bilinear form on M λ as follows    1 ({t1 }, {t2 }) =  0  if {t1 } = {t2 } otherwise. Suppose that charF = p is a prime and λ is p-regular. Deﬁne Dλ = S λ /(S λ ∩ S λ⊥ ). Theorem 2.26 ([24], Theorem 11.5). {Dλ | λ a p-regular partition} is a complete system of inequivalent irreducible FSn -modules. Moreover, Dλ with λ p-regular, is self-dual and absolutely irreducible, and every ﬁeld is a splitting ﬁeld for Sn . We record here some results on the decomposition of the Specht modules S λ and the degrees of Dλ for some small partitions. Lemma 2.27 ([24], Theorems 24.1, 24.15). Let F be a ﬁeld of characteristic p. Then (i) S (n−1,1) = D(n−1,1) + εp (n)D(n) , and dimD(n−1,1) = n − 1 − εp (n). (ii) if p > 2, then S (n−2,1 ) = D(n−2,1 ) + εp (n)D(n−1,1) and   1  (n − 1)(n − 2), 2 =  1 (n − 2)(n − 3),  2 2 2 dimD(n−2,1 2) if p n if p | n. (iii) if p > 2, then S (n−2,2) = D(n−2,2) + εp (n − 2)D(n−1,1) + εp (n − 1)D(n) , and   1  n(n − 3),  2   1 = (n2 − 3n − 2),  2    1 2  (n − 5n + 2), 2 if n ≡ 1, 2 (mod p) if n ≡ 1 (mod p) if n ≡ 2 (mod p) dimD(n−2,2) 29 (iv) If p = 2, then S (n−2,2) = D(n−2,2) + εp (n − 2)D(n−1,1) + δD(n) , and   1 (n − 1)(n − 4),  2       1 (n2 − 3n − 2), 2 =  1 (n2 − 5n + 2),   2     1  n(n − 3), 2 if n ≡ 0 (mod 4) if n ≡ 1 (mod 4) if n ≡ 2 (mod 4) if n ≡ 3 (mod 4) dimD(n−2,2) where δ=    1, if n ≡ 1, 2 (mod 4)  0, otherwise.  For k 1, denote by Rn (k) the set of irreducible Sn -modules D such that D ∼ Dλ or = n, with λ1 n − k. 17 if p = 2. n D ∼ Dλ ⊗ S (1 ) for some p-regular partition λ = Theorem 2.28 ([25], [39] Proposition 2.2). Let n 12 if p = 2, and n Then any irreducible FSn -module D either belongs to Rn (2) or dimD > (n − 2)(n − 3). We will need information about the restrictions to Sn−1 , An−1 of modular irreducible representations of symmetric groups Sn , and alternating groups An . The result will be useful in determine the maximality of members of S, whose socles are alternating groups. ai +1 For i ∈ {1, . . . , h}, we deﬁne λ(i) = (λa1 , λa2 , . . . , λai −1 , λai −1 , λi − 1, λi+1 , . . . , λah ), a 1 2 i−1 i h partition of n − 1. With above notations, we can state the branching rules for symmetric groups. Theorem 2.29 ([31], Theorem 0.3). Let λ n be a p-regular partition. Then Dλ ↓Sn−1 is irreducible if and only if λ is a JS-partition. If this is the case, then Dλ ↓Sn−1 = Dλ(1) . Next we consider the restriction to An of Dλ . In order to determine Dλ ↓An , we need to ﬁnd the ﬁxed points under the Mullineux map m(λ), originally deﬁned in [40]. Following Ben Ford in [10], we will use the p-modular Frobenius symbol to ﬁnd the ﬁxed points of m(λ). Let λ = (λ1 , λ2 , . . . , λk ) n be a partition of n. The rim of the Young diagram [λ] 30 of λ is the collection of nodes which are either at the bottom of a column, at the right end of a row, or both. The p-rim of [λ] is deﬁned as follows: Beginning at the top right-hand corner of [λ], the ﬁrst p nodes of the rim are in the p-rim. Then skip to the next row, and take the next p nodes of the rim. Continue until we reach the end of the rim. The last p-segment may contain fewer than p nodes. Let h1 be the number of nodes in the p-rim of λ, and let r1 be the number of rows in λ. Delete the p-rim and repeat the process to get h1 , r1 , . . . , hk , rk , where hk+1 = rk+1 = 0, but hk = 0 = rk . The Mullineux symbol is a 2 × k matrix,  h1 h2 · · · M (λ) =  r1 r2 · · · Now, the p-regular partition m(λ) of n is deﬁned via  h1 h2 · · · M (m(λ)) =  s1 s2 · · · where  hk  , sk  hk  . rk    0, εi =  1,  if p | hi if p hi and si = hi −ri +εi . Note that the partition m(λ) can be reconstructed from the Mullineux symbol M (m(λ)). The following notations will be useful if we just want to know whether a given partition is ﬁxed under the Mulllineux map. The p-modular Frobenius symbol for λ, denoted by F rp (λ), is a 3 × k matrix   a1 a2 · · ·  F rp (λ) =  b1 b2 · · ·   ε1 ε2 · · · 31  ak   bk    εk constructed as follows:    ai = hi − ri  b = r −ε  i i i If λ has p-modular Frobenius symbol F rp (λ) as constructed above then the Mullineux map m is deﬁned by   b1 b2 · · ·  F rp (m(λ)) = a1 a2 · · ·   ε1 ε2 · · ·  bk   ak    εk which means that we interchange the ﬁrst two rows of F rp (λ). Therefore, we see that λ is a ﬁxed point of m if and only if the ﬁrst two rows of F rp (λ) are the same (see [10]). Denote by sgnn , the sign character of Sn , which takes value 1 at even permutation and −1 at odd permutation. Now, if λ is a p-regular partition of n then Dλ sgnn = Dm(λ) . This is the Mullineux conjecture, which was ﬁnally proved in [11]. Let λ be a 3-regular partition of n with λ1 n − 2. Then λ is rarely a ﬁxed point of Mullineux map. 5 and λ is a p-regular partition of n. Suppose that n 6 and λ = (n − 2, 12 ). Lemma 2.30 Assume that p = 3, n λ1 n − 2. Then m(λ) = λ if and only if 5 Proof. As λ1 n − 2, the possibilities for λ1 are n, n − 1 or n − 2. It follows that λ = (n), (n − 1, 1), (n − 2, 2) or (n − 2, 12 ). If one of the ﬁrst three cases holds then the result follows by Lemma 1.8 in [30]. Assume that λ = (n − 2, 12 ). We ﬁrst compute the Mullineux symbol M (λ) of λ. We have  5 3 · · · M (λ) =  3 1 ···  3 a , 1 1 32 where 3, and so 1, occurs t times with t = Hence n−2 ] 3 − 1, and 0  2 a − 1  1 0 .   0 1 a = n − 2 − 3(t + 1) 2.  2 2 · · ·  F r3 (λ) = 2 1 · · ·   1 0 ··· 1, or equivalently n If t 8, then clearly, the ﬁrst two rows of F r3 (λ) cannot be equal, n 7. Then   2 a − 1   F r3 (λ) = 2 0 .     1 1 so that λ = m(λ). Thus 5 Observe that the ﬁrst two rows of F r3 (λ) are equal if and only if a = 0 or a = 1. Since t = 0, a = n − 2 − 3 = n − 5, so that n = 5 or 6. Theorem 2.31 ([10], Theorem 2.1). Let F be a splitting ﬁeld for An of characteristic p > 2, and λ n be a p-regular partition. Then (i) If λ = m(λ), then Dλ ↓An = Dm(λ) ↓An is irreducible. (ii) If λ = m(λ), then Dλ ↓An is a sum of two irreducible, non-equivalent representations λ λ of An , say D+ and D− , interchanging by an odd permutation. Finally λ λ {Dλ ↓An |λ = m(λ)} ∪ {D+ , D− | λ = m(λ)} is a complete system of inequivalent irreducible FAn -modules. We next prove a gap result between the minimal module and the second minimal module for Alternating groups in characteristic 3. Lemma 2.32 Let F be a splitting ﬁeld for An of characteristic p = 3. Suppose that n and V is an irreducible FAn -module with dimV > n. Then dimV 33 1 (n2 2 12 − 5n + 2). λ Proof. It follows from Theorem 2.31 that either V = Dλ ↓An with m(λ) = λ or V = D± , where m(λ) = λ. Now, let U = Dλ for partition λ obtained above. By Theorem 2.28, either dimU > (n − 2)(n − 3), or U ∈ Rn (2). Observe that dimV = dimU if m(λ) = λ 1 and if m(λ) = λ then dimV = 2 dimU. Thus, if dimU > (n − 2)(n − 3), then clearly, dimV 1 dimU 2 1 1 > 2 (n − 2)(n − 3) > 2 (n2 − 5n + 2). Therefore, we can assume that U ∈ Rn (2). Thus there exists a 3-regular partition µ with µ(1) n − 2 such that λ = µ or λ = m(µ). As n > 10 and dimV > n, it follows from Proposition 2.23 that µ is not (n) nor (n − 1, 1), and so µ = (n − 2, 2), or (n − 2, 12 ). By Lemma 2.30, µ is not ﬁxed under the Mullineux map. Also, as Mullineux map is an involutory map, m(m(µ)) = µ = m(µ). We conclude that Dµ and Dm(µ) are irreducible upon restricted to An . Since dimDm(µ) = dimDµ ⊗ sgnn = dimDµ , we have dimV = dimDλ = dimDµ . Finally, the result follows from Lemma 2.27. Continue the hypotheses in Theorem 2.31, let λ ◦ ◦ n be a p-regular partition with p odd. λ λ Deﬁne Dλ = Dλ ↓An if λ = m(λ) and Dλ ∈ {D+ , D− }, if λ is a ﬁxed point of the Mullineux map. The branching rule for An is given in the following theorem. Theorem 2.33 ([4], Theorem 5.10). Let n 2, p an odd prime, and λ be a p-regular partition of n. Then the following are equivalent: (i) Dλ ↓An−1 is irreducible; (ii) One of the following holds: (a) Dλ ↓Sn−1 is irreducible; (b) Dλ ↓Sn−1 = Dλ(1) Dm(λ(1)) and λ = m(λ) but λ(1) = m(λ(1)). ◦ 34 2.5 Representations of Finite groups of Lie type in cross characteristic The irreducible cross characteristic representations of L2 (q) are given in Table 2.7. These are taken from [19]. For notations in this table, the columns d are the degrees of the representations, the columns are the characteristic of the ﬁelds of representations, and the ‘ﬁeld’ columns give the irrationalities of the Brauer characters. Landazuri and Seitz [32] and Seitz and Zalesskii [43] obtained the lower bound for the irreducible cross characteristic representations of ﬁnite simple Chevalley groups. This result has been improved by many other authors including Guralnick, Magaard, Saxl and Tiep [16], Guralnick and Tiep [15], [17], and Hoﬀman [20]. Table 2.6 is taken from [20]. Theorem 2.34 Assume that L is a ﬁnite simple Chevalley group in characteristic p. Then Rp (L) e(L), where e(L) is as in Table 2.6. 2.6 Representations of Finite groups of Lie type in deﬁning characteristic Simple algebraic groups Let k be an algebraically closed ﬁeld of characteristic p. A simple algebraic group over k is a linear algebraic group over k which has no proper, closed, connected normal subgroups. The simple algebraic groups over k were classiﬁed and they are the groups of types A (k), B (k), C (k), D (k), E (k)(l = 6, 7, 8), F4 (k) and G2 (k). For each type, there is a unique simply connected group whose center Z is as large as possible, and an adjoint group with trivial center. Let G be a simple algebraic group over k, and let q = pf . Regard G as a subgroup of GLn (k) for some n. Let σq be a map from G to G sending (aij ) 35 Table 2.6: Bounds for Cross Characteristic Representations of Chevalley Groups. L e(L) exceptions P SL2 (q) (q − 1)/(2, q − 1) e(P SL2 (4)) = 2, e(P SL2 (9)) = 3 P SLn (q), n 3 (q n − q)/(q − 1) − 1 e(P SL3 (2)) = 2, e(P SL3 (4)) = 4, e(P SL4 (2)) = 7, e(P SL4 (3)) = 26 n P Sp2n (q), n 2 (q − 1)/2, q odd (q n − 1)(q n − q)/(2(q + 1)), q even e(P Sp4 (2) ) = 2 n P SUn (q), n 3 (q − q)/(q + 1), n odd (q n − 1)/(q + 1), n even e(P SU4 (2)) = 4, e(P SU4 (3)) = 6 + n n−1 2 P Ω2n (q), n 4 (q − 1)(q + q)/(q − 1) − 2, q > 3 (q n − 1)(q n−1 − 1)/(q 2 − 1), q 3 e(P Ω+ (2)) = 8 8 P Ω− (q), n 4 (q n + 1)(q n−1 − q)/(q 2 − 1) − 1 e(P Ω− (2)) 32, 2n 8 e(P Ω− (4)) 1026, 8 e(P Ω− (2)) 151, 10 e(P Ω− (3)) 2376 10 P Ω2n+1 (q), n 3 (q 2n − 1)/(q 2 − 1) − 2, q > 3, q odd (q n − 1)(q n − q)/(q 2 − 1), q = 3 e(P Ω7 (3)) = 27 E6 (q) qΦ8 (q)Φ9 (q) − 2 E7 (q) qΦ7 (q)Φ12 (q)Φ14 (q) − 3 E8 (q) qΦ4 (q)2 Φ8 (q)Φ12 (q)Φ20 (q)Φ24 (q) − 4 F4 (q) q 4 (q 6 − 1), q odd q 7 (q 3 − 1)(q − 1)/2, q even e(F4 (2)) = 52 2 9 2 E6 (q) q (q − 1) G2 (q) q(q 2 − 1) e(G2 (3)) = 14, e(G2 (4)) = 12 3 D4 (q) q 3 (q 2 − 1) 2 F4 (q) q 4 q/2(q − 1) Sz(q) q/2(q − 1) e(Sz(8)) = 8 2 G2 (q) q(q − 1) 36 Table 2.7: Irreducible representations of L2 (q). (a) L2 (q), q ≡ 1(mod 4), q degree G ﬁeld ind q−1 √ L2 (q) 2 q − 2 q−1 √ 2.L2 (q) =2 q − 2 q+1 √ L2 (q) =2 q + 2 −j j q−1 L2 (q) ζq+1 + ζq+1 + j −j q − 1 2.L2 (q) ((q + 1)/2) = 1 ζq+1 + ζq+1 − q L2 (q) (q + 1) + 2j −2j q+1 L2 (q) ((q − 1)/4) = 1 ζq−1 + ζq−1 + j −j q + 1 2.L2 (q) =2 ζq−1 + ζq−1 − (b) L2 (q), q ≡ 3( mod 4), q degree G ﬁeld ind √ q−1 L2 (q) −q ◦ 2 √ q+1 2.L2 (q) =2 −q ◦ 2 2j −2j q−1 L2 (q) ζq+1 + ζq+1 + j −j q − 1 2.L2 (q) =2 ζq+1 + ζq+1 − q L2 (q) (q + 1) + j −j q+1 L2 (q) ((q − 1)/2) = 1 ζq−1 + ζq−1 + j −j q + 1 2.L2 (q) ((q − 1)/2) = 1, = 2 ζq−1 + ζq−1 − (c) L2 (q), q ≡ 0( mod 2), q degree G ﬁeld ind j −j q − 1 L2 (q) ζq+1 + ζq+1 + q L2 (q) (q + 1) + j −j q + 1 L2 (q) (q − 1) = 1 ζq−1 + ζq−1 + 37 to (aq ). Then σq is called a standard Frobenius map. In general, a map σ : G → G is said ij to be a Frobenius map if some power of σ is a standard Frobenius map. Steinberg ([46], Corollary 10.13) showed that if σ : G → G is surjective then either σ is an automorphism or Gσ is ﬁnite. Now let σ be a Frobenius map and let Gσ be the ﬁxed point group of σ in G. Clearly, σ is a surjective map but it is not an automorphism. Hence Gσ is ﬁnite, and the ﬁnite groups Op (Gσ ) are called the ﬁnite groups of Lie type in characteristic p. A torus of the simple algebraic group G over k is a subgroup which is isomorphic to a direct product of copies of k ∗ . Every torus lies in a maximal torus and all maximal tori are conjugate in G. Each maximal torus T is isomorphic to k ∗ , where is the Lie rank of G, and CG (T ) = T. The group W = NG (T )/T is ﬁnite and is called the Weyl group of G. A Borel subgroup of G is a maximal closed, connected, solvable subgroup of G. It is shown that all Borel subgroups are conjugate in G, and the maximal tori of G are those of the Borel subgroups of G. Fix a Borel subgroup B and a maximal torus T of G. Let U be the unipotent radical of B, which consists of all unipotent elements of the largest connected normal solvable subgroup of B. Then B is self-normalizing, B = NG (U ) = U : T, and there exists a unique Borel subgroup of G, denoted by B − , such that B ∩ B − = T, called opposite B (relative to T ). (cf. [22], 21.3, 23.1, 26.2). Modules and weights Let L be a simply connected group of Lie type over Fq , where q = pf . Thus L = Op (Gσ ), where G is a simply connected, simple algebraic group over k, and σ is a suitable Frobenius map. Fix a maximal torus T and a Borel subgroup B of G. Let X = X(T ) be the character group of T, the set of algebraic group homomorphisms from T to k ∗ . If M is any irreducible rational kG-module, that is, the corresponding map G → GL(M ) is an algebraic group homomorphism, then M = µ∈X Mµ , where for µ ∈ X, we have Mµ = {v ∈ M | vt = µ(t)v for all t ∈ T }. If Mµ = 0, then µ is said to be a weight of M, and Mµ the µ-weight space of M. The Weyl group W = NG (T )/T acts on X and induces 38 a group of permutations on the weights of M. As X is abelian, it is a Z-module. Thus we may form the tensor product E = R Z X, and the action of W on X yields an action of W on E. Choose a positive deﬁnite, bilinear, symmetric W -invariant R-form (., .) on E. Assume that L is the adjoint module for G, that is, L is the Lie algebra of G over k, with the natural G-action. The set Φ of weights of L is called the set of roots of G. Select a system Φ+ of positive roots from Φ, with corresponding set Π = {α1 , . . . , α } of fundamental roots, giving the Dynkin diagram as in Figure 2.6. Deﬁne a partial order on 2α X by writing µ λ if and only if λ − µ is a sum of positive roots. For α ∈ Φ, α∗ = (α, α) is the co-root corresponding to α. As L is simply connected, there is a basis {λ1 . . . , λ } ∗ ∗ of E which is dual to {α1 , . . . , α∗ }, that is, with (λi , αj ) = δij . The λi form a Z-basis for X, and they are called the fundamental dominant weights. Deﬁne X + = { ci ∈ Z, ci 2 i=1 ci λi | 0} and Xq = { i=1 ci λi | ci ∈ Z, 0 ci q − 1}. Let L be a group of type B2 ,2 G2 or 2 F4 , over Fp2a+1 , where p = 2 or 3. For a root α, set   a  p , q(α) =  pa+1 ,  if α is a long root, if α is a short root. ci q(αi ) − 1}. Each member of X + is called a We deﬁne Xq = { dominant weight. i=1 ci λi | ci ∈ Z, 0 The irreducible rational G-modules are characterized by their highest weights. Theorem 2.35 ([22], 31.1). Let M be an irreducible (rational) kG-module. (a) There is a unique B-stable 1-dimentional subspace, spanned by a vector v + ∈ M, (a maximal vector), with dominant weight λ, whose multiplicity is 1, and dim(Mλ ) = 1. For all other weights µ of M, µ = λ − i=1 ci λi λ, where ci ∈ Z, ci 0. These weights are permuted by the Weyl group W, and W -conjugate weights have the same multiplicity. (b) Irreducible kG-modules are determined up to isomorphism by their highest weights. 39 i We show the Dynkin diagrams with the node of αi labeled by i. (This is the labeling use in all the databases of the CHEVIE [5] project.) s s s s p p p p p p F4 i i Al 1 Dl G2 1 s1 d 3 ds s2 s> 2s s 3 2 3 l s s Bl 1 s< 2s s> 2s 1 3 s s s s p p p p p p 4 l s s s s 4 l Cl 1 3 l 1 E7 1 s 4 s s2 5 s 6 s s 2 7 s s> 3s 4 s E8 E6 1 s 3 s s s s2 3 s s 4 5 6 s2 5 s s 6 7 8 Figure 2.1: Dynkin diagram. 146 (c) For each dominant weight λ ∈ X, there exists an irreducible kG-module L(λ) of highest weight λ. The unique dominant weight λ in this theorem is called the highest weight of M. The following theorem of Steinberg shows us how to get kL-modules from kG-modules. Theorem 2.36 ([46],§13). Let Λ = Xq if L is of type 2 B2 ,2 G2 or 2 F4 , and Λ = Xq otherwise. Then for λ ∈ Λ, the modules L(λ) remain irreducible and inequivalent upon restriction to L, and exhaust the irreducible kL-modules. Let M be a kL-module aﬀording a representation ρ : L → GLn (k), and ν be the automorphism of GLn (k) induced by the action of the ﬁeld automorphism t → tp on matrix entries. For r 1, let M (r) be the space M with L-action given by the representation ρν r . A dominant weight is called p-restricted if it lies in Xp . Steinberg’s twisted tensor product theorem shows how to construct all highest weight modules L(λ) from those of p-restricted highest weights. 40 Theorem 2.37 ([44], Steinberg’s Twisted Tensor Product Theorem ). Let λ = µ0 + pµ1 + · · · + pe−1 µe−1 ∈ Xq , where µi are p-restricted for all i, and q = pe . Then L(λ) = L(µ0 ) ⊗ L(µ1 )(1) ⊗ · · · ⊗ L(µe−1 )(e−1) . Let w0 be the longest element of the Weyl group W of G, w0 can be characterized as the unique element of W such that w0 (Φ+ ) = Φ− , where the root system Φ is the disjoint union of Φ+ and Φ− . Elements of Φ− are called negative roots. Let L be of type A , D , E6 or D4 over Fq , L possesses a graph automorphism τ◦ of order 2, 2, 2, or 3, respectively, which induces a symmetry τ on the Dynkin diagram.    −1 w0 =  −τ  for types B , C , D ( even), G2 , F4 , E7 , E8 , for types A , D ( odd), E6 . Proposition 2.38 ([42], 1.8). L(λ)∗ ∼ L(−w0 λ). = Proposition 2.39 ([42], 1.11 ). Let λ = q1 µ1 +· · ·+qk µk , where µ1 , . . . , µk are p-restricted dominant weights and q1 , . . . , qk are pairwise distinct powers of p. Then G leaves invariant a non-degenerate bilinear form on L(λ) if and only if G leaves such a form invariant on each of L(µ1 ), . . . , L(µk ). Proposition 2.40 ([45]) Let L be a group of Lie type over Fq . (i) If L is untwested or of type 2 B2 ,2 G2 or 2 F4 , then Fq is a splitting ﬁeld for L. (ii) If L is of type 2 A ,2 D or 2 E6 , then Fq2 is a splitting ﬁeld for L. (iii) If L is of type 3 D4 , then Fq3 is a splitting ﬁeld. Proposition 2.41 ([29], Theorem 5.4.5). Let L be simply connected of Lie type over Fpe , and suppose that V is an absolutely irreducible Fpf L-module which is realized over no proper subﬁeld of Fpf . 41 (i) If L is untwisted, then f | e and there is an irreducible kL−module M such that V ⊗ k ∼ M ⊗ M (f ) ⊗ · · · ⊗ M (e−f ) . = In particular, dim(V ) = dim(M )e/f . (ii) If L is of type 2 A , 2 D or 2 E6 , then one of the following occurs. (a) f | e, V ∼ V τ◦ , there is an irreducible kL-module M such that M ∼ M τ◦ and = = V ⊗ k ∼ M ⊗ M (f ) ⊗ · · · ⊗ M (e−f ) . = In particular, dim(V ) = dim(M )e/f . (b) f | 2e but f does not divide e, V ∼ V τ◦ , V (f /2) ∼ V τ◦ , there is an irreducible = = kL-module M such that M ∼ M τ◦ and = V ⊗ k ∼ M ⊗ (M τ◦ )(f /2) ⊗ M (f ) ⊗ (M τ◦ )(3f /2) ⊗ · · · ⊗ (M τ◦ )(e−f ) ⊗ M (e−(e/f )) . = In particular, dim(V ) = dim(M )2e/f . (iv) If L has type 3 D4 , then either f | e and (ii)(a) holds or f | 3e, but f does not divide e, and V ⊗ k is a tensor product of 3e/f modules M ⊗ (M τ◦ )(f /2) ⊗ M (f ) ⊗ (M τ◦ )(3f /2) ⊗ . . . and dimV = (dimM )(3e/f ) . (v) If L is of type 2 B2 ,2 G2 or 2 F4 over Fq , where q = p2a+1 and p is 2, 3 or 2, respectively, then f | 2a + 1 and dimV and 2 F4 , respectively. Rp (L)(2a+1)/f , where Rp (L) is 4, 7 and 26 for types 2 B2 ,2 G2 42 Chapter 3 Rank 3 permutation characters and maximal subgroups 3.1 Higman rank 3 parameters and the equation Let G be a ﬁnite group acting transitively on a non-empty ﬁnite set E. Let P be a stabilizer of a point x ∈ E in G. Recall that the action is primitive if and only if P is maximal in G. Also the rank of G is the number of orbits of P on E. Now suppose that G is of even order and acts primitively rank 3 on E. So P has exactly three orbits on E, namely, {x}, ∆(x) and Γ (x). Deﬁne k = |∆(x)|, l = |Γ (x)| and    µ if y ∈ Γ (x)  λ if y ∈ ∆(x)     λ1 if y ∈ Γ (x) |Γ (x) ∩ Γ (y)| =  µ if y ∈ ∆(x).  1 Suppose that k l. |∆(x) ∩ ∆(y)| = Lemma 3.1 ([18], Lemma 5 and 7). Let G act primitively rank 3 on E. Then (i) |E| = k + l + 1; 43 (ii) µl = k(k − 1 − λ); (iii) D = (λ − µ)2 + 4(k − µ) is a square; (iv) λ1 = l − k + µ − 1; (v) µ1 = l − k + λ + 1. Let V be the permutation module for G on E over C, hence E is a basis for V. Further ∆ and Γ can be viewed as linear maps on V, via the corresponding x → x→ y∈Γ (x) y∈∆(x) y and y, for x ∈ E, and extend linearly. We have Σy∈E y is an eigenvector for ∆ and Γ, with eigenvalues k and l, respectively. Other eigenvalues for ∆ and Γ are as follows: Lemma 3.2 ([18], Lemma 6). The eigenvalues for ∆ are: λ−µ+ s= 2 √ D λ−µ− and t = 2 √ D . Lemma 3.3 ([2], 1.4.3). For r ∈ {s, t}, the eigenvalues for Γ are −(r + 1). Let Vr , r = s or r = t, be the irreducible modules for G on V which is the eigenspace for ∆ with eigenvalue r. Let fr = dim(Vr ). Lemma 3.4 ([18]). fs = k + t(k + l) k + s(k + l) and ft = . t−s s−t Let M be any subgroup of G. Fix x ∈ E, and let P = Gx . By identifying E with G/P, x corresponds to the coset P in G/P. Consider the action of G on the right cosets G/M and form the permutation module VM . Denote by y the coset M as a point in G/M. Deﬁne d = dx = |xM ∩ ∆(x)| and c = cx = |xM ∩ Γ (x)|. As G acts transitively on E and G/M, xG = E and G/M = yG. Deﬁne α : V → VM by α(xg) = Σp∈P ypg for g ∈ G 44 and β : VM → V by β(yg) = Σm∈M xmg for g ∈ G. Then α and β are CG maps and θ = 1 .β |P ||P ∩M | ◦ α ∈ EndCG (V ). The map θ can be written in terms of linear maps ∆ and Γ as follows: Lemma 3.5 ([2], (2.1)). θ = I + Proof. For any g ∈ G, we have 1 1 1 .β(α(xg)) = . β(ypg) = . xmpg. |P ||P ∩ M | |P ||P ∩ M | p∈P |P ||P ∩ M | p∈P,m∈M u∈O d∆ cΓ + k l θ(xg) = Let O be one of the three orbits of P on E and vO = u ∈ V. As P acts on O, xmp ∈ O if and only if xm ∈ O, in which case since P is transitive on O, |P | vO . |O| xmp = p∈P Moreover there are |Mx | = |P ∩ M | elements in M ﬁxing x and |P ∩ M |d, |P ∩ M |c elements m ∈ M with xm ∈ O for O = ∆, Γ, respectively. Therefore dv∆ cvΓ d∆ cΓ + )(x)g = (I + + )(xg). k l k l θ(xg) = (vx + This proves the lemma. For r = s, t, let πr be the projection of V on Vr . Lemma 3.6 ([2], (2.2) and (2.3)). Let r = s or t, then (i) θ ◦ πr = 0 if and only if Vr ker(θ); (ii) If θ ◦ πr = 0 then α : Vr → VM is an injective CG map; 45 (iii) For r = s, t : θ ◦ πr = 0 if and only if dr (r + 1)c = . k l 1+ (3.1) Proof. (1) is clear as πr (V ) = Vr . By Lemma 3.4, Vs is not CG-isomorphic to Vt , so that θ : Vr → Vr . As Vr is irreducible CG-module, θ is an isomorphism if θ is non-zero. Thus (2) follows from (1). For (3), let r = s, t and v ∈ V. Let Σ = ∆ or Γ. Then (Σ ◦ πr )(v) = e(Σ, r)πr (v), where e(Σ, r) is the eigenvalue of Σ on Vr . Thus Σ ◦ πr = e(Σ, r)πr . From deﬁnition, e(r, ∆) = r and by Lemma 3.3, e(r, Γ ) = −(r + 1). Therefore by Lemma 3.5, we obtain: dr (r + 1)c d∆ cΓ + ) ◦ πr = (1 + − )πr . k l k l θ ◦ πr = (I + Thus θ ◦ πr = 0 if and only if dr (r + 1)c − = 0. k l 1+ This ﬁnishes the proof. As G is a primitive rank 3 group on E with P the stabilizer of a point x in E, the permutation character 1G has a decomposition 1G = 1 + χs + χt , where 1 is the trivial P P character, and χs , χt are irreducible characters of G, aﬀorded by the irreducible modules Vs , Vt , with degrees fs , ft , respectively. We say that 1G P 1G if 1G − 1G is a character M M P of G. This is equivalent to (χr , 1G ) > 0 for any r ∈ {s, t}. By Lemma 3.6, for r ∈ {s, t}, M (χr , 1G ) = 0 if and only if equation (3.1) holds for any x ∈ E. Note that the parameters M c, d depend on x, or equivalently, on the conjugate of P. When we pick diﬀerent conjugate 46 of P, parameters c and d change. Thus for r ∈ {s, t}, if equation (3.1) does not hold for some point x ∈ E or some conjugate of P, then by Lemma 3.6, there is an injective CG map from Vr to VM and hence 1G P 1G . Otherwise we need to change to diﬀerent M conjugate of P or diﬀerent point in E. See Proposition 3.20 for such an example. The following result will be used frequently to show that there is no containment if M has at most 2 orbits on E. Corollary 3.7 Let G be a primitive rank 3 group acting on a ﬁnite set E. Let P be the stabilizer of a point in E, and M be any subgroup of G. If M has at most two orbits on E, then 1G P 1G . M 1G . Then φ = 1G − 1G is a character M M P 3, since φ, 1G are P Proof. By way of contradiction, suppose that 1G P of G. Thus (1G , 1G ) = (1G , φ + 1G ) = (1G , 1G ) + (1G , φ) = 3 + (1G , φ) P M P P P P P P characters of G. By Lemma 2.21, (1G , 1G ) is the number of orbits of M on G/P. Now, by M P identifying E with G/P, M has (1G , 1G ) M P 3 orbits on E. A contradiction. 3.2 An example Let L = Ω5 (3). By Theorem 2.16, we have isomorphisms: Ω5 (3) ∼ P Sp4 (3) ∼ P SU4 (2). = = Information on the maximal subgroups of L, extracted from [9], is given in Table 3.1. Let x be a minus point and P the stabilizer in L of x. From Table 3.1, we have P ∼ S6 = and the permutation character of P in L can be decomposed as 1L = 1a + 15b + 20a. It P follows that L has exactly 3 orbits on the cosets L/P. Since P is maximal in L, L acts as a primitive rank 3 permutation group on L/P. Let M be any maximal subgroup of L which is not conjugate to P. For example, take M to be the stabilizer of an isotropic point. Then M ∼ 33 : S4 , and 1L = 1a + 15b + 24a. We have 1L − 1L = 24a − 20a, = M M P which is not a character of L. Thus 1L P which is not conjugate to P, 1L P 1L . In fact, for any maximal subgroup M of L M 1L . Note that L is also a primitive rank 3 permutation M 47 Table 3.1: Maximal Order Index Structure 960 27 24 : A5 720 36 S6 1+2 648 40 3+ : 2A4 648 40 33 : S4 576 45 2 · (A4 × A4 ).2 subgroups of Ω5 (3) Character Speciﬁcation 1a + 6a + 20a base 1a + 15b + 20a minus point 1a + 15a + 24a isotropic line 1a + 15b + 24a isotropic point 1a + 20a + 24a plus point group on the cosets L/M for any maximal subgroup M of L. In general, we expect more containments. 3.3 Main Hypothesis and Notations 2 or m 3, respectively. Let V be the natural From now on, we assume the following set up. Let L be one of the following quasi-simple groups Ω2m+1 (3), or Ωε (3) with m 2m module for L over F = F3 with a non-degenerate quadratic form Q. Denote by E(V ) an L-orbit of non-singular points in V. Let G be a group such that G acts as a primitive rank 3 group on E(V ), and L G, G/Z(L) Aut(L/Z(L)). We say that G is a nearly G I(V ), where I(V ) is the simple primitive rank 3 group of type L. Observe that L full isometric group of V. Denote by P the stabilizer of a point in E(V ). The letter M will be reserved for the maximal subgroup of G. If M ∈ C(G) then for any subgroup X satisfying Ω(V ) X Ξ(V ), the X-associate of M is denoted by MX . If M ∈ S(G) then we denote the socle of M by S. Then S is a non-abelian ﬁnite simple group and the full covering group S of S acts absolutely irreducibly on V and is not realizable over a proper subﬁeld of F. Moreover S ﬁxes a non-degenerate quadratic form on V so that the Frobenius-Schur indicator ind(V ) is +. (See Section 2.2 and 2.3). 48 3.4 3.4.1 Ω2m+1(3) Parameters for Ω2m+1 (3) 2. We now assume the hypothesis and notations in section 3.3 with L = Ω2m+1 (3), m There are two types of non-singular points in V, namely + and − points. For ξ ∈ {±}, denote by Eξ (V ) the set of all non-singular points of type ξ. For any x ∈ Eξ (V ), deﬁne ∆(x) = E(V ) ∩ x⊥ and Γ (x) = E(V ) ∩ (V − x⊥ − { x }). Then P has exactly three orbits { x }, ∆(x) and Γ (x) on Eξ (V ). For ξ = ±, we write ξ instead of ξ1 to denote ±1. Recall the Higman rank 3 parameters deﬁned in section 3.1. Lemma 3.8 Let ξ ∈ {±} and x ∈ Eξ (V ). Then 1 (i) |Eξ (V )| = 2 3m (3m + ξ); (ii) k = 1 3m−1 (3m − ξ); 2 (iii) l = (3m − ξ)(3m−1 + ξ); 1 (iv) µ = λ = 2 3m−1 (3m−1 − ξ); √ (v) D = 2.3m−1 ; (vi) s = 3m−1 ; (vii) t = −3m−1 . Proof. Let γ = Q(x). We have ρV (x) = sgn(x⊥ ) = ξ. Since q = 3, by Lemma 2.10, V Eξ (V ) = { v | v ∈ Vγ }. Observe that each point in V has q − 1 representatives. Therefore 1 1 by Lemma 2.11, |Eξ (V )| = 2 S(2m + 1, x) = 2 (32m + ξ3m ) = 1 3m (3m + ξ), which gives 2 (i). From deﬁnition k = |∆(x)| = |Eξ (V ) ∩ x⊥ | = |{ v ⊆ x⊥ | Q(v) = Q(x)}|. By 1 Lemma 2.11 again, k = 2 S ξ (2m, γ) = 1 S ξ (2m, γ) = 1 (q 2m−1 − ξq m−1 ). The parameter l 2 2 can be computed easily by the relation 1 + k + l = |Eξ (V )|. This proves (ii) and (iii). To compute λ, take y ∈ ∆(x). Then ρ(y) = ρ(x) = ξ, (x, y) = 0 and Q(y) = Q(x). 49 We have λ = |∆(x) ∩ ∆(y)| = where z ∈ W := x⊥ ∩ y ⊥ = x, y 1 |{v 2 ⊥ ∈ x⊥ ∩ y ⊥ | Q(v) = Q(x)}| = 1 S(2m 2 − 1, z), with Q(z) = Q(x). We need to determine ρW (z) = ⊥ ⊥ sgnzW . Since x⊥ = y ⊥ (y ⊥ ∩ x⊥ ) = y ⊥ W and W = zW ⊥ z , we deduce ⊥ ⊥ that x⊥ = y, z ⊥ zW , where dimzW = 2m − 2, dimW = 2m − 1 and dim y, z = 2. Now, as D y, z = det(diag(−γ, −γ)) = , by Proposition 2.6, sgn y, z = (−)1 = −. ⊥ It follows from Proposition 2.7 that sgnzW = −sgnx⊥ = −ξ. Thus by Lemma 2.11, 1 λ = 2 (32m−2 − ξ3m−1 ), which is (iv). The remaining parameters follow from Lemma 3.1 and Lemma 3.2. Corollary 3.9 Let M be a subgroup of G and ξ ∈ {±}. Suppose that equation (3.1) holds for some r ∈ {s, t}, and for some M -orbit x M with x ∈ Eξ (V ). Then (i) If (ξ, r) = (+, s) or (−, t) then equation (3.1) has the form c − 2d = ξ3m − 1. (3.2) (ii) If (ξ, r) = (+, t) or (−, s) then equation (3.1) has the form (ξ3m−1 + 1)(ξ3m − 1 + c − 2d) = 2c (3.3) (iii) If m 2 then 1+c+d 3m + 1 2 3m−1 . (3.4) Proof. From deﬁnitions c 0. Let A = 1 + c + d. Multiply both sides of drl drl equation (3.1) by l, we have (r + 1)c = l + . Subtract from both sides, k k (r + 1)c − drl = l. k (3.5) 0 and d (1) If ξ = + and r = s, then by Lemma 3.8, we have r = 3m−1 , l = (3m − 1)(3m−1 + 1), l and = 2(3m−1 + 1)/3m−1 . From (3.5), (3m−1 + 1)c − 2d(3m−1 + 1) = (3m − 1)(3m−1 + 1). k 50 Divide both sides by 3m−1 + 1, we get c − 2d = 3m − 1 = ξ3m − 1. Hence c = 2d + 3m − 1. Thus A = 3m + 3d 3m 1 m (3 2 + 1). (2) If ξ = − and r = t, then by Lemma 3.8, we have r = −3m−1 , l = (3m +1)(3m−1 −1), l and = 2(3m−1 − 1)/3m−1 . From (3.5) (−3m−1 + 1)c + 2d(3m−1 − 1) = (3m + 1)(3m−1 − 1). k Divide both sides by 3m−1 − 1, we get c − 2d = −3m − 1 = ξ3m − 1. In this case, 2d = 3m + 1 + c. Since c 0, 2A = 2 + 2c + 3m + 1 + c 3m + 3 > 3m + 1. (3) If ξ = + and r = t, then by Lemma 3.8, we have r = −3m−1 , l = (3m −1)(3m−1 +1), l and = 2(3m−1 + 1)/3m−1 . From (3.5) (−3m−1 + 1)c + 2d(3m−1 + 1) = (3m − 1)(3m−1 + 1) k or 2c = (3m−1 + 1)(3m − 1 + c − 2d) = (ξ3m−1 + 1)(ξ3m − 1 + c − 2d), and 2d(3m−1 + 1) = (3m−1 + 1)(3m − 1) + (3m−1 − 1)c. As c 2d 3m − 1. Now 2A = 2 + 2c + 2d 0, 2d(3m−1 + 1) 3m + 1. (3m−1 + 1)(3m − 1), or 2 + 2c + 3m − 1 (4) If ξ = − and r = s, then by Lemma 3.8, we have r = 3m−1 , l = (3m + 1)(3m−1 − 1), l and = 2(3m−1 − 1)/3m−1 . From (3.5) (3m−1 + 1)c − 2d(3m−1 − 1) = (3m + 1)(3m−1 − 1), k or equivalently 2c = (3m−1 − 1)(3m + 1 − c + 2d) = (ξ3m−1 + 1)(ξ3m − 1 + c − 2d). Therefore (3m−1 +1)c = (3m−1 −1)(3m +1)+2d(3m−1 −1). As d Now A 0, c (3m−1 −1)(3m +1)/(3m−1 +1). 1 + (3m−1 − 1)(3m + 1)/(3m−1 + 1) = (32m−1 − 3m−1 )/(3m−1 + 1). However 0. 1 (32m−1 − 3m−1 )/(3m−1 + 1) − 2 (3m + 1) = (3m − 1 + 3m+1 (3m−2 − 1))/(2(3m−1 + 1)) Therefore A 1 m (3 2 1 + 1). Finally 2 (3m + 1) > 1 3m > 3m−1 . The proof is completed. 2 3.4.2 Permutation characters of maximal subgroups in C By Proposition 2.6.2 in [29], A = I = S × −1 and S = Ω r r . Let M ∈ C(G) be a maximal subgroup of G. Let MΞ ∈ C(Ξ) be such that M = G∩MΞ . Then MI = M∆ = MΞ , and MΩ M MI . The reducible subgroups C1 The reducible subgroups C1 are all the groups MΞ of the forms NΞ (W ), where dimW = α, n 1 α and W is either non-degenerate or totally singular. The corresponding 2 51 ε2 ε subgroups are of type Oα1 (q) ⊥ On−α (q) or Pα , where n = dimV, ε1 = sgn(W ), ε2 = sgn(W ⊥ ), and (α, ε1 ) = (n − α, ε2 ). The subgroup MΞ is maximal in Ξ except when MΞ + ± is of type O2 (3) ⊥ On−2 (3), as it is contained in subgroups of type O1 (3) ⊥ On−1 (3). ε2 ε Proposition 3.10 Assume M is of type Oα1 (3) ⊥ O2m+1−α (3). There is an M -orbit on ε2 Eξ (V ) such that equation (3.1) does not hold unless M is of type O1 (3) ⊥ O2m (3). In this case M is in Table 1.1. ε2 ε Proof. As M is of type Oα1 (3) ⊥ O2m+1−α (3), there exists a non-degenerate subspace W V of dimension α such that M = NG (W ). Put W1 = W, W2 = W ⊥ , εi = sgnWi . Write Xi = X(Wi ), where X ranges over the symbols Ω, S and I. By Lemma 4.1.1 in [29], we have MI = I1 × I2 and Ω1 × Ω2 dimW1 = 2b+1, where 0 MΩ . Without loss of generality, we can suppose that 1. b < m. If b = 0, then the proposition holds. Assume that b Then W1 contains non-singular points of both types. Let xi , i = 1, 2 be non-singular points in W1 of diﬀerent types. By Lemma 2.3, xi Ω1 = xi I1 , i = 1, 2. Moreover, as I2 centralizes xi , i = 1, 2, we have xi MΩ = xi MI , so that xi MΩ = xi M. Thus it is suﬃcient to compute parameters c, d for subgroup MΩ in L. Let x ∈ {x1 , x2 }, η = ρW1 (x) and ξ = ρV (x). Since x MΩ = Eη (W1 ), it follows that c = l(W1 ) and d = k(W1 ), the 1 parameters for Eη (W1 ). By Lemma 3.8, d = 2 3b−1 (3b − η), c = (3b − η)(3b−1 + η), and so c − 2d = η3b − 1. Suppose that equation (3.1) holds for some r = s, t and any M -orbits on Eξ (V ). If (3.2) holds then η3b − 1 = ξ3m − 1. This implies that b = m, a contradiction. Thus (3.3) holds. Then (3m−1 + ξ)(3m − 2ξ + ξη3b ) = 2(3b − η)(3b−1 + η). Observe that m−1 b. Assume ﬁrst that m − 1 = b and ξ = −η. We have 3m−1 + ξ = 3b − η and 3m − 2ξ + ξη3b = 3b+1 + 2η − 3b = 2(3b + η). Clearly 2(3b + η) > 2(3b−1 + η), and hence equation (3.3) does not hold in this case. Next assume that m − 1 = b and ξ = η. Then (3m−1 +ξ)(3m −2ξ+ξη3b ) = (3b +ξ)(3b+1 −2ξ+3b ) = 2(3b +η)(2.3b −η) > 2(3b−1 +η)(3b −η), a contradiction. Finally assume that m − 1 > b so that m − 1 3b+1 + ξ > 3b − η and 3m − 2ξ + ξη3b b + 1. Then 3m−1 + ξ 9.3b − 2ξ − 3b = 8.3b − 2ξ > 2(3b−1 + η). Multiplying 52 these two inequalities side by side will lead to a contradiction. Thus equation (3.3) does not hold. Therefore b = 0 and so W is a point or a hyperplane. Let W be a totally singular subspace of V of dimension α. By Witt’s Lemma, we can assume that W has a basis {e1 , . . . , eα }, where the vectors ei are taken from a standard basis of V. Let Y = f1 , . . . , fα , and X = eα+1 , . . . , em , fα+1 , . . . , fm , a . Then V = (W Y ) ⊥ X, W ⊥ = W ⊥ X. Finally let U = CI(V ) (W, W ⊥ /W, V /W ⊥ ), and N = NI(V ) (W, Y, X). The relation among these groups is given in the following lemmas. Lemma 3.11 ([29], Lemma 4.1.9). Assume that n = 2m is even and that case U, S or O+ holds. Let β = {e1 , · · · , em , f1 , · · · , fm } be the corresponding standard basis. Let W1 = e1 , · · · , em , W2 = f1 , · · · , fm , and T0 = NI(V ) (W1 , W2 ), where V = β . Then (i) T0 ∼ GLm (q u ) and T0 acts naturally on W1 . = (ii) As T0 -modules we have W2 ∼ W1 , where α is an element of order u in Aut(F). = α∗ (iii) T0 ∩ Ω(V ) = {x ∈ T0 | detW1 (x) ∈ (F∗ )z }, where z = q + 1, 1, 2 in cases U, S and O+ , respectively. Lemma 3.12 ([29], Lemma 4.1.12). Let W be a totally singular subspace of V. Keeping the notations above, we have: (i) MI = U : N ; (ii) N = T0 × I(X), where GLα (q u ) as T0 -modules we have Y ∼ W ; = (iii) U is a p-group and U Ω(V ). 53 T0 I(W ⊕ Y ), and T0 acts naturally on W ; and It follows from (i) and (iii) of Lemma 3.12 that MΩ = U (N ∩ Ω(V )). Proposition 3.13 Assume M is of type Pα . Then M has at most two orbits in Eξ (V ) so that 1G P 1G by Corollary 3.7 and so M is in Table 1.1. M Proof. We will show that MΩ has at most two orbits on E(V ), and hence we deduce that M also has at most two orbits on E(V ). Assume the notation above and denote by f the associated bilinear form of Q. From deﬁnition M stabilizes a totally singular subspace W of dimension α. Let Y and X be deﬁned as above. By Proposition 2.8, X has a basis βX = {x1 , . . . , xs }, with s = 2m + 1 − 2α = 2(m − α) + 1, such that [f↓X ]βX = λIs , where D(X) ≡ λ(mod (F∗ )2 ). Let β = {e1 , . . . , eα , x1 , . . . , xs , f1 , . . . , fα }. Let x ∈ X be a non-singular point with ξ = ρX (x). As sgn(W ⊕ Y ) = +, ξ = ρV (x). If α < m, then s = dim(X) = 2(m−α)+1 3, so that X contain both plus and minus points. Otherwise, MΩ , and U MΩ , we have x(Ω(X)U ) ⊆ xMΩ . For X has no minus points. As Ω(X) any v ∈ xΩ(X), w ∈ W, we will show that there exists u ∈ U such that vu = v + w, which implies that | x Ω(X)U | = |Eξ (X)|.|W |. Thus 1 1 1 |xΩ(X)U | = |W |.|Eξ (X)| = 3α . 3m−α (3m−α + ξ)) = 3m (3m−α + ξ). 2 2 2 | x MΩ | Therefore | x MΩ | Let B = [f ]β . Then    0 0 Iα    B =  0 λIs 0  .     Iα 0 0 We have [v]β = (0, a, 0) and [w]β = (b, 0, 0), where a, b are row vectors in Fs and Fα , respectively. Since v is non-singular, v is non-zero. Choose B ∈ Ms×α (F) such that 54 1 m m−α 3 (3 + ξ). 2 (3.6) aB = b. Let C = −λ−1 B t , A + At = −λ−1 B t B = −λCC t , and   Iα 0 0    u =  B Is 0  .     A C Iα Then uBut = B, u centralizes W, W ⊥ /W and V /W ⊥ . Thus u ∈ U and vu = v + w. Let z = ηe1 + f1 ∈ V, where η = Q(x). Then Q(z) = Q(x) so that x and z 1 belong to the same Ω-orbit of non-singular points in V. Let T = 2 T0 . By Lemma 3.11, SLα (3) T N ∩Ω MΩ , and so T U MΩ , where T0 , N are as in Lemma 3.12. Thus z(T U ) ⊆ zMΩ . We ﬁnd the stabilizer (T U )z in T U of vector z. For any g ∈ T U, there exist h ∈ T, u ∈ U such that g = hu. We have     D 0 0  Iα 0 0      [h]β =  0 Is 0  , and [u]β =  B Is 0  ,         ∗ 0 0 D A C Iα where D = (dij ) ∈ SLα (3), D∗ its inverse transpose, and C = −λ−1 B t , A + At = −λ−1 B t B = −λCC t . Thus [g]β = [hu]β = [h]β [u]β . Suppose g ∈ (T U )z . Write E1 = (η, 0, . . . , 0) and F1 = (1, 0, . . . , 0). α α 55 Then [z]β = (E1 , 0, F1 ). As zg = z, we have   0 0  D   (E1 , 0, F1 )  B Is 0  = (E1 , 0, F1 ),     D∗ A D∗ C D∗ or (E1 D + F1 D∗ A, F1 D∗ C, F1 D∗ ) = (E1 , 0, F1 ), hence    F1 D∗ = F1    F D∗ C = 0  1     E1 D + F1 D∗ A = E1 Since F1 D∗ = F1 ,     t t 1 b  1 −b D1  1 0  −1 D∗ =  ,  , and D =  ,D =  −1 ∗ 0 D1 0 D1 b D1   where D1 ∈ SLα−1 (3) and b is a column vector of size α−1. As F1 D∗ C = 0 and F1 D∗ = F1 , we have F1 C = 0. Hence     0 0  a11 a  C= ,A =  , c0 C 1 a0 A 1 where C1 ∈ Mα−1,s−1 (3), c0 is a column vector of size α − 1, A1 ∈ Mα−1 (3), a11 ∈ F and a, a0 are row, column vectors, respectively, of size α − 1. Now as E1 D + F1 D∗ A = E1 D + F1 A = E1 , 56 we have     t 1 −b D1  a11 a  (ξ, 0)   + (1, 0)   = (η, 0). 0 D1 a0 A 1 It follows that (η + a11 , a − ηbt D1 ) = (η, 0). Therefore a11 = 0 and a = ηbt D1 . Finally as A + At = −λCC t , we have  0   t a0 + ηD1 b at 0 t    0 0 + ηb D1   −1  ,  = −λ  t 0 c0 ct + C 1 C 1 A1 + At 0 1 t t hence a0 = −ξD1 b, A1 + At = −λ(c0 ct + C1 C1 ) and 1 0   A= 0 t  ηb D1  . t −ηD1 b A1 In summary, for any g ∈ (T U )z , we have      0 0  D 0 0  Iα 0 0   D      [g]β =  B Is 0  =  0 Is 0   B Is 0  ,           D∗ A D∗ C D∗ 0 0 D∗ A C Iα where         t ξbt D1   0 0 0  1 −b D1  ∗ 1 0  ,A =  D= ,D =  ,C =   ,   t ∗ c0 C 1 −ξD1 b A1 b D1 0 D1 B = −λC t ∈ Ms,α (3), t with C1 ∈ Mα−1,s−1 (3), A1 ∈ Mα−1 (3), b, c0 ∈ Mα−1,1 (3), A1 + At = −λ−1 (c0 ct + C1 C1 ), 1 0 A ∈ Mα (3), D, D∗ ∈ SLα (3), C ∈ Mα,s (3). We see that the subgroup of SLα (3) generated 57 by all matrices D is isomorphic to U0 : SLα−1 (3), where U0 is an elementary abelian subgroup of order 3α−1 . Given such D, there are 3α−1+(α−1)(s−1) choices for C and 3 2 (α−1)(α−2) choices for A. Therefore 1 1 |(T U )z | = 3α−1 .|SLα−1 (3)|.3α−1+(α−1)(s−1)+ 2 (α−1)(α−2) . 1 Since |U | = 3α.s+ 2 α(α−1) , we have |z(T U )| = |T U : (T U )z | = 3s+α−1 (3α − 1). Thus 1 s+α−1 α 3 (3 − 1). 2 | z MΩ | (3.7) It follows from (3.6) and (3.7) that 1 m m−α 1 1 3 (3 +ξ)+ 32m−α (3α −1) = 3m (3m +ξ) = |Eξ (V )|. 2 2 2 |Eξ (V )| | x MΩ |+| z MΩ | Therefore | x MΩ | = | x Ω(X)U |, | z MΩ | = | z MΩ U |, so that Eξ (V ) = x MΩ ∪ z MΩ . Hence MΩ has at most two orbits on Eξ (V ). Clearly as MΩ M, x MΩ ⊆ x M and similarly z MΩ ⊆ z M. Since Eξ (V ) = x MΩ ∪ z MΩ and x M ∪ z M ⊆ Eξ (V ), it implies that Eξ (V ) = x M ∪ z M. Thus M has at most two orbits on Eξ (V ). The imprimitive subgroups C2 A subspace decomposition D = {V1 , . . . , Vb } of V is a set of subspaces V1 , . . . , Vb of V with b 2 such that V = V1 ⊕ V2 ⊕ · · · Vb . Let G be a subgroup of GL(V ). The stabilizer in G of D is the group NG {V1 , . . . , Vb }, which is the subgroup of G, permuting the spaces Vi amongst themselves and denoted by GD . The centralizer in G of D, is the group G(D) = NG (V1 , . . . , Vb ), which is a subgroup of G ﬁxing each Vi . We also deﬁne GD = GD /G(D) . If the spaces Vi in the subspace decomposition D all have the same dimension α, then D is called an α-decomposition. If the Vi s are non-degenerate and pairwise orthogonal, then D is said to be non-degenerate. For any vector v ∈ V, v can be written in the 58 form v = v1 + v2 + · · · + vb , where vi ∈ Vi . We deﬁne the D − length of v to be the k number of non-zero vectors vi appearing in v, and denote by Db , the number of all points of D−length k, 1 k b. The members of C2 (Γ On (q)) are the stabilizers in Γ On (q) of α-decomposition D of V such that D is non-degenerate and if α = 1 then q = p, a prime. Lemma 3.14 ([29], Proposition 4.2.11, 4.2.15, 4.2.15). Let MΩ be the stabilizer in Ω(V ) of a non-degenerate α-decomposition D and n = bα. Then (i) If α > 1, then MΩ ∼ Ω(V )(D) .Sb . = (ii) If α = 1, and q ≡ ±3 (mod 8), then MΩ ∼ Ω(V )(D) .An . = Lemma 3.15 Assume that D is a 1-decomposition of V with q odd. For 1 n k |Dn | = (q − 1)k−1 . k k n, Proof. Without loss of generality, we can assume that V has an orthonormal basis β = {v1 , · · · , vn }. If v ∈ V has D-length k then v is a linear combination of a set of k basic n vectors taken from the basis β, with coeﬃcients in F∗ . Clearly, there are choices for q k k-sets, and for each k-set, there are (q − 1)k−1 points of length k. The result follows. Lemma 3.16 Assume G is nearly simple primitive rank 3 of type Ωε (3) with ε ∈ {◦, ±} n and M is of type O1 (3) Sn . Let β = {x1 , . . . , xn } be an orthogonal basis for V. Then (1) If z = x1 then d1 = dz = n − 1 and c1 = cz = 0; (2) If z = x1 + x2 then d2 = dz = n2 − 5n + 7 and c2 = cz = 4n − 8. Proof. By multiplying a suitable non-zero constant to the quadratic form Q, we can assume that V has an orthonormal basis β = {x1 , . . . , xn }. Setting ri = rxi , the reﬂection along vector xi . Then we have n ∼ 2n , and Ω(V )(D) = ri rj |1 = n ∼ 2n−1 . = I(V )(D) = ri |1 i i, j 59 For 1 i=j n, we see that rxi −xj permutes {xi , xj }, and ﬁxes xt for any t ∈ {i, j} and so rxi −xj acts as a transposition (i, j). Thus if we denote by J, the group generated by all reﬂections rxi −xj , where 1 i=j n, then J ∼ Sn and hence J1 , the subgroup = of J generated by rxi −xj rxr −xs , with i = j, r = s is isomorphic to An . By Lemma 3.14(ii), MΩ = Ω(V )(D) J1 . For any 1 i=j n, as (xi , xj ) = 0, rj ﬁxes xi . Hence Ω(V )(D) leaves invariant point x1 , and x1 + x2 Ω(V )(D) = { x1 + x2 , x1 − x2 }, as (x1 + x2 )r2 r3 = x1 − x2 . By (4.2.17) in [29], we have MI = MΩ r3 , rx3 −x4 . Thus x MΩ = x MI for any x ∈ {x1 , x1 + x2 }, so that it suﬃces to compute the parameters for MΩ in L. Now, since n 5, An acts transitively on the set {1, 2, · · · , n}. Thus x1 MΩ = { x1 , · · · , xn }. Hence 1+c1 +d1 = n. Moreover, as (xi , x1 ) = 0 for any i > 1, we have d1 = |x⊥ ∩ x1 MΩ | = 1 |{ x2 , · · · , xn }| = n − 1, and so c1 = 0. Similarly, as An acts doubly transitively on {1, · · · , n}, we have x1 + x2 MΩ = x1 + x2 Ω(V )(D) J1 = { x1 + x2 , x1 − x2 }J1 . 2 Thus 1 + c2 + d2 = | x1 + x2 MΩ | = Dn = n(n − 1), by Lemma 3.15. For any v ∈ x1 + x2 ⊥ ∩ x1 + x2 MΩ , v = xi ± xj for some i = j ∈ {1, · · · , n} and (v, x1 + x2 ) = 0. Clearly, v is generated by x1 − x2 or xi ± xj for some i < j ∈ {3, · · · , n}. By Lemma 3.15 again, d2 = | x1 + x2 ⊥ 2 ∩ x1 + x2 MΩ | = Dn−2 + 1 = n2 − 5n + 7, and so c2 = 4n − 8. Proposition 3.17 Assume M is of type O1 (3) Sn , with n = 2m + 1. There is an M orbit on Eξ (V ) such that equation (3.1) does not hold unless (n, ξ, r) = (5, +, t), (7, +, t) or (5, −, s), in which cases M has 2 orbits on Eξ (V ) so that 1G P and hence M is in Table 1.1. Proof. Retain the notations in previous lemma. By Proposition 2.6, sgn(x⊥ ) = (−1)m 1 and sgn(x1 + x2 )⊥ = (−1)m+1 , as the discriminant of the corresponding subspaces is square or non-square respectively. When m is even x1 is a plus vector and x1 + x2 is a minus vector and vice versa when m is odd. Let x ∈ {x1 , x1 + x2 }. We consider the following cases: 60 1G by Corollary 3.7, M (i) x is a plus point. If m is even then we choose x = x1 . By Lemma 3.16 d = d1 = n − 1, c = c1 = 0. Then k = −rd, so r must be t and so 3m − 1 = 4m. This equation holds only when m = 2 and hence n = 5. If m is odd then choose x = x1 + x2 , and hence d = d2 = (n − 2)(n − 3) + 1, c = c2 = 4n − 8 by Lemma 3.16 again. Then c − 2d = 14n − 2n2 − 22. As n 5, 2n2 + 22 > 14n so that c − 2d < 0. Therefore, equation (3.2) cannot hold. If equation (3.3) holds then (3m−1 + 1)(3m − 1 + c − 2d) = 2c. It follows that (3m−1 +1)(3m −1+14n−2n2 −22) = 8(2m−1). If m hence this equation cannot hold. For 2 m 5, then 3m−1 +1 > 8(2m−1), 4, the equation occurs only when m = 3. Thus equation (3.1) holds only when r = t and m = 3 or n = 7. (ii) x is a minus point. If m is odd then x = x1 and d = n − 1, c = 0. Then k = −rd. It follows that r = t, and hence 3m + 1 = 4m. Since m 2, 3m + 1 > 4m, so that this equation cannot hold. If m is even, x = x1 + x2 and d = (n − 2)(n − 3) + 1, c = 4n − 8. We have 2d − c = 2n2 + 22 − 14n. If equation (3.3) holds then (3m−1 − 1)(3m + 1 + 2d − c) = 2c, hence (3m−1 − 1)(3m + 1 + 2n2 + 22 − 14n) = 8(2m − 1). Since 2n2 + 22 − 14n > 0, (3m−1 − 1)(3m + 1 + 2n2 + 22 − 14n) > (3m−1 − 1)(3m + 1) > 8(2m − 1) for any m 3, and when m = 2, (3m−1 − 1)(3m + 1 + 2n2 + 22 − 14n) = 24 = 8(2m − 1). If equation (3.2) holds then 2d − c = 2n2 + 22 − 14n = 3m + 1. We can check that 3m + 1 > 2n2 + 22 − 14n for any m 2. Thus equation (3.1) holds only when m = 2 or n = 5 and r = s. To ﬁnish the proof, we need to verify that when these cases happen then equation (3.1) also holds for all x ∈ Eξ (V ). In view of Corollary 3.7, we will show that there are only two orbits of non-singular points of speciﬁed types. Firstly, suppose that n = 5. 1 Then m = 2, and |Eξ | = 2 3m (3m + ξ). In this case, x1 MΩ and x1 + x2 + x3 + x4 MΩ are two orbits of plus points with orbit sizes 5 and 23 E+ (V ) = 1 2 2 3 (3 2 5 4 = 40, respectively. However, + 1) = 45 = 5 + 40. Hence there are only two orbits of plus points. Similarly, x1 + x2 MΩ and x1 + x2 + x3 + x4 + x5 MΩ are two orbits of minus points with orbit sizes 2 5 2 = 20 and 24 5 5 = 16, respectively. Since |E− (V )| = 1 32 (32 − 1) = 2 61 36 = 20 + 16, there are exactly two orbits of minus points. Finally, suppose that n = 7, 1 and ξ = +. Then m = 3 and E+ (V ) = 2 33 (33 + 1) = 378. In this case, x1 + x2 MΩ and x1 + x2 + x3 + x4 + x5 MΩ are two orbits of plus points with orbit sizes 2 24 7 5 7 2 = 42 and = 336. Clearly, as 336 + 42 = 378, these are only two orbits of plus points. This completes the proof. We next consider the case when α > 1. Since dimV is odd, it follows that α and b are both odd. Write α = 2a + 1, b = 2b1 + 1. Proposition 3.18 Assume M is of type Oα (3) Sb , with α > 1 odd. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. We can assume that V has an orthonormal basis which is the union of orthonormal bases of all Vi . Let N = Ω1 × Ω2 × · · · × Ωb MI = I1 Sb . Thus Ω1 Sb MΩ MΩ . By Lemma 4.2.8 in [29], we have 3 and so V1 contains both MI . Since α > 1 is odd, α plus and minus points. Let xξ ∈ V1 be a non-singular vector of type ξ ∈ {±}. Clearly xξ Ω1 Sb = xξ I1 Sb , we conclude that xξ MΩ = xξ MI = xξ N.Sb . Thus we only need to compute the parameter for MΩ in L. Since Sb centralizes V1 , for all i > 1, xξ MΩ = xξ N.Sb = xξ Ω1 .Sb = Eξ (V1 )Sb = ∪b Eξ (Vi ). i=1 1 Thus A = | xξ MΩ | = b. 2 3a (3a + ξ) by Lemma 3.8(i). Hence A 1 .b.3a (3a 2 MΩ permutes the Vi s, and Ωi + 1). In view of inequality (3.4), it suﬃces to show that 1 3m 2 inequality is equivalent to 3ba+b1 1 b.3a (3a 2 + 1). Since m = ba + b1 , this 1, 3ba 33a = 3a .32a b.3a (3a +1), b.3a .(3a + 1). As b b−1 2 3 and a 3.32a > 32a +3a . Now, if we can prove that 3b1 = 3 and we are done. To show that 3b1 then 3 b−1 2 b, then clearly 3ba 3b1 b, we will argue by induction on b b−1 2 = 3 3 = b. Suppose that 3 b. Then 3 (b+1)−1 2 3. When b = 3 √ √ b−1 = 3 2 3 b. 3, by 62 induction assumption. We have 3b2 = b2 + 2b2 b2 + 6b > b2 + 2b + 1 = (b + 1)2 , as b √ b+1−1 Thus b. 3 b + 1. Hence 3 2 b + 1. The result follows. The ﬁeld extension subgroups C3 3. Let F be a ﬁeld extension of F = F3 of degree α, where α is a prime divisor of n = dimV. Then V acquires the structure of an F -vector space in a natural way. Write V for V regarded as a vector space over F . Denote by T the trace map from F to F. If Q is a quadratic form on (V , F ) then Q = T Q is a quadratic form on (V, F). Write f for the associated bilinear form of Q . Denote by N = NF /F the norm map of F over F. Let µ, ν be the generators for F∗ and Gal(F /F), respectively. Also the trace map from F to F deﬁnes a non-degenerate bilinear form on F . Let QT : F → F be a map deﬁned by QT (x) = −T (x2 ) for x ∈ F . Then QT is a quadratic form on F and fT (x, y) = T (xy) is the bilinear form associated to QT . Then (F , QT , F) is an orthogonal geometry. Lemma 3.19 Let βT = {ζ1 , ζ2 , · · · , ζα } be an F-basis of F = Fpα , where p = 3, and v ∈ V be such that f (v , v ) = λ ∈ F∗ . Then (i) D(F ) ≡ det(fβT ) ≡ (−1)α−1 (mod (F∗ )2 ); (ii) spanF (v ) is a non-degenerate α-subspace in V with discriminant D(F )N (λ). Proof. From deﬁnition, we have  T (ζ1 ζ2 ) · · · T (ζ1 ζα )     T (ζ2 ζ1 ) T (ζ 2 ) · · · T (ζ2 ζα ) 2   = . . . . ...  . . .  . .   .   2 T (ζα ζ1 ) T (ζα ζ2 ) · · · T (ζα ) 2 T (ζ1 )  f βT 63 Let ζ2 ··· ζα   ζ1   p p p  ζ ζ2 ··· ζα    1 X= . . .  ..  . . .  . . .   .   α−1 α−1 p p pα−1 · · · ζα ζ2 ζ1 and E = diag(λ, λp , · · · , λp α−1   ). As T (a) = α−1 i=0 ap for any a ∈ F , X t X = fβT , hence i det(fβT ) = det(X t X) = det(X)2 . Since det(fβT ) ∈ F∗ and detX ∈ F , if α is odd then clearly detX ∈ F∗ , as F does not have any subﬁeld of degree 2 over F. Thus (detX)2 ∈ (F∗ )2 = {1}, so det(fβT ) = (detX)2 = 1. Now suppose that α = 2. Let ζ be a root of x2 − x − 1 in F, and let η = ζ + 1. Then η 2 = −1, F = F(η) and T (η) = 0. Choose βT = {1, η}. Then     T (1) T (η)  2 0  f βT =  . = 0 −2 T (η) T (η 2 ) Hence det(fβT ) = −1. This proves (i). Let W = v F and β = {ζ1 v , ζ2 v , · · · , ζα v }. As (ζi v , ζj v ) = T (f (ζi v , ζj v )) = T (ζi ζj f (v , v )) = T (λζi ζj ), we have  T (λζ1 ζ2 ) · · · T (λζ1 ζα )     T (λζ2 ζ1 ) T (λζ 2 ) · · · T (λζ2 ζα ) 2   fβ =  . . . . ...   . . . . . .     2 T (λζα ζ1 ) T (λζα ζ2 ) · · · T (λζα )  2 T (λζ1 ) Obviously X t EX = fβ . Therefore, det(fβ ) = det(X)2 N (λ) = D(F )N (λ). n Proposition 3.20 Assume M is of type O α (3α ) with n = 2m + 1 and α | n. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold unless α = 3. In this case M has 3 orbits on Eξ (V ) and equation (3.1) holds for all M -orbits on Eξ (V ) with r = s, and 64 hence M is in Table 1.1. Proof. Let q = 3α , and µ, ν be the generators for F∗ and Gal(F /F), respectively. As n is n odd, it follows that α is also odd. Write α = 2α1 + 1 and = 2b + 1. Then m = bα + α1 , α where n = 2m + 1. Multiplying by a suitable constant to the quadratic form Q , we can assume that D(Q ) = . By Proposition 2.8, there exists a basis β = {w1 , w2 , · · · , w2b+1 } of (V , Q ) such that fβ = I2b+1 . Deﬁne φ = φQ ,β = φβ (ν). Then o(φ ) = α. We will show that φ ∈ Ω. Let βn = {ζ, ζ 3 , . . . , ζ 3 βi = βn ⊗ wi , Since (ζ 3 wi )φ = ζ 3 j j+1 α−1 } be a normal basis of F over F, and wi , we obtain 1 0 ··· 0 . . . 0 0  0   1 · · · 0  . .. . . . . . . .   0 · · · 1  0 ··· 0  0   0  . (φ )βi =  . .   0  1 Clearly det(φ )βi = (−1)α−1 = 1 hence det(φ ) = 1. Therefore φ ∈ S. As [S : Ω] = 2 and o(φ ) = α is odd, φ ∈ Ω. Let I = I(V, F, Q), I = I(V , F , Q ), and Ω = Ω(V , F , Q ). Then by (4.3.11) in [29] MI = I φ Ω MΩ NL (F ) ∼ I .Zα . Since Ω is perfect, Ω = L, hence MI = I φ . By Proposition 4.3.17 in [29], [MΩ : Ω ] = α. As φ ∈ Ω ∩ I φ , φ ∈ MΩ , and hence MΩ = Ω φ . It follows from Lemma 2.3 that z Ω = z I for any non-singular point z in (V , F , Q ), so that z MΩ = z MI , hence we only need to compute parameters for MΩ in L. We ﬁrst claim the following: (1) Q (wφ ) = Q (w)ν for any w ∈ V . (2) z MΩ = { w ∈ V | Q (w) ∈ {γ, γ ν , . . . , γ ν For (1), assume that w = hence Q (wφ ) = ( 2b+1 i=1 2b+1 i=1 α−1 }}, where z ∈ V with γ = Q (z). 2b+1 i=1 (λi wi )φ λi wi ∈ V . Then wφ = = 2b+1 i=1 λν w i , i −λ2 )ν = Q (w)ν . For (2), from (1) we have Q (zφ ) = γ ν , hence i α−1 {Q (zφj )}α = {γ, γ ν , · · · , γ ν j=1 }. Thus z MΩ = { w ∈ V | Q (w) ∈ {γ, γ ν , . . . , γ ν 65 α−1 }}. For a non-zero vector w ∈ V , consider spanF (w) as an α-subspace in V. We consider the case when α = 3 and α > 3 separately. (a) Case α > 3. Let z ∈ {w1 , w1 + w2 }. Then Q (z) = α = 0 so z is non-singular in V. Also as Q (z) = 1 and Q(z) = T Q (z) = 1 is ﬁxed under ν, by (2) we have 1 ⊥ z MΩ = z Ω , and by Lemma 2.11, | z MΩ | = | z Ω | = 2 (q 2b + ε.q b ), with ε = sgn(zV ). We have z MΩ ∩ z ⊥ = z Ω ∩ z ⊥ = {v ∈ V | Q (v) = Q (z), T f (v, z) = 0}. For w ∈ ⊥ ⊥ z MΩ ∩zV , write w = ϕf (z, z)−1 z +w0 , where w0 ∈ zV , and T (ϕ) = 0. Then f (w, z) = ϕ and Q (w0 ) = Q (z)−1 (Q (z)2 − ϕ2 ). As T (±Q (z)) = T (±1) = 0, Q (w0 ) = 0 for any ⊥ ⊥ ϕ ∈ F with T (ϕ) = 0. When ϕ ∈ KerT is ﬁxed, as dimF (zV ) = 2b and sgnzV = ε, by Lemma 2.11, there are q 2b−1 − εq b−1 vectors w0 with Q(w0 ) = Q (z)−1 (1 − ϕ2 ) = 0. 1 ⊥ Also dimF (KerT ) = α − 1, we conclude that dz = | z MΩ ∩ zV | = 2 3α−1 (q 2b−1 − εq b−1 ) = 1 2b (q − εq b ). 6 1 Thus cz = 3 (q 2b + 2εq b ) − 1 = (ε3bα − 1)(ε3bα−1 + 1), and cz − 2dz = εq b − 1 = ε.3bα − 1. Assume that equation (3.2) holds. Then cz − 2dz = ε.3bα − 1 = ξ3m−1 − 1, ⊥ where ξ = sgn(zV ). The last equation yields m = bα + 1. Recall that m = bα + α1 , hence 1 α1 = 2 (α − 1) = 1, this forces α = 3, a contradiction. Now suppose that equation (3.3) holds. Then (3m−1 + ξ)(3m − ξ + ξ(c − 2d)) = 2c, hence (3m−1 + ξ)(3m − εξ3bα − 2ξ) = 2(3bα −ε)(3bα−1 +ε). We will show that 3m−1 +ξ > 2(3bα −ε) and 3m −εξ3bα −2ξ > 3bα−1 +ε, so that after multiplying these two inequalities side by side, we get a contradiction. For the ﬁrst inequality, we have 3m−1 + ξ α1 2. Now 2(3bα − ε) 3bα+α1 −1 − 1 3α1 −1 3bα − 1 3.3bα − 1, as 2(3bα + 1). Thus, it suﬃces to show that 3.3bα − 1 > 2(3bα + 1). This inequality is equivalent to 3bα > 3. This is true because bα > 1. For the second inequality, we have 3m − εξ3bα − 2ξ (3bα + 1) + (3bα − 3) > 3bα−1 + 1 3bα+α1 − 3bα − 2 = (3α1 − 1)3bα − 2 > 2.3bα − 2 = 3bα−1 + ε, as 3bα − 3 > 0. (b) Case α = 3. Let ω be a root of x3 − x + 1 in F. Then ω = F∗ , and KerT has a basis {1, ω} with T (ω 2 ) = −1. We have m = 3b+1, q = 33 . Let x1 = ωw1 , x2 = ω 2 w1 , x3 = ω 4 (w1 + w2 ) and y1 = ω(w1 + w2 ), y2 = ω 2 (w1 + w2 ), y3 = ω 4 w1 . For i = 1, · · · , 3, we have 66 Q (xi ) = 0, Q (yi ) = 0, and Q(xi ) = 1, Q(yi ) = −1 and so xi , yi are non-singular in both V and V. Also all xi s (yi s) belong to diﬀerent Ω -orbits but they are in the same Ω-orbits. For each i = 1, 2, we have xi ⊥ = w2 , · · · , w2b+1 and yi ⊥ = w1 − w2 , w3 , · · · , w2b+1 , V V so that D(xi ⊥ ) = V , D(yi ⊥ ) = V , and hence by Proposition 2.6, sgn(xi ⊥ ) = (−)b V and sgn(yi ⊥ ) = (−)b−1 , where dimxi ⊥ = dimyi ⊥ = 2b. For i = 3, as computation V V V above, we have sgn(x3 ⊥ ) = (−)b−1 and sgn(y3 ⊥ ) = (−)b . We now determine the type V V of xi and yi in (V, F, Q). Let U = spanF (w3 ) ⊥ · · · ⊥ spanF (w2b+1 ) V be an F- subspace. We have x1 ⊥ = w1 , ω 2 w1 ⊥ spanF (w2 ) ⊥ U, x2 ⊥ = ωw1 , (ω 2 − ω)w1 ⊥ V V spanF (w2 ) ⊥ U, x3 ⊥ = (ω + 1)(w1 + w2 ), (ω 2 − 1)(w1 + w2 ) ⊥ spanF (w1 − w2 ) ⊥ U, V and similarly y1 ⊥ = (w1 + w2 ), ω 2 (w1 + w2 ) ⊥ spanF (w1 − w2 ) ⊥ U, y2 ⊥ = ω(w1 + V V w2 ), (ω 2 − ω)(w1 + w2 ) ⊥ spanF (w1 − w2 ) ⊥ U, y3 ⊥ = (ω + 1)w1 , (ω 2 − 1)w1 ⊥ V spanF (w2 ) ⊥ U. By Lemma 3.19, D(spanF (wi )) = N (f (wi , wi )) = N (1) = 1 = D(spanF (w1 −w2 )) = N (f (w1 −w2 , w1 −w2 )) = N (−1) = −1 = D(yi ⊥ ) = V . Thus D(xi ⊥ ) = V and and for all i = 1, · · · , 3, and so by Proposition 2.7, sgn(xi ⊥ ) = (−)m−1 = (−)b V and sgn(yi ⊥ ) = (−)m = (−)3b+1 = (−)b−1 . Let z ∈ {xi , yi } and γ = Q (z). We have V z MΩ = { w ∈ V | Q (w) ∈ {γ, γ 3 , γ 9 }} = 3 j=1 zφj Ω . Observe that in (V , F , Q ) all ⊥ vectors zφj , j = 1, · · · , 3 have the same type, say ε = sgn(zV ). It follows from Lemma 1 1 ⊥ 2.11 that 1 + cz + dz = 3. 2 (q 2b + ε.q b ) = 2 (36b+1 + ε33b+1 ). For any w ∈ z MΩ ∩ zV , Q (w) ∈ {γ, γ 3 , γ 9 } and T (ϕ) = 0, with ϕ = f (w, z). Write w = ϕf (z, z)−1 z + w0 , where ⊥ w0 ∈ zV . We have f (w, z) = ϕ and Q (w0 ) = γ −1 (γQ (w) − ϕ2 ). ⊥ ⊥ Assume that i = 1, 2. Let z ∈ {xi , yi } and γ = Q (z). Then sgn(zV ) = sgn(zV ) = ε. We will show that Q (w0 ) = 0 for any ϕ ∈ kerT. By way of contradiction, suppose that Q (w0 ) = 0. Then ϕ2 ∈ {γ 2 , γ 4 , γ 10 }, hence ϕ ∈ {±γ, ±γ 2 , ±γ 5 } or ϕ ∈ {±ω 2 , ±ω 4 , ±ω 8 , ±ω 20 , ±ω 10 }. As the trace map is non-zero on these values, we get a contradiction. Thus Q (w0 ) = 0. By Lemma 2.11, dz = 3 1 (q 2b−1 − ε.q b−1 ).3α−1 = 1 (q 2b − ε.q b ) 2 2 ⊥ ⊥ as |kerT | = 3α−1 = 32 , dimF (zV ) = 2b, and ε = sgn(zV ). Then cz = q 2b + 2ε.q b − 1. Hence 67 cz − 2dz = ε.3q b − 1 = ε.3m − 1. Therefore equation (3.2) holds. ⊥ ⊥ Assume that z ∈ {x3 , y3 }. Then sgn(zV ) = −sgn(zV ) = −ε and γ = ±ω 8 . Also Q (w0 ) = γ −1 (γQ (w) − ϕ2 ). For any w ∈ z MΩ , Q (w) ∈ {γ, γ 3 , γ 9 }. If Q (w) = γ then Q (w0 ) = 0 as T (ϕ) = T (±γ) = 0. If Q (w) = γ 3 then Q (w0 ) = 0 if and only if ϕ ∈ {±γ 2 }, and similarly, if Q (w) = γ 9 then Q (w0 ) = 0 if and only if ϕ ∈ {±γ 5 }. By Lemma 2.11, we have 2dz = 32 .(q 2b−1 − εq b−1 ) + 2[2(q 2b−1 + ε(q b − q b−1 )) + 7(q 2b−1 − εq b−1 )] or 2dz = 36b + ε33b+1 , and hence cz = 36b − 1, so that cz − 2dz = −ε33b+1 − 1 = (−ε)3m − 1. Thus equation (3.2) holds. Let ξ = (−)b = sgn(xi ⊥ ) and η = sgn(yi ⊥ ), i = 1, · · · , 3. As | x1 MΩ | + | x2 MΩ | + V V 1 1 1 1 | x3 MΩ | = 2 (36b+1 + ξ33b+1 ) + 2 (36b+1 + ξ33b+1 ) + 2 (36b+1 − ξ33b+1 ) = 2 33b+1 (33b+1 + ξ) = 1 m m 3 (3 2 + ξ) = |Eξ (V )|, MΩ has exactly three orbits on Eξ (V ). Thus equation (3.2) holds for all points in Eξ (V ). Similarly MΩ has three orbits on Eη (V ), and so equation (3.2) holds for all points in Eη (V ). The tensor product subgroups C4 Let Vi be vector spaces over Fq of dimension ni , i = 1, · · · , t. Let V = V1 ⊗ · · · ⊗ Vt . For gi ∈ GL(Vi ), the element g1 ⊗ · · · ⊗ gt ∈ GL(V ) acts on V as follows: (v1 ⊗ · · · ⊗ vt )(g1 ⊗ · · · ⊗ gt ) = v1 g1 ⊗ · · · ⊗ vt gt (vi ∈ Vi ), and extend linearly. Now suppose that q is odd, and that fi is a non-degenerate bilinear form on Vi , so that (Vi , F, fi ) is either symplectic or orthogonal geometry. We next deﬁne the bilinear form f = f1 ⊗ · · · ⊗ ft on V1 ⊗ · · · ⊗ Vt by f (v1 ⊗ · · · ⊗ vt , w1 ⊗ · · · ⊗ wt ) = t i=1 fi (vi , wi ) and extend linearly. We write (V, f ) = (V1 ⊗ · · · ⊗ Vt , f1 ⊗ · · · ⊗ ft ) 68 for such a structure and call tensor decomposition and denote by D. The members of C4 (Γ ) is the stabilizer of tensor decomposition D such that (a) (V, f ) = (V1 ⊗ V2 , f1 ⊗ f2 ), (b) (V1 , f1 ) is not similar to (V2 , f2 ), (c) fi are symmetric, (n1 , ε1 ) = (n2 , ε2 ), where ni = dimVi symplectic, dimV1 < dimV2 . Proposition 3.21 Assume M is of type On1 (3) ⊗ On2 (3), with n1 < n2 . There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. Let v = v1 ⊗ v2 ∈ V, where vi ∈ Vi are non-singular vectors. Then v is a nonsingular point. By (4.4.14)[29], MI = I1 ⊗I2 . We have Ω1 ×Ω2 MΩ I1 ⊗I2 = MI . As Ωi 3, εi = sgnVi , or fi are act transitively on E(Vi ), it follows that v MI = v MΩ = v1 ⊗ v2 (Ω1 × Ω2 ) = v1 Ω1 ⊗ v2 Ω2 . Therefore v MΩ = v M since MΩ Then | v MΩ | and b > a | v1 Ω1 |.| v2 Ω2 | M MI . Write n1 = 2a + 1, n2 = 2b + 1. < 32(a+b) (as 3a + 3b + 1 < 3a+b 1 a a .3 (3 + 1).3b (3b + 1) 2 1). Since dimV = 2m + 1 = (2a + 1)(2b + 1), m = 2ab + a + b. Then 0 so that 3m−1 32(a+b) > | v MΩ |. This m − 1 − 2(a + b) = a(b − 2) + b(a − 1) + a − 1 violates (3.4) so that equation (3.1) cannot hold. The tensor product subgroups C7 Let V1 be an α-dimensional vector space over F = Fq , and assume that f1 is either 0, a non-degenerate bilinear form, or a non-degenerate unitary form. For i = 1, · · · , b, let (Vi , fi ) be a classical geometry which is similar to (V1 , f1 ). For each i, denote by ηi the similarity from (V1 , f1 ) to (Vi , fi ) satisfying fi (vηi , wηi ) = λi f1 (v, w) for all v, w ∈ V1 , where λi ∈ F∗ is independent of v and w. Thus we obtain a tensor decomposition D given by (V, κ) = (V1 , f1 ) ⊗ · · · ⊗ (Vb , fb ), where V = V1 ⊗ · · · ⊗ Vb and κ = Q(f1 , · · · , fb ) if q is even and each fi is symplectic and κ = f1 ⊗ · · · ⊗ fb otherwise. Let Xi = X(Vi , fi ) for 69 X ∈ {Ω, S, I, Λ, Ξ, A}. Deﬁne ΞD = Ξ(D) Sb . The members of C7 (Ξ) are the groups ΞD with b 2 described as above. 5. There is an M -orbit on Proposition 3.22 Assume M is of type Oα (3) Sb with α Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. We have Ωα (3) Sb MΩ Oα (3) Sb . Let v1 = v ⊗ v ⊗ · · · ⊗ v and v2 = v1 + w ⊗ w ⊗ · · · ⊗ w ∈ V, where v = w belong to some orthogonal basis of V1 . As Sb ﬁxes vi , vi MΩ = vi (Πb Ωα (3)). Also vi MΩ = vi MI = vi M. For i = 1, 2, the i=1 stabilizers of vi in Πb Ωα (3) contain a subgroup which is isomorphic to Πb Ωα−2 (3). i=1 i=1 Thus | vi MΩ | = | vi (Πb Ωα (3))| i=1 [Ωα (3) : Ωα−2 (3)]b < 3(4a−3)b , where a = α−1 2 2. 1 Now m − 1 = 2 ((2a + 1)b − 3). Consider the following function in variable x ∈ [2, +∞), where b 2, 1 f (x) := ((2x + 1)b − 3) − (4x − 3)b. 2 We have f (x) = b(2x + 1)b−1 − 4b where g(b) = 5b − 10b − 3 and b b(2x + 1) − 4b b > 0. Hence f (x) f (2) = 1 g(b), 2 2. By induction on b 2, g(b) > 0. Thus 3m−1 > 3b(4a−3) > | vi MΩ |. This contradicts (3.4). Thus equation (3.1) cannot hold. 3.4.3 Permutation characters of maximal subgroups in S In this section, we consider the maximal subgroup M ∈ S(G). By Deﬁnition 2.12, M is an almost simple group and the socle S of M is a non-abelian simple group. Then the full covering group S of S acts absolutely irreducible on V, the natural module for G and preserves a non-degenerate quadratic form on V. Recall the construction of the fully deleted permutation module for alternating groups as in the discussion before Proposition 2.23. Let {ε1 , . . . , εn } be a standard basis for Fn , p ⊥ and let w0 = ε1 + · · · + εn ∈ Fn . Put U = w0 , W = Fp w0 , and V = U/(U ∩ W ). Deﬁne p ei = εi − εi+1 , i = 1, . . . , n − 1. Then {ei }n−1 is a basis for V if p does not divide n, and i=1 70 {ei + U ∩ W }n−2 is a basis for V if p|n. Let Q be the quadratic form on V induced from i=1 the quadratic form associated to the natural bilinear form on Fn . Then (V, Fp , Q) is a p classical orthogonal geometry and An ei instead of ei + U ∩ W. Lemma 3.23 Assume M is almost simple of type An with n 10 and V is the fully Ω(V ). To simplify the notation, we always write deleted permutation module for An in characteristic 3. Let v = ε1 −ε2 , w = ε1 +ε2 −ε3 −ε4 ∈ V. Then 1 1 (1) | v M | = 2 n(n − 1), dv = 2 (n − 2)(n − 3), and cv = 2n − 4; (2) | w M | = 1 n(n − 1)(n − 2)(n − 3), cw = 2n3 − 25n2 + 111n − 172 and 8 1 dw = 2 + 4(n − 4)2 + 8 (n − 4)(n − 5)(n − 6)(n − 7). Proof. As An acts transitively on V, xM = xAn = xSn for any x ∈ V. Since v = ε1 − ε2 , it is clear that if g ∈ Sn and (ε1 − ε2 )g = ε1 − ε2 , then g must ﬁx indices 1 and 2. Thus (Sn )v Sn−2 . Similarly, (ε1 + ε2 − ε3 − ε4 )g = ε1 + ε2 − ε3 − ε4 implies that g must ﬁx the 1 partitions {1, 2}, {3, 4}. Thus g ∈ S2 × S2 × Sn−4 . Therefore |M : M v | = 2 [Sn : Sn−2 ] = 1 n(n 2 − 1), and |M : M w | = 1 [Sn : S2 × S2 × Sn−4 ] = 1 n(n − 1)(n − 2)(n − 3). 2 8 (i) Parameters for v. We have u ∈ v M ∩ v ⊥ if and only if u = εi − εj ∈ Fp v, i = j and (εi − εj , ε1 − ε2 ) = 0. This happens only if {i, j} ∩ {1, 2} = ∅, or i, j ∈ {3, 4, . . . , n}. There are n−2 2 1 such points u . Thus dv = | v M ∩v ⊥ | = 2 (n−2)(n−3), and cv = 2n−4. (ii) Parameters for w. 1 We will show that d = | w M ∩ w⊥ | = 2 + 4(n − 4)2 + 8 (n − 4)(n − 5)(n − 6)(n − 7) and c = 2n3 − 25n2 + 111n − 172. For any u ∈ w M ∩ w⊥ , u = εi + εj − εr − εs , where |{i, j, r, s}| = 4, and (u, w) = 0. Denote by supp(u) the set of non-zero indices of εi appearing in u. We consider the cases: (1) supp(u) ∩ supp(w) = ∅. It follows that supp(u) ∈ {5, 6, . . . , n}. Hence there are (n − 4)(n − 5)(n − 6)(n − 7)/8 points. (2) |supp(u) ∩ supp(w)| = 1. There are no such u, since (u, w) = 0. 71 (3) |supp(u)∩supp(w)| = 2. Suppose that i, j ∈ {1, 2, 3, 4}. Then either u = εi −εj +εr −εs , where εi − εj ∈ {ε1 − ε2 , ε3 − ε4 }, or u = εi + εj − εr − εs , where εi + εj ∈ {ε1 + ε3 , ε1 + ε4 , ε2 + ε3 , ε2 + ε4 }; and r, s ∈ {5, . . . , n}. There are 2(n − 4)(n − 5) and 4 respectively. Thus there are 4(n − 4)(n − 5) points in this case. (4) |supp(u) ∩ supp(w)| = 3. Suppose that i, j, r ∈ {1, 2, 3, 4}. Then u = ±εi ± εj ± εr ± εs , where s ∈ {5, . . . , n}, εi , εj , εr with their signs appearing exactly as in w, and sign of εs is chosen so that there are 2 minuses and 2 pluses. There are 4 3 n−4 2 points (n − 4) = 4(n − 4). (5) |supp(u)∩supp(w)| = 4. There are just 2 points in this case: {ε1 −ε2 +ε3 −ε4 , ε1 −ε2 − ε3 +ε4 }. Therefore dw = 1 (n−4)(n−5)(n−6)(n−7)+2(n−4)(n−5)+4(n−4)(n−5)+2 = 8 1 (n − 4)(n − 5)(n − 6)(n − 7) + 2(n − 4)2 8 1 + 2, and cw = 8 n(n − 1)(n − 2)(n − 3) − d − 1. Proposition 3.24 Assume M is almost simple of type An , with n 10, and V is the fully deleted permutation module for An in characteristic p = 3. Further assume that n − 1 − ε3 (n) = 2m + 1. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. Let v = ε1 − ε2 , w = ε1 + ε2 − ε3 − ε4 ∈ V. Then Q(v) = 1, and Q(w) = −1. Hence v, w are non-singular vectors in V. We see that n − 1 − ε3 (n) is odd if and only if 1 n = 6k+2, n = 6k+3 or n = 6k+4. By Lemma 3.23(1) | v M | = 2 n(n−1). If n 1 (n − 5) 2 1 > log3 ( 1 n(n − 1)). Since m − 1 = 2 (n − 1 − ε3 (n)) − 1 2 1 (n − 5), 2 13, then 2. It as ε3 (n) 1 follows that 1 + c + d = 2 n(n − 1) < 3m−1 . This violates (3.4) and so equation (3.1) cannot hold. Hence, we only need to consider 10 n 12. Since n = 6k + 2, 6k + 3 or 6k + 4, it follows that n = 10. Then n−1−ε3 (n) = 9, dv = 28, cv = 16, m = 4. If equation (3.1) holds, then either cv −2dv = ξ34 −1 = 81ξ−1 = −40, or (27ξ+1)(81ξ−41) = 32. These equations clearly cannot hold with ξ = ±1. By Lemma 3.23(2), | w M | = n(n − 1)(n − 2)(n − 3)/8. If n 23, then (3m + 1)/2 > n(n − 1)(n − 2)(n − 3)/8 so that equation (3.1) cannot hold n 22. Then n ∈ {10, 14, 15, 16, 20, 21, 22}. by (3.4). Thus we can assume that 10 (a) n = 10. Then n − ε3 (n) = 9, m = 4, d = 191, c = 438 and c − 2d = 56. 72 Table 3.2: Low degree representations of small Alternating groups. An A6 A7 A7 A8 A8 A8 A9 A9 λ dimD 9 13 15 7 13 21 21 7 2 2 2 λ (4, 2) (5, 2) (5, 1 ) (7, 1) (6, 2) (6, 1 ) (7, 1 ) (8, 1) m(λ) (22 , 12 ) (3, 2, 12 ) (3, 22 ) (4, 3, 1) (32 , 12 ) (32 , 2) (4, 3, 2) (42 , 1) (b) n = 14. Then n − ε3 (n) = 13, m = 6, d = 1032, c = 1970 and c − 2d = −94. (c) n = 15. Then n − ε3 (n) = 13, m = 6, d = 1476, c = 2618 and c − 2d = −334. (d) n = 16. Then n − ε3 (n) = 15, m = 7, d = 2063, c = 3396 and c − 2d = −730. (e) n = 20. Then n − ε3 (n) = 19, m = 9, d = 6486, c = 8048 and c − 2d = −4924. (f ) n = 21. Then n − ε3 (n) = 19, m = 9, d = 8298, c = 9656 and c − 2d = −6940. (g) n = 22. Then n − ε3 (n) = 21, m = 10, d = 10478, c = 11466 and c − 2d = −9490. We can check that equation (3.1) cannot hold in any of these cases. Proposition 3.25 Assume M is almost simple of type An with n 12, and V is not the fully deleted permutation module for An in characteristic p = 3. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. As n m 12, by Lemma 2.32, we have dim(V ) = 2m + 1 However when n 12, 3m−1 > 3 n2 −5n−4 4 1 (n2 2 − 5n + 2) so that 1 (n2 −5n). 4 > n! = |Aut(An )|. Thus equation (3.1) cannot hold which in view of (3.4). Proposition 3.26 Assume M is almost simple of type S = An with 5 n 11. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold unless n = 9 and V ∼ D(8,1) , = in which case M has at most 2 orbits on Eξ (V ) so that 1G P hence (L, S) = (Ω7 (3), A9 ) is in Table 1.2. Proof. Using the information on the p-modular representations of alternating groups and their covering groups in [26], we only need to consider the cases given in Table 3.2. (i) Let λ = (8, 1). Then m(λ) = (42 , 1) = λ. By Theorem 2.31, Dλ ↓A9 is irreducible. Thus 73 1G by Corollary 3.7 and M A9 S9 Ω7 (3), and there is two classes of S9 in Ω7 (3). As 8 − 1 + 1 + 1 = 9 ≡ 0(mod 3), λ is a JS-partition (see Deﬁnition 2.22), and hence by Theorem 2.29, Dλ ↓S8 = Dλ(1) = D(7,1) . Then since (7, 1) = m(7, 1) = (4, 3, 1), we have: A8 S8 S9 . In this case, Dλ is the fully deleted permutation module for S9 over F3 . Then n − 1 − ε3 (n) = 7, and m = 3. Let v = ε1 − ε2 , w = ε1 + ε2 − ε3 − ε4 ∈ V. There are only two orbits of type ρV (v), with representatives v = ε1 − ε2 and ε1 + ε2 + ε3 + ε4 − ε5 = ε1 + ε2 + ε3 + ε4 − ε5 − ε6 − ε7 − ε8 , and one orbit of type ρV (w). Thus equation (3.1) holds for both types of points. (ii) If λ = (7, 1) or λ = (4, 3, 1), then A8 < S8 < S9 < Ω7 (3), since D(8,1) ↓S8 = D(7,1) and D(7,1) ↓A8 is irreducible. (iii) If λ = (6, 12 ) or (32 , 2), then A8 < S8 < S9 < Ω21 (3), since D(7,1 ) ↓S8 = D(6,1 D(6,1 ) ↓A8 is irreducible. (iv) If λ = (5, 2) or (3, 2, 12 ), then A7 < S7 < S8 < Ω13 (3). (v) Now, if λ = (n − 2, 12 ), where n = 7 or 9, then Dλ = ∧2 (D(n−1,1) ). As D(n−1,1) is the fully deleted permutation module for Sn , we can apply the construction above for fully deleted module. Let v = e1 ∧ e3 = (ε1 − ε2 ) ∧ (ε3 − ε4 ), and w = (e1 − e2 ) ∧ (e3 − e4 ) = (ε1 +ε2 +ε3 )∧(ε3 +ε4 +ε5 ). Then v, w are non-singular points of diﬀerent types in Dλ . We have |vSn | = n! 1 2 2.(n−4)! 2 2 2) and and |wSn | = 1 n! . 2 2.(n−5)! We then get contradictions by using (3.4). S6 Ω9 (3). (vi) If λ = (4, 2) then dimDλ = 9, m = 4, and we have an embedding A6 However, as A6 ∼ L2 (9) < A10 < Ω9 (3), hence M = NG (A6 ) is not maximal in G. = (vii) If λ = (6, 2), then dimV = 13, m = 6. We have A8 S8 Ω13 (3). Using GAP, S8 has two points w1 , w2 of diﬀerent types with the same orbit sizes 315. We have (c, d) = (194, 120) or (c, d) = (212, 102). We see that equations (3.2) and (3.3) cannot hold. Groups of Lie type: Representations in cross-characteristic Proposition 3.27 Assume M is almost simple of type S, where S is a ﬁnite simple Chevalley group in cross-characteristic. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold unless (L, S) = (Ω7 (3), P Sp6 (2)), in which case M has only two orbits 74 Table 3.3: Small groups in cross-characteristic. e(S)−1 S (3 2 + 1)/2 |Aut(S)| L2 (q) L2 (q), 3 q 68 Ln (q), n 3 L3 (2), L4 (2), L5 (2), L3 (4) P Sp2n (q), n 2 S4 (5), S4 (7), S6 (5) S4 (2), S6 (2), S8 (2), S4 (4) Un (q), n 3 Un (2), 3 n 7, U3 (4), U3 (5) P Ω+ (q), n 4 Ω+ (2) 2n 8 P Ω− (q), n 4 Ω− (2) 2n 8 Ω2n+1 (q), n 3, q odd E6 (q) E7 (q) E8 (q) F4 (q) F4 (2) 2 E6 (q) G2 (q) 3 3 D4 (q) D4 (2) 2 2 F4 (q) F4 (2) Sz(q) Sz(8) 2 G2 (q) on Eξ (V ) so that 1G P 1G by Corollary 3.7, and so (L, S) is in Table 1.2. M Proof. Suppose equation (3.1) holds for some r ∈ {s, t} and for some M -orbit x M with x ∈ Eξ (V ). Then | x M | 1 m (3 2 + 1), by (3.4). On the other hand, using the lower bounds for degrees of cross-characteristic representations of ﬁnite Chevalley groups given in Table 2.6, we have 2m + 1 | x M| |M | e(S), so that (3m + 1)/2 e(S)−1 2 (3(e(S)−1)/2 )/2. Moreover |Aut(S)|. By this condition, |Aut(S)|. It follows that (3 + 1)/2 we get a ﬁnite list as in Table 3.3. Next we shorten the list by using information on cross-characteristic representations of small groups in the preliminaries together with [26] and [13]. Further V is an absolutely irreducible F3 S-module which is self-dual with ind(V ) = +1, and dimV is odd. (i) Case S = L2 (q), 2 q 68. As L2 (2), L2 (3) are not simple, L2 (4) ∼ L2 (5) ∼ A5 , and = = q 75 67. q is a prime power, we can assume that 7 Case q ≡ 1 (mod 4). By Table 2.7(a) and the fact that dimV is odd, we have dimV ∈ {(q + 1)/2, q}. Using (3.4) again, we only need to consider the following cases: (1) q ∈ {13, 17, 29, 37, 41, 49} and dimV = (q + 1)/2; (2) q = 13, 17 and dimV = q. If dimV = (q + 1)/2, then q must be a square in F3 , so that q ≡ 1 (mod 3) and hence q = 13, 37, 49. If dimV = q, then 3 (q + 1), so that q = 13. If (S, dimV ) = (L2 (13), 7) then L2 (13) < G2 (3) < Ω7 (3) by [9] so M is not maximal in G. If (S, dimV ) = (L2 (13), 13) then L2 (13) < Ω13 (3). Let w be a non-singular eigenvector of an element of order 13. Then | w S| = 14 and hence | w M | |Out(S)|.| w S| = 2.14 < 3m−1 = 35 so that equation (3.1) cannot hold. For other type of point, if M = S then there exists a point u with | u S| = 1092 and (c, d) = (734, 357). If M = S.2 then there exists a point v with | v M | = 2184 and (c, d) = (1469, 714). We check that equation (3.1) cannot hold in any of these cases. If (S, dimV ) = (L2 (37), 19) then m = 9 and the eigenvector w of an element of order 37 is non-singular and | w S| = 38, |Out(S)| = 2 and hence | w M | 38.2 < 3m−1 , so that equation (3.1) cannot hold for this point. For other type of point, there exists a point u with | u S| = |S| = 25308 and (c, d) = (16919, 8388). We see that equation (3.1) cannot hold. If (S, dimV ) = (L2 (49), 25) then m = 12 and there exist two non-singular vectors of diﬀerent type u+ , u− ∈ V which are eigenvectors of an element of order 7 in S such that | uξ S| 8400, ξ = ±. As |Out(S)| = 4 and | uξ M | |Out(S)|.| uξ S| 4.8400 < 3m−1 . In view of (3.4), equation (3.1) cannot hold. Case q ≡ 3 (mod 4). As in previous case, by Table 2.7(b), we have dimV = q and 3 q + 1. Using (3.4) again, we get q ∈ {7, 19}. If (S, dimV ) = (L2 (7), 7) then L2 (7) < Ω7 (3) but Ω7 (3) has no maximal subgroup 76 with socle L2 (7) by [9]. If (S, dimV ) = (L2 (19), 19) then m = 9 and there exist two non-singular vectors of diﬀerent type u+ , u− ∈ V which are eigenvectors of an element of order 5 in S such that | uξ S| 342, ξ = ±. As |Out(S)| = 2 and | uξ M | |Out(S)|.| uξ S| 2.342 < 3m−1 . In view of (3.4) equation (3.1) cannot hold. Case q ≡ 0 (mod 2). As in previous case, by Table 2.7(c), we have dimV ∈ {q−1, q+1}. Using (3.4) again, we have q = 8, 16. By [26], the only possibility is q = 8 and dimV = 7. However by [9], L2 (8) G2 (3) Ω7 (3). (ii) Case Ln (q), (n, q) ∈ {(3, 2), (4, 2), (5, 2), (3, 4)}. As L3 (2) ∼ L2 (7), and L4 (2) ∼ A8 , = = which have been done above, we can exclude these groups. If S = L3 (4), then Out(S) ∼ 2 × S3 ∼ D12 . By [26], dimV ∈ {15, 19, 45, 63}. By using = = (3.4), we only need to consider the representations of degrees 15 and 19. Assume ﬁrst that dimV = 15. Then m = 7. If M = L3 (4) then there exists two nonsingular vectors ui , i = 1, 2 with | ui L3 (4)| = 2016, (c1 , d1 ) = (1250, 765), and (c2 , d2 ) = (1350, 660). Similarly, if M = L3 (4).21 then (c1 , d1 ) = (2600, 1431), (c2 , d2 ) = (1250, 765), if M = L3 (4).22 or L3 (4).23 then (c1 , d1 ) = (2720, 1311), (c2 , d2 ) = (2810, 1221), if M = L3 (4).22 then (c1 , d1 ) = (1250, 765), (c2 , d2 ) = (5327, 2736). We check that equation (3.1) cannot hold. Assume that dimV = 19. Then m = 9. If M = L3 (4) then there exist two nonsingular vectors of diﬀerent type ui , i = 1, 2 which are the eigenvectors of elements of order 7, 5, respectively and (c1 , d1 ) = (707, 252), (c2 , d2 ) = (1190, 825). For the remaining extensions of L3 (4), there exist two non-singular vectors of diﬀerent type ui , i = 1, 2, which are the eigenvectors of an element of order 5 such that the parameters (ci , di ), i = 1, 2, are as follows: If M = L3 (4).22 or M = L3 (4).23 then (c1 , d1 ) = (1190, 825), (c2 , d2 ) = (1100, 915). If M = L3 (4).23 then (c1 , d1 ) = (2561, 1470), (c2 , d2 ) = (1100, 915). If M = L3 (4).S3 or L3 (4).D12 then (c1 , d1 ) = (3842, 2205), (c2 , d2 ) = (4220, 1827). If M = L3 (4).3 77 then (c1 , d1 ) = (3932, 2115), (c2 , d2 ) = (4220, 1827). We can check that equation (3.1) cannot hold in any of these cases. Finally if S = L5 (2), then dimV so that equation (3.1) cannot hold. (iii) Case S ∈ {S4 (5), S4 (7), S6 (5)}. If S ∼ S4 (5) or S6 (5) then the smallest odd degree = non-trivial irreducible representations of S has degree 13, and 63, respectively. However since the smallest ﬁeld of deﬁnitions of these representations are quadratic extensions of F3 , (cf. [19]), L can not embed in Ω13 (3) and Ω63 (3). By Theorem 2.1, in [16], if Φ is a representation of S which is not the smallest representation, then dimΦ (q n − 1)(q n − 155. But (3m + 1)/2 (377 + 1)/2 > |Aut(L5 (2))| q)/(2(q + 1)), which are 40 and 1240, respectively. But then inequality (3.4) cannot hold. If S = S4 (7), then the smallest non-trivial representation in characteristic 3 of S is a Weil representation of degree 25. However, the Frobenius -Schur indicator is ◦, (cf. [19]), which means that S4 (7) ﬁxes no quadratic form. Thus S4 (7) cannot embed in Ω25 (3). If Φ is a non-trivial representation of S4 (7) which is not the smallest representation of S, then dim(Φ) 126, but this again violates (3.4). Thus equation (3.1) cannot hold. (iv) Case S ∈ {S4 (2), S6 (2), S8 (2), S4 (4)}. By the isomorphism S4 (2) ∼ S6 , it follows = that A6 ∼ S4 (2) . Thus we can exclude this case. If S = S4 (4) then dimV = (3m + 1)/2 51 and (325 + 1)/2 > |Aut(S4 (4))| so equation (3.1) cannot hold. For S6 (2), and S8 (2), we need to consider the following cases (S6 (2), 7), (S6 (2), 21), (S6 (2), 27) (S8 (2), 35). If S = S6 (2) and dim(V ) = 7, then by [9], S6 (2) is a maximal subgroup of Ω7 (3) and it has only two orbits on E(V ) so that equation (3.1) holds for both types of points. If (S, dimV ) = (S6 (2), 21) then m = 10, Out(S) = 1 and S6 (2) Ω21 (3). There exist two non-singular vectors of diﬀerent type ui , i = 1, 2 which are the eigenvectors of elements of order 5, 12, respectively and (c1 , d1 ) = (212, 165), (c2 , d2 ) = (2132, 1647). If (S, dimV ) = (S6 (2), 27) then m = 13 and S6 (2) Ω27 (3). There exist two non- singular vectors of diﬀerent type ui , i = 1, 2 which are the eigenvectors of an element of 78 order 5 and (c1 , d1 ) = (96968, 48183), (c2 , d2 ) = (47912, 24663). If (S, dimV ) = (S8 (2), 35) then m = 17, Out(S8 (2)) = 1 and S8 (2) Ω35 (3). There exist two non-singular vectors of diﬀerent type ui , i = 1, 2 which are the eigenvectors of an element of order 5 and (c1 , d1 ) = (119, 0), (c2 , d2 ) = (256094, 129465) and so equation (3.1) cannot hold. (v) Case S ∈ {Un (2), 3 n 7, U3 (4), U3 (5)}. As U3 (2) ∼ 32 .Q8 and U4 (2) ∼ S4 (3), = = we can rule out these cases. If S = U5 (2), then by [26], the smallest odd degree non-trivial 3-modular representation of S has degree 55. Thus (3m + 1)/2 If S = U3 (4), then dimV ∈ {13, 39, 75}. However if dimV (327 + 1)/2 > |Aut(S)|. 39 then (3.4) cannot hold, and if dimV = 13, then by [19], U3 (4) ﬁxes no quadratic form. If S = U7 (2), then by [19], we have dimV > 250 and so (3.4) cannot hold. For the remaining cases, using [26], we need to consider the following cases: (U3 (5), 21) and (U6 (2), 21). If (M, dimV ) = (U3 (5), 21) then we have m = 10, Out(S) = S3 and M Ω21 (3). For each extension M of S, there exist two non-singular vectors of diﬀerent type with parameters (ci , di ), i = 1, 2, as follow: if M = U3 (5) then (c1 , d1 ) = (7033, 3466), (c2 , d2 ) = (27145, 14854); if M = U3 (5).2 then (c1 , d1 ) = (55437, 28562), (c2 , d2 ) = (55371, 28628). If (S, dimV ) = (U6 (2), 21) then we have m = 10, M Ω21 (3) and Out(S) = S3 . For each extension M of S, there exist two non-singular vectors of diﬀerent type with parameters (ci , di ), i = 1, 2, as follow: if M = U6 (2) then (c1 , d1 ) = (7033, 3466), (c2 , d2 ) = (27145, 14854); if M = U6 (2).2 then (c1 , d1 ) = (55437, 28562), (c2 , d2 ) = (55371, 28628). We can check that equation (3.1) cannot hold in any of these cases. + (vi) Case S = O8 (2). By [26], the smallest odd degree non-trivial 3-modular represen- tation of S has degree 35, and the second smallest odd degree one has degree 147. Using + (3.4) again, we only need to consider the 3-modular representation of O8 (2) of degree 35. We have m = 17 and there are two non-singular points vi , i = 1, 2, of diﬀerent type with | v1 S| = |120 and | v2 S = |90720. Then | vi M | |Out(S)|| vi S| < 3m−1 and so 79 equation (3.1) cannot hold. − (vii) Case S = O8 (2). By [26], dimV 203. Then inequality (3.4) cannot hold. (viii) Case S = F4 (2). By [19], dimV > 255. Then inequality (3.4) cannot hold. (ix) Case S = G2 (4). By [26], dimV = 2m + 1 649. Clearly, 3m−1 > 415 |Aut(S)|. (x) Case S = Sz(8). By [26], dimV = 35. Then inequality (3.4) cannot hold. (xi) Case S = 3 D4 (2). By [26], either dimV = 25 or dimV hold then m 174 and clearly 3m−1 3174 > 3.229 F4 (3) 351. If the latter case |Aut(S)|. When dimV = 25, the Ω25 (3).(see[35]). 337 > 227 > |Aut(S)|. group S is not maximal in Ω25 (3) as 3 D4 (2) (xii) Case S = 2 F4 (2) . By [26], 2m + 1 77. Then 3m−1 Groups of Lie type: Representations in deﬁning characteristic Let k be an algebraically closed ﬁeld of characteristic p. Let G be a simply connected, simple algebraic group over k. Fix a maximal torus T and a Borel subgroup B containing T. Let U be the unipotent radical of B. Then B = U T. Let Φ be the root system of G, select a system of positive roots Φ+ from Φ, with corresponding fundamental roots Π = {α1 , . . . , α }. Let {λ1 , . . . , λ } be the fundamental dominant weights and X + be the set of dominant weights. Let L be the simple Lie algebra over C of the same type as G. For each dominant weight λ ∈ X + , there exists an irreducible L-module V (λ) of highest weight λ, and a maximal vector v + (unique up to scalar multiplication). Let U = U(L) be the universal enveloping algebra of L, and UZ be the Kostant Z-form of U (see [21], 26.3). Now, UZ v + is the minimal admissible lattice in V (λ), and UZ v + ⊗Z k is a kG-module of highest weight λ, also denoted by V (λ), and called a Weyl module for G ([21], 27.3). The Weyl module V (λ) has a unique maximal submodule J(λ) and L(λ) = V (λ)/J(λ) is an irreducible kG-module of highest weight λ. Assume that S is simply connected of type A or 2 A over Fq , where q = pf , and G be the corresponding simply connected, simple algebraic group over k, such that S = Gσ for some suitable Frobenius map σ. Let N = N (S) be the natural module for S. We collect 80 here some information about L(λ) for some special dominant weights λ. (1) Let 0 < c < p, and λ = cλ1 or λ = cλ . Then L(λ) has all weight spaces of dimension 1. L(λ) is isomorphic to the space of homogeneous polynomials of degree c, ( + c)! that is, L(λ) ∼ S c (N ). In particular, dimL(λ) = ([42], 1.14). = !c! (2) If > 1, then L(λi ) ∼ i N and dimL(λi ) = +1 (see [7]). = i (3) Let λ = n1 λ1 + n2 λ2 + . . . n λ be a dominant weight. Then L(λ) preserves a non-degenerate bilinear form if and only if n1 = n , n2 = n −1 , . . . . Thus if L(λi ) leaves invariant no non-degenerate bilinear form, and if does not preserve any such form. Let λ = λ( +1)/2 . If is even then +1 2 is odd then L(λi ), i = ≡ −1 (mod 4) then L(λ) ﬁxes a ≡ 1 (mod 4). (see symmetric bilinear form and it ﬁxes an alternating bilinear form if [5] Chapter VIII §13 Table 1, p. 217). The following constructions for adjoint modules of groups of type A and 2 A are taken from [36], pp.491 − 492. (4) We construct the irreducible module L(λ1 + λ ) as follows: Let V := V1 /(V1 ∩ V2 ), where V1 = {A ∈ M +1 (q) | T r(A) = 0}, V2 = {aI +1 | a ∈ Fq }. Let S act on V1 by conjugation. Then V is an irreducible S-module of dimension 2 + 2 − εp ( + 1). The bilinear form on V1 is deﬁned as follows: for any A, B ∈ V1 , (A, B) = T r(AB). We can check that S preserves this bilinear form. Also, V has a basis consisting of Ei,j , 1 j + 1, Eii − Ei+1,i+1 , i = 1, · · · , − εp ( + 1). (5) Let S = SUn (q) and λ = λ1 + λ , where n = t i< + 1. Let V2 = {aIn | a ∈ Fq2 }, V1 = {A ∈ Mn (q 2 ) | T r(A) = 0, A = A }, and set V := V1 /(V1 ∩ V2 ), where the map A → A is the map that raises each entry to its q th -power. Let S act on V1 as in (4). The bilinear form on V1 is also deﬁned as in (4). We can check that S preserves this bilinear form and L(λ1 + λ ) ∼ V. Moreover ﬁx a generator µ of F∗2 , V has a basis consisting of = q Ei,j + Ej,i , µEi,j + µEj,i 1 i<j + 1, Eii − Ei+1,i+1 , i = 1, · · · , − εp ( + 1). Proposition 3.28 Assume M is almost simple of type S, where S is simply connected of 81 type A or 2 A over F3f . There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). We consider the case f = 1 and f > 1 separately. Case f = 1. We can assume that 2. By Theorems 2.35 and 2.36, there exists a 3- restricted dominant weight λ ∈ X3 such that V ∼ L(λ). As |Aut(S)| = |Aut(Lε+1 (3))| = 3( +1) , where ε = ±. It follows from (3.4) that 3m−1 < 3( +1) . Hence m < ( + 1)2 + 1 and dimV < 2( + 1)2 + 3. We need to look for all dominant weights λ ∈ X3 such that L(λ) is self-dual, has dimension less than 2( + 1)2 + 3 and of odd degree. If 3 2 2 18, 2( + 1)2 + 3, and so by Theorem 5.1 in [38], λ is one of the following 38 restricted dominant weights {λ1 , λ , λ2 , λ −1 , 2λ1 , 2λ , λ1 + λ }. Since L(λ) is self-dual, the then only possibility for λ is λ1 + λ . If A21 in [38], either λ = 2λ2 when Suppose that < 18, then by Theorem 4.4, Appendix A6 through 17. = 3 or λ = λ1 + λ for 2 2 4. We have dimL(λ1 +λ ) = +2 −εp ( +1). As dimL(λ1 +λ ) is odd, 1. Consequently 7. As constructed it follows that = 6b1 +1, 6b1 +2 or 6b1 +3 for b1 above, L(λ1 +λ ) ∼ V := V1 /(V1 ∩V2 ). Let U be the subgroup of S consisting of all matrices = of the form diag(I2 , A), where A ∈ SLε−1 (3). Then U ∼ SLε−1 (3). For ξ = ±1, let xξ = = E1,2 +ξE2,1 +V1 ∩V2 , when ε = +, and x+ = E1,2 +E2,1 +V1 ∩V2 , x− = µE1,2 +µE2,1 +V1 ∩V2 when ε = −. Then xξ ∈ V and Q(xξ ) = 0. It follows that xξ is a non-singular point in V, of plus or minus type depending on ξ and . As V1 ∩ V2 is ﬁxed under natural action of U, and clearly, U centralizes xξ , it follows that U 1+c+d S xξ , the stabilizer of xξ in S. We have 2.32 −1 (3 + 1)(3 +1 + 1) < 34 +2 . 1 2 ( 2 |Aut(S) : U | = [Aut(Lε+1 (3)) : SLε−1 (3)] 2 As 2m + 1 = 1 2 ( 2 + 2 − ε3 ( + 1) 1 ( 2 2 + 2 − 1, m − 1 + 2 − 4). We have + 2 − 4) − (4 + 2) = ( − 6) − 8). If 2 8 then ( − 6) − 8 > 0, hence m − 1 > 4 + 2. If = 7 then 2m + 1 = Thus m − 1 4 + 2 for any + 2 = 63, or m = 31 and m − 1 = 30 = 4 + 2. 34 +2 > 1 + c + d. This contradicts to 4. Hence 3m−1 82 inequality (3.4). Therefore, equation (3.1) cannot hold in this case. We are left with the cases = 2, λ = λ1 + λ2 , = 3, λ = λ1 + λ3 , and = 3, λ = 2λ2 . If the ﬁrst case holds then S = SLε (3), and dimL(λ1 + λ2 ) = 7. However, by [9], Ω7 (3) 3 has no maximal subgroup with socle Lε (3). 3 Assume that (S, L) = (L4 (3), Ω15 (3)). For each extension of S, using [13], we can ﬁnd two non-singular points of diﬀerent type with parameters (c, d) as follow: if M = L4 (3) then (c, d) = (42524, 20655), (1160, 945), if M = L4 (3).2 then (c, d) = (311768, 154791), (505196, 252963). Assume that (S, L) = (U4 (3), Ω15 (3)). As in case L4 (3), for each extension M of U4 (3), we can ﬁnd two non-singular points of diﬀerent type with the parameters (c, d) as follow: If M = U4 (3) or U4 (3).2 then (c, d) = (435212, 217971), (2780, 1755); if M = U4 (3).22 then (c, d) = (217970, 108621), (435212, 217971); if M = U4 (3).4 or U4 (3).D8 then (c, d) = (217970, 108621). If (S, L) = (L4 (3), Ω19 (3)) and M = L4 (3), L4 (3).2 then there exists two non-singular points of diﬀerent types with (c, d) = (2600, 1611), (1070, 1035). Assume (S, L) = (U4 (3), Ω19 (3)). For each extension M of U4 (3), we can ﬁnd two nonsingular points of diﬀerent type with the parameters (c, d) as follow: if M = U4 (3), U4 (3).2 then (c, d) = (2690, 1845), (217700, 108891); if M = U4 (3).4 then (c, d) = (435752, 217431), (217700, 108891); if M = U4 (3).22 , U4 (3).D8 then (c, d) = (435752, 217431), (2420, 2115). We can check that equation (3.1) cannot hold in any of these cases. Case f > 1. First consider case = 1. As SL2 (q) ∼ SU2 (q), we can assume that = ε = +. If f = 2, then S = SL2 (9). Then S = L2 (9) ∼ A6 . Thus, we can assume = that f 3. If λ is any 3-restricted dominant weight then λ = cλ1 , where 0 c 2, dimL(cλ1 ) = c + 1 and L(cλ1 ) is self-dual. By Proposition 2.41, dimV = (dimΨ)f , for some irreducible k S-module Ψ. As dimV = 2m + 1 is odd, dimΨ is odd and hence dimΨ 3 = dimL(2λ1 ). It follows that 2m + 1 3f and hence m − 1 (3f − 3)/2. 4f, with As |Aut(L2 (3f ))| = f · 3f (32f − 1) < 34f , it follows from (3.4) that 3f − 3)/2 83 f 3. However by induction on f Consider case 3, this is not true. Thus equation (3.1) cannot hold. 2. It is shown in case f = 1 that if λ ∈ X3 such that L(λ) is self- dual and has smallest odd degree then λ = λ1 + λ . By Propositions 2.41 and 2.39 again, 2m + 1 = (dimΨ)f , for some self-dual irreducible k S-module Ψ of odd degree. Hence dimΨ dimL(λ1 + λ ). It follows that 2m + 1 ( 2 + 2 − ε3 ( + 1))f . We will show 2 that 3m−1 > |Aut(Lε+1 (3f ))|. Then (3.4) cannot hold. As |Aut(Lε+1 (3f ))| < 3f ( +1) and m−1 (( 2 + 2 − ε3 ( + 1))f − 3)/2 (( 2 + 2 − 1)f − 3)/2, it suﬃces to show that (( 2 + 2 − 1)f − 3)/2 > f ( + 1)2 . This is true by induction. The proof is now completed. Let S be a simply connected group of type B over Fq , where q = pf , and G be the corresponding simply connected, simple algebraic group over k, such that S = Gσ for some suitable Frobenius map σ. Let N be the natural module for S with the standard basis β = {e1 , . . . , e , x, f1 , . . . , f }. Multiplying some suitable constant to the symmetric bilinear form, we can assume that the representing matrix of the symmetric bilinear form on N has the form   0 0 I    B = 0 1 0.     I 0 0 Let T be the set of all matrices of the form diag(d, 1, d−1 ), where d = diag(t1 , . . . , t ) ∈ GL (k). As T ∼ (k ∗ ) , T is a maximal torus of S. For i = 1 . . . , deﬁne γi : T → k ∗ , = by γi (diag(t1 , . . . , t , 1, t−1 , . . . , t−1 )) = ti . Then {γi }i=1 form an orthonormal basis for E. 1 Also deﬁne α +1−i = γ +1−i − γ −i , for i = 1, . . . , − 1, and α1 = γ1 . Then {α1 , . . . , α is a fundamental root system of type B , and the corresponding Z-basis of the fundamental 1 dominant weights is {λ1 , . . . , λ }, deﬁned as following: λ1 = 2 (γ1 + · · · + γ ), and λ +1−i = γ + γ −1 + · · · + γ +1−i , for i = 1 . . . , − 1. Proposition 3.29 Assume M is almost simple of type S, where S is simply connected of type B over F3f . There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold 84 so that M is not in Tables 1.1-1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). Case f = 1. First we claim that if λ is a 3-restricted dominant weight such that dimL(λ) is odd and greater than dimN then λ must be one of the following weights: (i) λ = λ −1 , 3, odd, and dimL(λ) = 2 2 + ; (ii) λ = 2λ , = 6k + 3, 6k + 4 or 6k + 5, for some non-negative integer k, and dimL(λ) = 2 2 + 3 ,2 2 + 3 − 1, 2 2 + 3 , respectively; (iii) = 3, λ = 2λ1 , and dimL(λ) = 35. |Aut(S)| = |Aut(Ω2 +1 (3))| = 3 2 2 From (3.4), we have 3m−1 Hence dimL(λ) = 2m + 1 5 2 2i i=1 (3 3 − 1) 2 32 2+ . 4 + 2 + 3. Notice that if 3 5 then 2 − (4 + 2 + 3) > 11, then −4 2 − 2 − 3 = ( − 1)2 − 4 > 0, and so 4 2 > 4 + 2 + 3. If dimL(λ) For 2 +2 +3 < 3 , and hence by Theorem 5.1 in [38], λ is either λ −1 or 2λ . 11, by Theorem 4.4 in [38] and the upper bound for dimension of L(λ) above, = 3, λ = 2λ1 , and dimL(2λ1 ) = 35. So case (iii) again, λ is one of the weights above or holds. It remains to get the restriction on in cases (i) and (ii). From the reference above, we also have dimL(λ −1 ) = (2 + 1) and dimL(2λ ) = 2 2 + 3 − ε3 (2 + 1). Now case (i) holds as dimL(λ −1 ) is odd if and only if is odd. Suppose that 3 | 2 + 1. then as 2 + 1 is odd, 2 + 1 = 3(2t + 1), hence = 3t + 1. Since dimL(2λ ) = 2 2 + 3 − 1 = (2 + 3) − 1 is odd, it follows that = 3t + 1 is even. Thus t = 2k + 1 and = 6k + 4. With the same argument, we can see that if 3 2 + 1, then = 6k + 3 or 6k + 4. Let Q be the non-degenerate quadratic form associated with the non-degenerate symmetric bilinear form on N. Then for v ∈ N, (v, v) = 2Q(v). On the tensor product N ⊗ N, we can deﬁne a non-degenerate symmetric bilinear form induced from the form on N as follows: for ui , wi ∈ N, i = 1, 2, (u1 ⊗ u2 , w1 ⊗ w2 ) = (u1 , w1 )(u2 , w2 ), and extend linearly on N ⊗ N. Recall that if u is a non-singular vector in N, then the reﬂection (v, u) u, for any v ∈ N. Let S act on N ⊗ N by ru : N → N deﬁned by vru = v − Q(u) 85 (u ⊗ v)g = (ug ⊗ vg). Then ((u1 ⊗ u2 )ru , (w1 ⊗ w2 )ru ) = ((u1 ⊗ u2 , w1 ⊗ w2 ) for any non-singular vector u. Thus, ∧2 N and S 2 (N ) leave invariant symmetric bilinear forms induced from the one on N ⊗ N. We have L(λ −1 ) ∼ ∧2 (N ) and L(2λ ) ∼ w⊥ /(w⊥ ∩ w ), = = where w = i=1 (ei ⊗ fi + fi ⊗ ei ) + x ⊗ x ∈ S 2 (N ). We now consider case (i). As L(λ −1 ) ∼ ∧2 N, dimL(λ −1 ) = (2 +1) and L(λ −1 ) has a = basis consisting of ei ∧ej , fi ∧fj , 1 i<j , ei ∧fj , 1 i, j and ei ∧x, x∧fi , 1 i . Also denote by Q the associated quadratic form on N ⊗ N. Then for ξ ∈ {±1}, let v = e1 ∧ x + ξx ∧ f1 = (e1 − ξf1 ) ∧ x. Since Q(e1 ∧ x) = 0 = Q(x ∧ f1 ), we have Q(v) = (e1 ∧ x, x ∧ f1 ) = ξ. Hence v is a non-singular point. Let N1 be the subspace of ⊥ N generated by {e1 − ξf1 , x}. As N1 is non-degenerate, N = N1 ⊥ N1 . Denote by H the centralizer of N1 in Ω(N ) ∼ Ω2 +1 (3). It follows that H ∼ Ω2 −1 (3), and H ﬁxes v. By = = (3.4) we have 3m−1 1+c+d 2 |Aut(Ω2 +1 (3)) : Ω2 −1 (3)| = 2 · 32 −1 (32 − 1) + = 2m + 1 < 8 + 3. As is odd and 34 . Hence m − 1 < 4 , so that 2 > 1, the above inequality holds only when = 3. In this case, we have Ω7 (3) Ω27 (3). Using GAP, there are two non-singular points of diﬀerent type xi , i = 1, 2, with (ci , di ) = (13040, 9072), (26324, 17901), we see that equation (3.1) cannot hold in this case. In case (ii), let v = e1 ⊗ e1 + ξf1 ⊗ f1 + w ∩ w⊥ , where ξ ∈ {±1}. Then Q(v) = ξ, hence v is non-singular in L(2λ ). Let N1 = e1 , f1 . Then N1 is a non-degenerate subspace of N. As in case (i), let H be the centralizer of N1 in Ω(N ), as H ∼ Ω2 −1 (3), we = have 3m−1 1+c+d |Aut(Ω2 +1 (3)) : Ω2 −1 (3)| < 34 , hence 2m + 1 < 8 + 3. As 2 2 +3 −ε3 (2 +1) < 8 +3. dimL(2λ ) = 2 2 +3 −ε3 (2 +1), it follows that 2 2 +3 −1 If 4 then 2 2 +3 −1 2·4 +3 −1 = 8 +(3 −1) > 8 +11 > 8 +3, and if = 3, then 3 in this case, (3.4) cannot hold. Finally Ω27 (3). Using GAP, there are two non- 2 2 +3 −ε3 (2 +1) = 27 = 8 +3. Thus since = 3 and λ = 2λ1 . In this case, we have Ω7 (3) singular points of diﬀerent type xi , i = 1, 2, with (ci , di ) = (13850, 8262), (26324, 17901), we see that equation (3.1) cannot hold in this case. 86 Case f 2. By Propositions 2.41 and 2.39, dimV = (dimΨ)f , for some self-dual 3. irreducible k S-module Ψ. As dimV is odd, so is dimΨ. Firstly, suppose that f Since dimΨ is at least 2 + 1, it follows that 2m + 1 3m−1 1 ((2 2 (2 + 1)f . By (3.4), we have (2 + 1)f , we have |Aut(Ω2 +1 (3f ))| < f · 3f (2 2+ ) 3f (2 2+ +1) . As 2m + 1 +1)f −3) < f ·(2 2 + +1). Clearing fraction, we get (2 +1)f −3−f ·(4 2 +2 +2) < 0. By induction, this inequality cannot happen. Secondly, suppose that f = 2 and dimΨ > 2 + 1. It follows from case f = 1 that dimΨ 2 2 + . Arguing as above, we have ( (2 + 1))2 − 3 < 2 · 2 · (2 2, f 2 + + 1) = 2(4 2 + 2 + 2). However, as 1)2 + 4 − 5 (2 + 1)f − 3 − 2(4 + 2 + 2) f 22 (2 + 1)2 − 2(2 + 2(2 + 1)2 + 3 > 0. Hence (2 + 1)f − 3 > 2(4 2 + 2 + 2), a contradiction. Finally, suppose that f = 2 and dimΨ = 2 + 1. In this case, we can assume that Ψ ∼ L(λ ) ∼ N, and so V = N ⊗ N (1) , where N is the natural module for Ω2 +1 (32 ), = = and N (1) denote the module received from the twist action of S on N. For any element v ∈ N, denote by v (1) the corresponding element in N (1) . Notice that if p is odd then for any a, b ∈ Ff , a + b = 0 or 1 if and only if ap + bp = 0 or 1, correspondingly. This holds p because ap + bp = (a + b)p . Fix a standard basis β = {e1 , · · · , e , x, f , · · · , f1 } of S. Let u = (ε1 +ξf1 ) ∈ N. Then for g ∈ S, in the basis β, we write g = (ai,j ). Assume that ug = g. Then a11 + a2 +1,1 = 1 = a1,2 +1 + a2 +1,2 +1 and a1i + a2 +1,i = 0, 1 < i < 2 + 1. Hence, by the notice above, we have ap + ap +1,1 = 1 = ap +1 + ap +1,2 +1 and ap + ap +1,i = 0, 1 < 11 1i 2 1,2 2 2 i < 2 + 1. Therefore, u(1) g = ug ν = u(ap ) = u = u(1) . This means that if g ∈ S ﬁxes ij u then g also ﬁxes u(1) . Let v = u ⊗ u(1) ∈ N ⊗ N (1) . Let H be the stabilizer of u in S. Then H ∼ Ωε (32 ) with ε = ±1, and H = 2 Ωε (32 )| 2 |Aut(Ω2 +1 (32 )) : Ω+ (32 )| 2 Sv . Hence by (3.4), 3m−1 |Aut(Ω2 +1 (32 )) : 34 +2 . Since 2m + 1 = (2 + 1)2 , it follows that (2 + 1)2 < 8 + 7. As 2, (2 + 1)2 − 8 − 7 = 4 2 + 4 + 1 − 8 − 7 = 4 ( − 1) − 6 > 0. This ﬁnal contradiction ﬁnishes the proof. Let S be a simply connected group of type C over Fq , where q = pf , and G be the 87 corresponding simply connected, simple algebraic group over k, such that S = Gσ for some suitable Frobenius map σ. Let N be the natural module for S with the standard basis β = {e1 , . . . , e , f1 , . . . , f }. The representing matrix of the non-degenerate symplectic    0 I form on N has the form B =  =  . From the isomorphisms Sp2 (q) ∼ SL2 (q), and −I 0 ∼ Ω5 (q) for q odd, we can assume that Sp4 (q) = 3. Proposition 3.30 Assume M is almost simple of type S, where S is simply connected of type C over F3f , with 3. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold unless ( , λ, dimV ) = (3, λ2 , 13) or (L, S, λ) = (Ω41 (3), P Sp8 (3), λ1 ). If the ﬁrst case holds then M has at most two orbits on Eξ (V ) so that 1G P 1G by Corollary M 3.7 and hence (L, S) = (Ω13 (3), P Sp6 (3)) is in Table 1.2. The last case is in Table 1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). Case f = 1. Let λ ∈ X3 be a 3-restricted dominant weight such that L(λ) ∼ V. We = ﬁrst get an upper bound for dimV. As |Aut(P Sp2 (3))| = f · 3 by (3.4), 3m−1 and so dimV < 32 3 2+ 2 2i i=1 (3 2 − 1) 32 2+ , , , and hence 2m + 1 4 2 + 2 + 3. If 3 5 then 4 +2 +3 < 3 . If > 11, then dimL(λ) < , and hence by Theorem 5.1, in [38], λ is either λ −1 or 2λ . For 2 11, by Theorem 4.4 in [38] and the upper bound = 4, λ = λ1 , and for dimension of L(λ) above, again, λ is one of the weights above or dimL(λ1 ) = 41. We have dimL(2λ ) = (2 + 1), L(2λ ) ∼ S 2 (N ), and dimL(λ −1 ) = = 2 2 − − 1 − εp ( ), L(λ −1 ) ∼ w⊥ /( w ∩ w⊥ ), where w = e1 ∧ f1 + · · · + e ∧ f . In these = cases, S leaves invariant a quadratic form Q induced from the symplectic form on N. In case λ = 2λ , let v = e1 ⊗ e1 + ξf1 ⊗ f1 ∈ L(2λ ). Since dimL(2λ ) = (2 + 1) is odd, must be odd. If = 3 then dimL(2λ ) = 21 and P Sp6 (3) Ω21 (3). Using [13], there are two non- singular points of diﬀerent type xi , i = 1, 2, with (ci , di ) = (7075430, 3538809), (26324, 17901), we see that equation (3.1) cannot hold in this case. 88 Thus we assume that 5. Let H be the centralizer in S of the subspace generated by |Aut(P Sp2 (3)) : P Sp2 −2 (3)| = {e1 , f1 }. Then H ∼ Sp2 −2 (3). By (3.4), we have 3m−1 = 2 2 2.3 (32 − 1)/3( −1) < 34 . Hence 2m + 1 < 8 + 3. As 2m + 1 = (2 + 1), it follows that (2 + 1) < 8 + 3. However as 5, (2 + 1) 5(2 + 1) = 10 + 5 > 8 + 3, a 2 contradiction. Next consider case λ = λ −1 . As dimL(λ −1 ) = 2 = 6k + 2, 6k + 3 or 6k + 4. − − 1 − εp ( ) is odd, Let v = e1 ∧ e2 + ξf1 ∧ f2 + w ∩ w⊥ , where ξ = ±1. Then v is non-singular in V. Let ⊥ N1 = e1 , e2 , f1 , f2 be a subspace of N. Since N1 is non-degenerate, N = N1 ⊥ N1 . Let ⊥ H, K be the centralizers in S of N1 , N1 , respectively. Then H ∼ Sp2( −2) (3), K ∼ Sp4 (3), = = and H, K commute. In the basis β1 = {e1 , e2 , f1 , f2 }, let 1 0  0 1  g= 0 0   0 −ξ  1 0 1   0 0 0   ,h =  0 1 0     0 0 1   1 1 0 0  0 0  0 0  . 1 0   −1 1 As gBg t = B, hBht = B, and det(g) = 1 = det(h), g, h ∈ Sp(V1 ). Furthermore, g, h are of order 3 and gh = hg, the subgroup generated by g and h are elementary abelian of order 9. Since vg = v and vh = h, it follows that E = g, h hence 1 + c + d Kv . Thus E × H Sv , |Aut(S) : (E × H)| = |Aut(P Sp2 (3)) : (E × P Sp2( −2) (3))| < 38 −7 . 2 Hence 2m + 1 < 16 − 11. Since 2m + 1 = 2 − − 1 − εp ( ) 2 2 − − 2, we have 2 2 − − 2 < 16 − 11, or equivalent 2 2 − 17 + 9 < 0. As = 6k + 2, 6k + 3, 6k + 4, if > 4 then k 1, and so 8. Then 2 2 −17 +9 4. = 3, 2 2 −16 −( −8)+1 = (2 −1)( −8)+1 > 0, a contradiction. Thus If = 4, then dimL(λ −1 ) = 27, by using GAP, equation (3.1) cannot hold. If then dimL(λ −1 ) = 13. Using GAP, there is only one orbit of minus points and two orbits of plus points. Hence equation (3.1) holds for both types of points by Corollary 3.7. We 89 are left with case Case f = 4, λ = λ1 , dimL(λ1 ) = 41. 2. It follows from case f = 1 that if λ is a 3-restricted dominant weight 2 2 such that L(λ) is self-dual and is of odd degree then dimL(λ) 2 2 − − 1 − εp ( ) − − 2. As V is an absolutely irreducible k S-module, by Propositions 2.41 and 2.39, 2m + 1 = (dimΨ)f , for some self-dual irreducible k S-module Ψ of odd degree. Thus 2m + 1 = (dimΨ)f (2 2 − − 2)f . By (3.4), 3m−1 |Aut(P Sp2 (3))| 3f (2 2+ +1) , so that 2m + 1 < 2f (2 2 + + 1) + 3. Combining these inequalities, we have (2 2 − − 2)f < 2f (2 2 + + 1) + 3, where 3, f 2. However by induction we see that this inequality cannot happen. The proof is now completed. Let S be a simply connected group of type D or 2 D over Fq , where q = pf , and G be the corresponding simply connected, simple algebraic group over k, such that S = Gσ for some suitable Frobenius map σ. Let N be the natural module for S with the standard basis β. The representing matrix of the non-degenerate symmetric bilinear form on N has   0 I  + − the form B =  = =  . From the isomorphisms Ω4 (q) ∼ SL2 (q)◦SL2 (q), Ω4 (q) ∼ L2 (q) I 0 and P Ω± (q) ∼ L± (q), we can assume that 4. = 6 4 Proposition 3.31 Assume M is almost simple of type S, where S is simply connected of type D or 2 D over F3f . There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). Case f = 1. Let λ ∈ X3 be a 3-restricted dominant weight such that L(λ) ∼ V. By = inequality (3.4), 3m−1 |Aut(P Ωε (3))| 2 32 2− +2 . Thus 2m + 1 4 2 − 2 + 7. By Theorem 5.1 and 4.4 in [38], either If λ = λ −1 , then dimL(λ) = 2 must be odd. Thus 2 4 and λ = λ −1 , 2λ or − = 4 and λ = 2λ1 , 2λ2 . and L(λ) ∼ ∧2 N. Since dimL(λ) is odd, = 5. Let a = e1 − ξf1 , b = e2 + f2 ∈ N and z = a ∧ b ∈ ∧2 N, 90 where ξ = ±1. Then z is non-singular in V = ∧2 N. Also, let N1 be a subspace of N generated by a and b. Clearly, N1 is non-degenerate, dimN1 = 2, and sgn(N1 ) = ξ. (For any v ∈ N1 , v = αa + βb, Q(v) = α2 Q(a) + β 2 Q(b) = −ξα2 + β 2 = 0 has non-zero ⊥ ⊥ solutions if and only if ξ = 1.) Since N = N1 ⊥ N1 , dimN1 = 2 − 2 and sgnN = ⊥ ⊥ sgnN1 · sgn(N1 ), it follows from Proposition 2.5.11 [29] that sgnN1 = εξ. Since the discriminant D(Q) ≡ detB = (−1) = −1 (mod(F∗ )2 ), as ⊥ in [29], H = Ω(N1 ) ∼ Ωεξ−2 (3) = 2 is odd, by Proposition 4.1.6 M z . Therefore, 34 −1 . Since Ω(N ) centralizes N1 . Hence H 1+c+d |Aut(P Ωε (3)) : Ωεξ−2 (3)| = 432 −2 (3 − ε1)(3 −1 + εξ1) 2 2 2− 5, 3m−1 = 3(2 −3)/2 > 34 −1 > 1 + c + d, a contradiction to (3.4). Next if λ = 2λ , then dimL(λ) = 2 2 + − 1 − ε3 ( ) and V = L(λ) is the heart of S 2 N, that is, w⊥ /(w⊥ ∩ w ), where w = w= −1 i=1 (ei ⊗fi +fi ⊗ei )+x⊗x+y⊗y i=1 (ei ⊗ fi + fi ⊗ ei ) ∈ S 2 N if S is of type D and otherwise. Let zξ = e1 ⊗e1 +ξf1 ⊗f1 + w ∩w⊥ , ξ = N and H be ±1, and N1 = e1 , f1 N. Then zξ is non-singular in V. Let N1 = e1 , f1 ⊥ the centralizer of N1 in Ω(N ) ∼ Ωε (3). Since sgnN1 = ε, H ∼ Ωε −2 (3) and H ﬁxes zξ . = 2 = 2 Thus 1 + c + d then 3m−1 3(2 |Aut(P Ωε (3)) : Ωε −2 (3)| 2 2 2+ 12.32 −2 (3 − ε1)(3 −1 + ε1) 34 . If 5, −5)/2 > 34 > 1 + c + d, contradicts to (3.4). If = 4, then 2m + 1 = 35, hence m − 1 = 16. As 3m−1 = 316 = 34 > 1 + c + d, we also get a contradiction to (3.4). Case f 2. It follows from case f = 1 that if λ is a 3-restricted dominant weight such 2 2 − . As V is an absolutely that L(λ) is self-dual and is of odd degree then dimL(λ) irreducible k S-module, by Propositions 2.41 and 2.39, 2m + 1 = (dimΨ)f , for some selfdual irreducible k S-module Ψ of odd degree. Thus 2m + 1 = (dimΨ)f By (3.4), 3m−1 |Aut(P Ωε (3))| 2 3.3f (2 2− (2 2 2 − )f . +1) 2 , so that 2m + 1 < 2f (2 − )f < 2f (2 2 − + 1) + 5. Combining these two inequalities, we have (2 4, f 2. By induction, this cannot happen. − + 1) + 5, where Assume that S is simply connected of exceptional type which deﬁnes over a ﬁeld of characteristic 3. Then S is one of the following types: G2 , F4 , E6 , E7 , E8 , 2 E6 , 3 D4 , 2 G2 . 91 Proposition 3.32 Assume M is almost simple of type S, where S is of exceptional type above and deﬁne over F3e with e 1. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold unless (L, S, λ) = (Ω7 (3), G2 (3), λi ), i = 1, 2, or (F4 (3), Ω25 (3), λ4 ), and M has one or at most two orbits on E(V ), respectively, so that 1G P 1G by Corollary M 3.7 and so they are in Table 1.2 or (L, S, λ) = (Ω77 (3), E6 (3), λ2 ), (Ω133 , E7 (3), λ1 ) and they are in Table 1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). (a) Case G2 . By (3.4), 3m−1 ﬁrst that e = 1. Then 2m + 1 |Aut(G2 (3e ))| 315e . Thus 2m + 1 30e + 3. Assume 33. Let λ ∈ X3 with V ∼ L(λ). From Appendix A.49 in = [38], λ is one of the following weights λ1 , λ2 , 2λ1 , 2λ2 . If λ is λi , i = 1, 2, then dimV = 7. In these cases, we have G2 (3) Ω7 (3). By [9], these are maximal embedding and G2 (3) has Ω27 (3). But only one orbit in E(V ). If λ = 2λ1 or 2λ2 then dimV = 27. We have G2 (3) this is not a maximal embedding as G2 (3) Ω7 (3) Ω27 (3), where the last embedding arises from the symmetric square of the natural module for Ω7 (3). (see Proposition 3.29). Assume that e 2. By Propositions 2.41 and 2.39, 2m + 1 = (dimΨ)e , for some self-dual 7e . Combining with irreducible k S-module Ψ of odd degree. It follows that 2m + 1 2m + 1 30e + 3, we have e = 2. Then dimV = 72 = 49. However G2 (9) is not maximal Ω7 (9) Ω72 (3) where the ﬁrst embedding arises as in previous in Ω49 (3) since G2 (9) case, while the second is the twisted tensor product embedding. (b) Case F4 . By (3.4), 3m−1 |Aut(F4 (3e ))| 353e . So 2m + 1 106e + 3. Assume that e = 1. From Appendix A.50 in [38], λ = λ4 and dimL(λ) = 25. In this case, we have an embedding F4 (3) Ω25 (3). By [8], F4 (3) has 5 orbits of points in V. But there are two orbits of singular points, and so there are at most two orbits for each types of non-singular points. Assume that e But this cannot happen for any e (c) Case ε E6 . By (3.4), 3m−1 2. |Aut(ε E6 (3e ))| 92 379e . Thus 2m + 1 158e + 3. 2. we have 2m + 1 25e , so that 25e 106e + 3. Assume that e = 1. From Appendix A.51 in [38], λ = λ2 and dimL(λ) = 77. (Note that dimL(λ1 ) = dimL(λ6 ) = 27 but these modules are not self-dual). In this case, we have 2 E6 (3) E6 (3) Ω77 (3), and V = L(E6 )/Z(L(E6 )), where L(E6 ) is the Lie algebra of 2, then 77e 158e + 3. However this cannot happen for any e |Aut(E7 (3e ))| 3134e . Thus 2m + 1 2.. E6 over F3 . If e (d) Case E7 . By (3.4), 3m−1 268e + 3. Assume that e = 1. From Appendix A.52 in [38], λ = λ1 and dimL(λ) = 133. We have E7 (3) Ω133 (3), and V = L(E7 ), the Lie algebra of E7 over F3 . Assume that e 268e + 3. This cannot happen for e |Aut(E8 (3e ))| 2. 3249e . Thus 2m + 1 498e + 3. Assume 2. We have 133e (e) Case E8 . By (3.4), 3m−1 that e = 1. From Appendix A.53 in [38], dimL(λ) that e 2. Clearly 3875e > 498e + 3 for any e 2. 3875 > 498.1 + 3 = 501. Assume (f ) Case 3 D4 . By (3.4), 3m−1 |Aut(3 D4 (3e ))| 330e . Thus 2m+1 60e+3. Assume that e = 1. From Appendix A.53 in [38], λ = 2λ1 , 2λ2 , 2λ4 and dimL(λ) = 35. Since the splitting ﬁeld for 3 D4 (3) is F33 , dimV = 3.35 = 105 > 63. Assume that e 35e > 60e + 3 for any e 2. |Aut(2 G2 (32e+1 ))| 38(2e+1) . Thus 2m + 1 32e + 19. 2. Clearly (g) Case 2 G2 . By (3.4), 3m−1 By Proposition 2.41, 2m + 1 72e+1 , and so 72e+1 32e + 19. This cannot happen. Embedding of Sporadic groups Let S be a sporadic simple group. Deﬁne g3 (S) := 2log3 (|Aut(S)|) + 4. Lemma 3.33 Let S be a simple sporadic group. The minimal degrees of irreducible faithful representations of S and its covering groups over F3 are given in Table 3.4 together with the value of the function g3 (S). Proposition 3.34 Assume M is almost simple of type S, where S is a simple sporadic group. There is an M -orbit on Eξ (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. 93 Table 3.4: Minimal degrees of representations for sporadic groups in characteristic 3. S R3 (S) [g3 (S)] S R3 (S) [g3 (S)] M11 5 20 Suz 64 54 M12 10 26 2.Suz 12 54 2.M12 6 26 ON 154 54 J1 56 25 Co3 22 53 M22 21 28 Co2 23 61 2.M22 10 28 F i22 77 63 J2 13 29 2.F i22 176 63 2.J2 6 29 HN 133 65 M23 22 33 Ly 651 74 HS 22 37 Th 248 75 2.HS 56 37 F i23 253 82 J3 18 37 Co1 276 82 M24 22 39 2.Co1 24 82 M cL 21 42 J4 1333 87 He 51 45 F i24 781 106 Ru 378 50 B 4371 144 2.Ru 28 50 2.B 96256 144 M 196882 229 Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). Then by (3.4), |Aut(S)| 3m−1 . It follows that 2m + 1 2log3 (|Aut(S)|) + 3 = g(S). By Lemma 3.33 and [19], we need to consider the following cases: (S, dimV ) = (M22 , 21), (M cL, 21), (Co2 , 23). If (S, dimV ) = (M22 , 21), then M22 < A22 < Ωε (3), 21 hence NG (S) is not maximal in G. If (S, dimV ) = (M cL, 21), then S has two orbits with representatives x , y and stabilizers S x = L3 (4) : 22 and S y = M11 which are in diﬀerent G-orbits. We also have cx = 12194, dx = 10080 and cy = 72809, dy = 40590, and we can check that (3.1) cannot hold in any of these cases. For S : 2, we also get the same result. Finally if (S, dimV ) = (Co2 , 23), then there exist two non-singular points in diﬀerent G-orbits with representative x , y with S x = 210 : M22 : 2 and S y = HS : 2 with sizes | x S| = 46575 < 3m−1 = 310 , | y S| = 476928. The parameters for y S are cy = 296450, dy = 180477. We can check that (3.1) cannot hold in this case. 94 3.5 3.5.1 Ω± (3) 2m Parameters for Ω± (3) 2m 2, and ε = ±. . Let Eε (V ) be ξ Assume the hypothesis and notations in section 3.3 with L = Ωε (3), m 2m For ξ ∈ F∗ = {±1}, we will identify +1 or + with , and −1 or − with the set of all points in V with norm ξ. Take x ∈ Eε (V ), and deﬁne ξ (x) = Eε (V ) ∩ x⊥ , Γ (x) = Eε (V ) ∩ (V − x⊥ − {x}). ξ ξ Then P has exactly three orbits { x }, ∆(x) and Γ (x) on Eε (V ). Recall that Ω ξ Lemma 3.35 Assume ξ = ±. We have (i) |Eε (V )| = 1 3m−1 (3m − ε) ξ 2 1 (ii) k = 2 3m−1 (3m−1 − ε) G I. (iii) l = 32(m−1) − 1 1 (iv) λ = 2 3m−2 (3m−1 + ε) 1 (v) µ = 2 3m−1 (3m−2 − ε) √ (vi) D = 4.3m−2 (vii) s = 3m−2 (ε + 2) (viii) t = 3m−2 (ε − 2) 1 Proof. For any ξ ∈ F∗ , by Lemma 2.11, |Vξ | = 3m−1 (3m − ε). Hence |Eε (V )| = 2 |Vξ |. This ξ proves (i). It also implies that the parameters for G do not depend on ξ. Let xε = e1 + εf1 be a non-singular point in V, where {e1 , f1 } is taken from a standard basis β for V. Clearly, x⊥ has a basis {x−ε } ∪ (β − {e1 , f1 }) and x−ε is an ε-type point in x⊥ . For ε ε any y ∈ x⊥ ∩ Eε (V ), Q(y) = Q(x) = ε = Q(x−ε ). Hence x⊥ ∩ Eε (V ) is the set of ε ε ε ε 1 all −ε-type points in x⊥ . Thus k = |E◦ (x⊥ )| = 2 3m−1 (3m−1 − ε). By Lemma 3.1, we ε −ε ε have l = |Eε (V )| − 1 − k = 32(m−1) − 1. To compute λ, take yε = e2 + εf2 ∈ ∆(xε ) = 95 x⊥ ∩ Eε (V ). Then ∆(xε ) ∩ ∆(yε ) = xε , yε ε ε ⊥ ∩ Eε (V ). As xε , yε ε ⊥ has a basis consisting of two orthogonal subsets {x−ε , y−ε } and β − {e1 , e2 , f1 , f2 } with sgn( x−ε , y−ε ) = − and sgn( β − {e1 , e2 , f1 , f2 } ) = ε, it follows that sgn( xε , yε ∆(yε )| = |E−ε ( xε , yε ε ⊥ ⊥ ) = −ε. Hence λ = |∆(xε ) ∩ = 2(m − 1). By Lemma 3.1, )| = 1 3m−2 (3m−1 + ε) as dim xε , yε 2 ⊥ 1 µl = k(k − 1 − λ), hence µ = 2 3m−1 (3m−2 − ε). The remaining parameters follow easily. Corollary 3.36 Let M be a subgroup of G. Suppose that equation (3.1) holds for some r ∈ {s, t} and for any x ∈ Eε (V ). Then (i) if (ε, r) = (+, s) or (−, t) then equation (3.1) has the form c − 2d = ε3m−1 − 1. (3.8) (ii) If (ε, r) = (+, t) or (−, s) then equation (3.1) has the form (ε3m−1 + 1)(ε3m − 3 + c − 2d) = 4c (3.9) (iii) If ε = + then 1+c+d (iv) If ε = − then 1+c+d 3m−1 + 3 2 3m−2 . (3.11) 3m−1 (3.10) Proof. We can write equation (3.1) in the form rl l = (r + 1)c − d . k If ε, r are as in (i) then r = ε3m−1 and (3.12) rl = 2(ε3m−1 + 1). Substitute all these to equation k m−1 m−1 (3.12), we have (ε3 −1)(ε3 +1) = (ε3m−1 +1)c−2d(ε3m−1 +1) = (ε3m−1 +1)(c−2d). rl Hence (i) follows. If ε and r are as in (ii), then r = −ε3m−2 and = −2(ε3m−1 + 1)/3. k 96 Now substitute these parameters into equation (3.12) and multiply both sides by 3, we have 3(ε3m−1 − 1)(ε3m−1 + 1) = 4c − (ε3m−1 + 1)c + 2d(ε3m−1 + 1), and (ii) follows. Let A = 1 + c + d. If (ε, r) = (+, s) then c = 2d + 3m−1 − 1. Then A = 3m−1 + 3d if (ε, r) = (+, t), then 2d(3m−1 +1) = (3m−1 −3)c+(3m−1 +1)(3m −3) Thus d 1 m (3 2 3m−1 . Next, (3m−1 +1)(3m −3). 3m−1 . This proves 1 m−1 (3 2 − 3), and A = 1 + c + d 1 1 + 2 (3m − 3) = 1 (3m − 1) 2 1 (iii). If (ε, r) = (−, t), then 2d − c = 3m−1 + 1. Hence d = 2 (c + 1 + 3m−1 ) + 1), and A = 1 + c + d 1 1 + 2 (3m−1 + 1) = 1 (3m−1 + 3) 2 3m−2 . Finally, if (ε, r) = (−, s), (3m + 3)(3m−1 − 1). Thus 3m−2 (3m + 1)/(3m−2 + 1). To 1 m−1 (3 2 then (3m−1 + 3)c = (3m + 3)(3m−1 − 1) + 2d(3m−1 − 1) c (3m + 3)(3m−1 − 1)/(3m−1 + 3), and A = 1 + c + d ﬁnish the proof, we need to show that 3m−2 (3m + 1)/(3m−2 + 1) fractions, this inequality is equivalent to 2.3m−2 (3m + 1) it is easy to see that 2.3m−2 3m−2 + 1 and 3m + 1 + 3). Clear (3m−2 + 1)(3m−1 + 3). However, 3m−1 + 3. Now the result follows by multiplying these two inequalities sides by sides. 3.5.2 Permutation characters of maximal subgroups in C The reducible subgroups C1 Proposition 3.37 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m ε2 ε of type Oα1 (3) ⊥ O2m−α (3). There is an M -orbit on Eε (V ) such that equation (3.1) does not hold unless M is of type O1 (3) ⊥ O2m−1 (3). In this case M is in Table 1.1. ε2 ε Proof. As M is of type Oα1 (3) ⊥ O2m−α (3), there exists a non-degenerate subspace W V of dimension α such that M = NG (W ). Put W1 = W, W2 = W ⊥ , εi = sgnWi . Write Xi = X(Wi ), where X ranges over the symbols Ω, S and I. By Lemma 4.1.1 in [29], we have MI = I1 × I2 and Ω1 × Ω2 MΩ . Suppose ﬁrst that dimW = 2b + 1 is odd. If b = 0 or m − 1 then the Proposition holds. So suppose that m − 2 b 1. Let x be a non-singular vector in W1 , with ρ(x) = ξ. Arguing as in Proposition 3.10, we have x MΩ = x MI . Thus we need to compute the 97 parameters for MΩ in L. Since Ω1 MΩ acts transitively on E◦ (W1 ), and Ω2 centralizes ξ x , we have x MΩ = x Ω1 , hence d = k(W1 ), c = l(W1 ), parameters for E◦ (W1 ). By ξ Lemma 3.8, d = 1 b−1 b 3 (3 2 − ξ), c = (3b − ξ)(3b−1 + ξ), and hence c − 2d = ξ.3b − 1. If (3.8) holds then ξ.3b − 1 = ε.3m−1 − 1. Hence b = m − 1, a contradiction. Thus (3.9) holds. Therefore (ε3m−1 + 1)(ε3m − 3 + ξ.3b − 1) = 4(3b − ξ)(3b−1 + ξ), whence (3m−1 + ε)(3m − ε.4 + ε.ξ.3b ) = 4(3b − ξ)(3b−1 + ξ). We will show that 3m−1 + ε > 2(3b−1 + ξ) 0 and 3m − ε.4 + ε.ξ.3b > 2(3b − ξ) 0, and by multiplying these two b, m − 1 (b − 1) + 2, hence 1 inequalities together, we get a contradiction. As m − 2 3m−1 + ε − 2(3b−1 + ξ) 32 .3b−1 − 2.3b−1 + ε − 2.ξ = 7.3b−1 + ε − 2.ξ. Since b 7 + ε − 2.ξ and ε, ξ ∈ {±}, 7.3b−1 + ε − 2.ξ (1 + ε) + (2 − 2.ξ) + 4 > 0. Thus b + 2, 3m − 4ε + εξ3b − 2(3b − ξ) 3m−1 + ε > 2(3b−1 + ξ). For the other inequality, as m 9.3b + (εξ − 2)3b + 2ξ − 4ε = (εξ + 1)3b + (6 + 2ξ − 4ε) + 6(3b − 1). As the ﬁrst two terms of the last expression are non-negative and the last one is positive since 3b − 1 (ε.ξ + 1).3b + (6 + 2ξ − 4ε) + 6(3b − 1) > 0, hence 3m + ε.ξ.3b > 2(3b − 2ξ). Secondly, assume that dimW = 2b is even, where 1 b m − 1. Note that if 2 > 0, dimW = 2, then sgn(W ) = −, otherwise, MΩ is contained in a stabilizer of a non-singular point, and so it is not maximal in L. Arguing as above, let x be a non-singular vector in W1 and let ξ = sgn(W1 ). Then x MI = x MΩ = x Ω(W1 ), and so d = k(W1 ), c = l(W1 ), 1 parameters for Eξ (W1 ). By Lemma 3.35, d = 2 3b−1 (3b−1 − ξ), and c = 32b−2 − 1. We Q(x) have c − 2d = ξ3b−1 − 1. If equation (3.8) holds then ξ.3b−1 − 1 = ε.3m−1 − 1. This implies that b = m, a contradiction. If equation (3.9) holds then (3m−1 + ε)(3m − ε.4 + ε.ξ.3b−1 ) = 4(3b−1 −ε)(3b−1 +ε). As m−1 b 1, 3m−1 +ε−2(3b−1 +ε) 3b −ε−2.3b−1 = 3b−1 −ε 0, and 3m −ε.4+ε.ξ.3b−1 −2(3b−1 −ε) (1+εξ)3b−1 +2(3b −ε) > 0. Thus 3m−1 +ε 2(3b−1 +ε) and 3m − ε.4 + ε.ξ.3b−1 > 2(3b−1 − ε). Finally, by multiplying these two inequalities, we get a contradiction. This completes the proof. Retain the notation as in Lemma 3.11 and Lemma 3.12. 98 Proposition 3.38 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m of type Pα . Then M has at most two orbits in E(V ) so that 1G P so M is in Table 1.1. Proof. It suﬃces to show that MΩ has at most 2 orbits on Eε (V ). From deﬁnition M stabilizes a totally singular subspace W of dimension α. By Witt’s Lemma, we can assume that W has a basis {e1 , · · · , eα }, where ei s are taken from a standard basis for Ωε (3). Let Y = f1 , · · · , fα and X = (W 2m Y )⊥ . By Proposition 2.6 (iv), X 1G by Corollary 3.7 and M has a basis {x1 , . . . , xs }, with s = 2(m − α), such that the matrix representing of the associated bilinear form of Q restricted to X is either Is or diag(λ, 1, · · · , 1), according as D(X) = or D(X) = . Let β = {e1 , . . . , eα , x1 , . . . , xs , f1 , . . . , fα }. Let U = CI (W, W ⊥ /W, V /W ⊥ ), T0 = NI (W, Y ) and N = NI (W, Y, X). By Lemma 3.12, U L, N = T0 × I(X) and MI = U : N, where GLα (3) ∼ T0 = I(W 1 Y ). Let T = 2 T0 . As (F∗ )2 = {1}, by Lemma 3.11, SLα (3) ∼ T = T0 ∩ Ω(V ), hence U (T × Ω(X)) = 3 ξ ∈ {±}. As sgn(W Y ) = +, and sgnV = sgn(W K. Fix Y ).sgn(X), we have sgn(X) = ε. Let x be a non-singular vector in X with ρ(x) = ξ, and let a = ξe1 + f1 ∈ V. Using the same argument as in Proposition 3.13 together with Lemma 3.35, we have | x MΩ | 1 |xΩ(X)U | 2 = |Eε (X)||W | = 1 3m−1 (3m−α − ε), and | a MΩ | ξ 2 | x MΩ | + | a MΩ | 1 m−1 m−α 3 (3 2 1 |aT U | 2 = 1 3s+α−1 (3α − 1). 2 Since |Eε | ξ 1 − ε) + 1 3s+α−1 (3α − 1) = 2 3m−1 (3m − ε) = 2 |Eε (V )|, as s = 2m − 2α. It follows that | x MΩ | = | x Ω(X)U |, | a MΩ | = | a T U | and ξ Eε (V ) = x MΩ ∪ a MΩ . Thus MΩ has at most two orbits on Eε (V ). ξ ξ The imprimitive subgroups C2 Proposition 3.39 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m of type O1 (3) S2m . There is an M -orbit on Eε (V ) such that equation (3.1) does not hold unless (n, ε, r) = (6, −, t) or (n, ε, ξ, r) = (10, −, , t), and so M is in Table 1.1. 99 Proof. Let n = 2m and let {x1 , . . . , xn } be an orthonormal basis for V. Argue as in Proposition 3.17, we have x MI = x MΩ for any x ∈ {x1 , x1 + x2 }. Therefore it suﬃces to compute the parameters for MΩ in L. We also have | x1 MΩ | = n, d1 = n − 1, c1 = 0, and | x1 +x2 MΩ | = n(n−1), d2 = n2 −5n+7, c2 = 4n−8 by Lemma 3.16. By Proposition 4.2.15 in [29], ε = (−)m . We consider the following cases: (i) Case m even. Then ε = +. If ξ = , then we can take x = x1 as Q(x1 ) = −1 = . 1 Then (3.1) becomes k = −rd1 . Since k = 2 3m−1 (3m−1 − 1) > 0, d1 = 2m − 1 > 0, r must be negative, hence r = t = 3m−2 (ε.1 − 2) = −3m−2 . Thus 3m = 4m + 1. This holds only when m = 2. As m 4, 3m > 4m + 1, and so this case cannot happen. If ξ = , we choose x = x1 + x2 . If m 6, then 1 + c2 + d2 = n(n − 1) < 3m−1 . This violates (3.10) so equation (3.1) cannot hold. Thus m = 4 and so c = 24, d = 31. We can check that (3.1) cannot hold in this case. (ii) Case m odd. Then ε = −. If ξ = , then we can take x = x1 . Then (3.1) becomes 1 k = −rd1 . Since k = 2 3m−1 (3m−1 + 1) > 0, d1 = 2m − 1 > 0, r = t = 3m−2 (ε.1 − 2) = −3m−1 . Thus 3m−1 = 4m−3. This holds only when m = 3. By [9], we see that MΩ has only 2 orbits on Eε (V ) so that (3.1) holds in this case. Finally, if ξ = Then 2d2 − c2 = 2(n(n − 7) + 11). If m , we choose x = x1 + x2 . 7, then 1 + c2 + d2 = n(n − 1) < 3m−2 . This 3 and m contradicts to (3.11) and hence equation (3.1) cannot hold. Thus since 7 > m is odd, m = 3, 5. If m = 3, using [9], (3.1) holds. Therefore we assume that m = 5. We see that 2d2 − c2 = 82 = 35−1 + 1. Thus (3.1) holds with r = t, ξ = points in V with generators 5 i=1 . Let v and w be two xi and 8 i=1 xi , respectively. Then v , w ∈ E− (V ), ξ 10 5 8 and | w MΩ | = |D10 | = 27 10 8 5 and by applying Lemma 3.15, | v MΩ | = |D10 | = 24 . As | x MΩ | + | v MΩ | + | w MΩ = |E− (V )|, we conclude that MΩ has only three orbits ξ on E− (V ) with representatives x , v and w . ξ 5 Parameters for v MΩ . By Lemma 3.15, v MΩ = D10 , the set of all points of length 5. We ﬁrst determine d = | v MΩ ∩v ⊥ |. Let y = n i=1 αi xi ∈ V, deﬁne supp(y) = {xj |αj = 0}. 100 Let y be a point in v MΩ ∩ v ⊥ with y = We consider the following cases: (1) supp(y) ∩ supp(v) = ∅. Then y = 10 i=6 5 i=1 αij xij . Then (v, y) = 0 and |supp(y)| = 5. αi xi . There are 24 such points in this case. (2) |supp(y) ∩ supp(v)| = 1. Then (v, y) = αj = 0, where αj is the coeﬃcient of the common vector. There are no such points y. (3) |supp(y)∩supp(v)| = 2. Then y = α1 xi1 +α2 xi2 +β1 y1 +β2 y2 +β3 y3 , where all coeﬃcients are non-zero, and xi1 , xi2 ∈ {1, 2, 3, 4, 5}, yi ∈ {6, 7, 8, 9, 10}. We have (v, y) = α1 +α2 = 0. Hence {α1 , α2 } = {1, −1}. There are 23 for α1 xi1 + α2 xi2 . Thus there are 1 2 2 1 5 2 5 3 choices for β1 y1 +β2 y2 +β3 y3 and 23 3 j=1 5 3 2 1 5 2 choices = 25 .52 points. 2 j=1 5 2 (4) |supp(y) ∩ supp(v)| = 3. Then y = α j xi j + βj yj , and (v, y) = 3 j=1 5 3 αj = 0. choices It follow that {α1 , α2 , α3 } = ±{1, 1, 1}. There are 22 for α1 xi1 + α2 xi2 + α3 xi3 , hence there are (5) |supp(y) ∩ supp(v)| = 4. Then y = 2 5 2 3 4 j=1 choices for β1 y1 +β2 y2 , 2 22 5 2 = 24 .52 points. 4 j=1 αj xij + β1 y1 . We have (v, y) = 5 1 αj = 0. 4 2 5 4 It follow that {α1 , α2 , α3 , α4 } = {1, 1, −1, −1}. As there are 2 choices for 4 j=1 choices for β1 y1 , 2 αj xij , there are 1 4 2 2 5 4 5 i=1 .2 5 1 = 6.52 points. 5 i=1 (6) supp(v) = supp(y). Then y = αi xi . Then (v, y) = αi = 0. This happens = 5 points. 10 5 only when {α1 , α2 , α3 , α4 , α5 } = ±{1, 1, 1, 1, −1}. Thus there are 5 1 5 Therefore d = 24 +25 .52 +24 .52 +6.52 +5 = 1371, and c = |D10 |−1−d = 24 −1−1371 = . 2660. Clearly, 2d − c = 2.1371 − 2660 = 82 = 34 + 1. Hence (3.1) holds with r = t, ξ = Parameters for w MΩ . For any y ∈ w K ∩ w⊥ , we have (w, y) = 0, |supp(y)| = 8 and |supp(v) ∩ supp(y)| 6. 6 j=1 (1) |supp(y)∩supp(v)| = 6. Write y = αj xij + 2 j=1 βj yj . Then (w, y) = 6 j=1 αj = 0. 8 6 It follows that {α1 , · · · α6 } = ±{1, 1, 1, 1, 1, 1} or {1, 1, 1, −1, −1, −1}. There are points in the ﬁrst case and 1 6 2 3 8 6 1 6 2 3 8 6 22 22 points in the last case. Thus, there are 8 6 22 + 22 = 1232 points y. 7 j=1 (2) |supp(y) ∩ supp(v)| = 7. Write y = αj xij + β1 y1 . Then (w, y) = 7 j=1 αj = 0. It 101 follows that {α1 , · · · , α7 } = ±{1, 1, 1, 1, 1, −1, −1}. There are (3) supp(w) = supp(y). Then y = 8 i=1 7 2 8 i=1 8 7 2 2 1 = 672 points. αi xi and (w, y) = αi = 0. This happens only when {α1 , α2 , · · · , α8 } = ±{1, 1, 1, 1, 1, 1, 1, −1} or {1, 1, 1, 1, −1, −1, −1, −1}. Then there are 8 1 + 1 8 2 4 = 43 points. 8 Thus d = 1232 + 672 + 43 = 1947, c = |D10 | − 1 − d = 3812, and 2d − c = 82 = 34 + 1. Hence (3.1) holds for r = t, ξ = . This ﬁnishes the proof. Lemma 3.40 Let (V, Q) be an orthogonal geometry with dimV = n. Fix a standard basis for V as in Deﬁnition 2.4. Let x, e ∈ V be a non-singular vector and singular vector respectively. Then   2m  3 − 1, (i) |{v ∈ V − {0}|Q(v) = 0}| =  32m−1 + 2ε3m−1 − 1,  ⊥ if n = 2m + 1 if n = 2m, sgn(V ) = ε. Assume n = 2m + 1 and ξ = ρ(x) = sgn(x  ).  2m−1  3 − ξ3m−1 , if λ = 0 (ii) |{v ∈ V − {0}|Q(v) = 0, (v, x) = λ}| =  32m−1 + 2ξ3m−1 − 1, if λ = 0.    2m−1  3 , if λ = 0 (iii) |{v ∈ V − {0}|Q(v) = ξ, (v, e) = λ}| =  3m (3m−1 + ξ), if λ = 0.    2m−1  3 , if λ = 0 (iv) |{v ∈ V − {0}|Q(v) = 0, (v, e) = λ}| =  32m−1 − 1, if λ = 0.  Proof. (i) is a corollary of Lemma 2.11 with q = 3. For (ii), when λ = 0, then the left side is the number of singular vectors in x⊥ , as dim(x⊥ ) = 2m, and sgn(x⊥ ) = ξ, the result follows from (i). When λ = 0, note that (v, x) = λ if and only if (−v, x) = −λ, and there 1 are only two non-zero values for λ ∈ F, the number in the left side is 2 [32m − 1 − (32m−1 + 2ε3m−1 − 1)] = 32m−1 − ξ3m−1 , this proves (ii). Let β = {e1 · · · , em , f1 , · · · , fm , a} be the standard basis for V. By Witt’s Lemma, we can assume that e = e1 . Then e⊥ = e where W = {e2 , · · · , em , f1 , · · · , fm , a} , and V = e1 102 W W, f1 . Now, suppose that v = α1 e1 + w + α2 f1 ∈ V, where w ∈ W, αi ∈ F and that (v, e) = 0, Q(v) = Q(x). Then as (v, e) = α2 = 0, v = α1 e + w and Q(v) = Q(w) = Q(x) = 0. By Lemma 3.8, there are 3m−1 (3m−1 +ξ.1) such non-singular vectors w ∈ W, where ρ(w) = ρ(x) = ξ. As α1 can take any values in F, there are 3.3m−1 (3m−1 + ξ.1) = 3m (3m−1 + ξ.1) such vectors v. Using the same argument as in (ii), we see that when (v, e) = λ = 0, then (v, e) = λ if and only if (−v, e) = −λ. By Lemma 3.8 there are 3m (3m +ξ.1) non-singular vectors u in V satisfying conditions ρ(u) = ρ(x), hence there are 1 [3m (3m + ξ.1) − 3m (3m−1 + ξ.1)] = 32m−1 such 2 vectors v. This proves (iii). Finally, if v = α1 e + w + α2 ∈ V, where w ∈ W, αi ∈ F and (v, e) = 0, Q(v) = 0. Then v = α1 e1 + w and Q(v) = Q(w) = 0. By Lemma 2.11, there are 32m−2 such w. As α1 can be chosen arbitrarily, there are 3 choices for α1 . Hence, after excluding 0, we have 3.32m−2 − 1 = 32m−1 − 1 such v. Using the same argument as above, 1 there are 2 [32m − 1 − (32m−1 − 1)] = 32m−1 vectors v such that (v, e) = λ = 0, Q(v) = 0. This ﬁnishes the proof. Suppose that α > 1 and D is a non-degenerate α-decomposition. By the Main Theorem in [29], if n 13, then M is not maximal in G unless α 4. Proposition 3.41 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m of type Oα (3) Sb , with α 4 and b 2. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold unless m is odd and (ε, r, α, b) = (−, t, m, 2). In this case M is in Table 1.1. Proof. Assume that V has an orthonormal basis which is the union of orthonormal bases of all Vi . Let N = Ω1 × Ω2 × · · · × Ωb . As α = dimV1 4, V1 contains non-singular points of both types. Let x ∈ V1 be a non-singular point, and ξ = sgn(V1 ) ∈ {±, ◦}. Argue as in Proposition 3.18, we only need to compute the parameters for MΩ in L. Since Sb permutes the set {Vi }b transitively and for i > 1, Ω(Vi ) centralize V1 , x MΩ = x N Sb = i=1 x Ω(V1 )Sb = Eξ (V1 )Sb = ∪b Eξ (Vi ). Thus A = | x MΩ | = b.|Eξ (V1 )|. Also, for i=1 ρ(x) ρ(x) ρ(x) 103 i > 1, Vi are all perpendicular to V1 , hence d = |x⊥ ∩ x MΩ | = k1 + (b − 1)|Eξ |, c = ρ(x) |Γ (x)| = l1 , where k1 , l1 are parameters for V1 . We consider the following cases: (a) dimV1 = 2a + 1 and b 3. Then as α 4 and α is odd, α 5, hence a 2. Also since n = 2m = (2a + 1)b, it follows that b = 2b1 is even, b1 1 A = b.|Eρ(x) (V1 )| = b. 2 3a (3a + ρ(x)) 2, and m = αb1 . We have b. 1 3a (3a + ρ(x)) < b.32a = b.3α−1 . By Corollary 2 3.36, it suﬃces to show that 3m−2 > b.3α−1 , or equivalently, 3m−2−α+1 = 3αb1 −α−1 > b. Since α > 4, b1 αb1 − α − 1 3αb1 −α−1 2, we have (α − 4)(b1 − 2) = αb1 − α − 1 − (α + 4b1 − 9) 4, α 5, α + 4b1 − 9 4. 5+b+b−9 0, and so b. Hence α + 4b1 − 9. As b = 2b1 3b . The results follows as 3b > b for all b 3. Then a (b) dimV1 = 2a and b 1 b. 2 3a−1 (3a − ξ) 2, and m = ab. We have A = b|Eξ (V1 )| = Q(x) b.32a−1 and 1 b. 2 3a−1 (3a + 1) < b.32a−1 . We will show that 3m−2 hence equation (3.1) cannot hold by Corollary 3.36. This inequality is equivalent to 3ab−2a−1 b. As b 3, a 2, we have (a − 2)(b − 3) = ab − 2a − 1 − (a + 2b − 7) b − 2. Clearly, 3b−2 b for any b 0, hence b. ab − 2a − 1 b + (a + b − 7) 3. Thus 3ab−2a−1 (c) dimV1 = 2a and b = 2. Then a 2, ε = ξ 2 = +, and m = α = 2a. By Lemma 1 1 3.35, d = k1 + |Eξ (V2 )| = 2 3a−1 (3a−1 − ξ) + 2 3a−1 (3a − ξ), c = l1 = 32a−2 − 1, and Q(x) c − 2d = −3m−1 − 1 + 2ξ3a−1 . We can check that equations (3.8) and (3.9) cannot hold. (d) dimV1 = α = 2a + 1 is odd and b = 2. Then m = 2a + 1. Let η be the type of x in V1 . By Proposition 4.2.14 in [29], we have ε = (−)(q−1)m/2 = −. By Lemma 3.8, | x MΩ | = 1 1 32a + η.3a , d = k1 + |Eη (V1 )| = 2 3a−1 (3a − η) + 2 3a (3a + η), c = l1 = (3a − η)(3a−1 + η), and c − 2d = −3m−1 − 1. Then c − 2d = −3m−1 − 1 = ε.3m−1 − 1. It follows that equation (3.1) holds with r = t for the point x ∈ E− (V ). Let yi ∈ Vi , i = 1, 2, be non-singular Q(x) vectors with Q(yi ) = −Q(x). As V1 , V2 are isometric, yi are of type −η for i = 1, 2. Put y = y1 + y2 ∈ V. Then y ∈ E− (V ) and y MΩ = y1 Ω(V1 ) + y2 Ω(V2 ). Hence Q(x) | y MΩ | = 2(1 + k1 + l1 )2 = 1 (32a − η.3a )2 . If v = v1 + v2 ∈ y MΩ ∩ y ⊥ , then vi ∈ yi Ω(Vi ) 2 ⊥ and (v1 , y1 ) + (v2 , y2 ) = 0. If (v1 , y1 ) = (v2 , y2 ) = 0, then vi ∈ yi Ω(Vi ) ∩ yi . Thus there 104 2 are 2k1 such points in this case. If (v1 , y1 ) = −(v2 , y2 ) = 0, then there are (l1 + 1)2 such 2 points. Hence d = 2k1 + (l1 + 1)2 , and c = | y MΩ | − 1 − d = (l1 + 1)2 − 1 + 4k1 (l1 + 1). Now c − 2d = −3m−1 − 1 = ε3m−1 − 1. Let z = x + e, where e ∈ V2 is a singular vector. Then Q(z) = Q(x), and hence z ∈ E− (V ). We have z MΩ = ( x Ω(V1 ) + e Ω(V2 )) ∪ Q(x) ( e Ω(V1 )+ x Ω(V2 )), where e is a singular vector in V1 , and x is a non-singular vector in V2 with Q(x ) = Q(x). Thus | z MΩ | = 1 .2.|{v ∈ V1 |Q(v) = Q(x)}|.|{v ∈ V1 |Q(v) = 0}|. 2 Hence, by Lemma 3.8 and 3.40 | z MΩ | = (32a + ξ.3a )(32a − 1). We compute parameters for z MΩ . For any v = v1 + v2 ∈ V1 ⊥ v2 which satisﬁes the following conditions (v, z) = (v1 , x) + (v2 , e) = 0, and v ∈ z MΩ . Then one of the following holds: (i). (v1 , x) = 0 = (v2 , e), Q(v1 ) = Q(x), Q(v2 ) = 0. By Lemma 3.40 and 3.8, there are 1 |{v 2 1 ∈ x⊥ |Q(v) = Q(x)}|.|{v ∈ e⊥ − {0}|Q(v) = 0}| = 2 (32a−1 − ξ3a−1 )(32a−1 − 1) points. 1 (ii). (v1 , x) = 0 = (v2 , e), with Q(v1 ) = 0, Q(v2 ) = Q(x). There are 2 |{v ∈ x⊥ |Q(v) = 0}|.|{v ∈ e⊥ − {0}|Q(v) = Q(x)}| = 1 (32a−1 + 2η3a−1 − 1)(32a−1 + ξ3a ) such points. 2 2 (iii). (v1 , x) = −(v2 , e) = 0, with Q(v1 ) = Q(x), Q(v2 ) = 0. There are 2 |{v ∈ V |Q(v) = Q(x), (v, x) = 1}|.|{v ∈ V − {0}|Q(v) = 0, (v, e) = −1}| = (32a−1 + 2η3a−1 ).32a−1 points. 1 (iv). (v1 , x) = −(v2 , e) = 0 with Q(v1 ) = 0, Q(v2 ) = Q(x). There are 2 .2.|{v ∈ V |Q(v) = 0, (v, x) = 1}|.|{v ∈ V }|Q(v) = Q(x), (v, e) = −1}| = (32a−1 − ξ3a−1 ).32a−1 such points. 1 Thus d = 2 (32a−1 − η3a−1 )(32a−1 − 1) + 1 (32a−1 + 2η3a−1 − 1)(32a−1 + η3a ) + (32a−1 + 2 2η3a−1 ).32a−1 +(32a−1 −η3a−1 ).32a−1 = 34a−1 +η.33a−1 −η.3a−1 , and so c = | z MΩ |−1−d = 2(34a−1 + η.33a−1 − η.3a−1 ) − 32a − 1. Then c − 2d = −32a − 1 = −3m−1 − 1 = ε3m−1 − 1. We have | x MΩ | + | y MΩ | + | z MΩ | = (32a + η.3a ) + 1 (32a − η.3a )2 + (32a + η.3a )(32a − 1) = 2 1 2a 2a .3 (3 2 + 1) = 1 m−1 m .3 (3 2 + 1) = |E− (V )|. Therefore MΩ has only three orbits of Q(x) non-singular points in E− (V ) with representatives x , y and z . Hence (3.1) holds Q(x) for r = t and for any points in E− (V ). Q(x) Proposition 3.42 Assume G is nearly simple primitive rank 3 of type Ω+ (3) and M is 2m of type GLm (3).2, so that D is a totally singular m-decomposition. Then M has only one 105 orbit of non-singular points on V so that 1G P Table 1.1. 1G by Corollary 3.7 and hence M is in M Proof. We may choose ei ∈ V1 and fj ∈ V2 such that β = {e1 , · · · , em , f1 , · · · , fm } is a standard basis for V. By Lemma 4.2.3 in [29], we have I(D) ∼ GLm (3), and hence MI = = I(D) .S2 ∼ GLm (3).2. Also by (4.2.6) and (4.2.7) in [29], we have MΩ = Ω(D) .2 when m is even = and MΩ = Ω(D) when m is odd, where Ω(D) ∼ SLm (3). For ξ = ±, let vξ = e1 + ξf1 ∈ V. = Then vξ is non-singular in V. Choose x ∈ I(D) with detV1 (x) = −1, and x ﬁxes vξ . Let y ∈ MI be the element which interchanges ei and fi for all i. Observe that x, y leaves vξ invariant. Thus vξ MI = vξ MΩ = vξ Ω(D) . Clearly | vξ Ω(D) | = 1 3m−1 (3m − 1) = 2 |E+ (V )|. Therefore MΩ has only one orbit of non-singular points in E+ (V ). Note that ξ ξ when m is odd, M is maximal in G if only if G contains some element that interchanges V1 , V2 and ﬁxes vξ . Proposition 3.43 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m of type Om (3)2 , so that D is non-degenerate and non-isometric, m 5 is odd. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. By Proposition 4.2.16 in [29], ε = (−)(3+1)/2 = +, MI = I1 × I2 and MΩ = S1 × S2 . Let vξ be a non-singular vector in V1 of type ξ. We have vξ MI = vξ MΩ = vξ Ω1 = Eξ (V1 ). As m is odd, m = 2a + 1. By Lemma 3.8, we have d = k(V1 ) = 1 3a−1 (3a − ξ) and 2 c = l(V1 ) = (3a − ξ)(3a−1 + ξ). Then c − 2d = ξ3a − 1. We see that (3.1) cannot hold The ﬁeld extension subgroups C3 Lemma 3.44 Assume case O± holds, with f a non-degenerate symmetric bilinear form, q odd and α = 2. Let v ∈ V satisfy f (v , v ) = λ ∈ F∗ , so that rv ∈ I (V ). (i) spanF (v ) is a non-degenerate 2-space in V with discriminant µN (λ), where µ = F∗ . 106 (ii)    Ω if λ ∈ (F∗ )2 / rv ∈  S \ Ω if λ ∈ (F∗ )2  Proof. This is Lemma 4.3.19 in [29]. Proposition 3.45 Assume G is nearly simple primitive rank 3 of type Ω+ (3) and M is 2m + of type O2b (3α ), where α is a prime divisor of 2m, and 2m α = 2b 4. There is an M -orbit on E+ (V ) such that equation (3.1) does not hold unless α = 2 or α = 3. If α = 2 then ξ M has only two orbits on E+ (V ) so that 1G P 1G by Corollary 3.7; if α = 3 then M has M three orbits on E+ (V ) and equation (3.1) holds for r = s and for all M -orbits on E+ (V ), ξ ξ and hence these cases are in Table 1.1. Proof. Let β = {e1 , . . . , eb , f1 , . . . , fb } be a standard basis for V over F . Deﬁne φ to be φβ ,Q (ν), where ν = Gal(F /F); and γ ν = γ 3 for any γ ∈ F . Denote by I, I the full isometry groups I(V, Q, F ), I(V , Q , F ), respectively. Also let Ω = Ω(V, F, Q), Ω = Ω(V , Q , F ) and S = S(V, F, Q). We ﬁrst prove the following: (1) Q (wφ ) = Q (w)ν for any w ∈ V ; (2) φ is of order α. Moreover, if α is an odd prime then φ ∈ Ω(V, Q, F). If α = 2 then φ ∈ Ω, if b is even and φ ∈ S \ Ω, if b is odd; (3) If φ ∈ MΩ , z MΩ = { w | Q (w) ∈ {γ, γ ν , · · · , γ ν For (1), write w = b ν i=1 (λi ei b i=1 α−1 }}, where z ∈ V , γ = Q (z). b i=1 (λi ei λi ei + µi fi , where λi , µi ∈ F . Then wφ = b ν i=1 (λi µi ) + µi fi )φ = + µν fi ), and Q (wφ ) = i α−1 = ( b i=1 λi µi )ν = Q (w)ν . For (2), let βn = {ζ, ζ 3 , . . . , ζ 3 } be a normal basis of F over F, βe = βn ⊗ {e1 , . . . , eb }, βf = βn ⊗ F. {f1 , . . . , fb }, and We = βe F , Wf = βf We see that We , Wf are two maximal totally singular subspaces in V and φ ﬁxes these two subspaces, so that φ ∈ NI(V ) (We , Wf ). We now determine whether or not φ ∈ Ω(V ). Let βi = βn ⊗ ei = {ζei , . . . , ζ 3 (ζ 3 ei )φ = ζ 3 j j+1 α−1 ei }. Since ei and (ζ 3 α−1 ei )φ = ζ 3 ei = ζei , 107 α  0   0  . (φ )βi =  . .   0  1 1 0 . . . 0 0  0 ··· 0   1 · · · 0  . .. . . . . . .   0 · · · 1  0 ··· 0 Since det(φ )βi = (−1)α−1 , det(φ )We = (−1)(α−1)b . Similarly, det(φ )Wf = (−1)(α−1)b . Thus det(φ )V = ((−1)(α−1)b )2 = 1, so that φ ∈ S. By Lemma 3.11, φ ∈ Ω(V ) if and only if det(φ )We ∈ (F∗ )2 . Hence (2) follows easily. We have MI = I φ Ω MΩ I φ . Assume that φ ∈ MΩ . Then Ω φ MΩ ∼ I .Zα , and = I φ . By Lemma 2.3, z Ω = z I . As z Ω φ ⊆ z MΩ ⊆ z I φ = z Ω φ , the equalities hold through out this chain. Hence z Ω φ = z MΩ = z MI . Finally (3) follows by applying (1). (a) Case α > 3. Let z = e1 + ξf1 , where ξ = ±1. Then Q (z) = ξ ∈ F∗ , Q(z) = T Q (z) = T (ξ) = ±α ∈ F∗ , hence z is a non-singular point in V, and since ξ is invariant under ν, z MΩ = z Ω = { w ∈ V | Q (w) = ξ}. By Lemma 2.11, | z MΩ | = 1 2b−1 (q 2 ⊥ − q b−1 ), where q = 3α . We see that w ∈ z MΩ ∩ zV if and only if w ∈ V , Q (w) = Q (z) = ξ, f (w, z) = ϕ ∈ kerT, where T = TF /F is the trace map. Write ⊥ w = ϕf (z, z)−1 z + z0 , where z0 ∈ zV . Then Q (z0 ) = ξ −1 (ξ 2 − ϕ2 ) = ξ(1 − ϕ2 ), as ξ 2 = 1. Since T (±1) = 0, Q (z0 ) = 0 for any ϕ ∈ kerT. Consider the classical ⊥ ⊥ orthogonal sub-geometry (zV , F , (Q )zV ) of (V , F , Q ) with dimzV = 2(b − 1) + 1. ⊥ ⊥ Let z+ , z− be plus and minus type vectors in (zV , F , (Q )zV ), respectively. Denote ⊥ by n+ , n− the number of ϕ ∈ kerT such that Q (z0 ) = Q (z+ ) and Q (z0 ) = Q (z− ), accordingly. As |kerT| = 3r−1 = 1 q, 3 n+ + n− = 1 q. 3 By Lemma 2.11 again, d = 1 1 1 ⊥ | z MΩ ∩ zV | = 2 n+ (q 2b−2 + q b−1 ) + 2 n− (q 2b−2 − q b−1 ) = 6 q 2b−1 + 1 (n+ + n− ).q b−1 . Thus 2 2d = 1 q 2b−1 + (n+ − n− )q b−1 . Notice that n+ − n− ∈ Z. Suppose that equation (3.8) holds. 3 1 1 Then c − 2d = 3m−1 − 1 = 3 q b − 1, so c = 2d + 1 q b − 1. As 1 + c + d = 2 (q 2b−1 − q b−1 ), 3 108 it follows that 2d = 1 2b−1 q 3 1 2b−1 q 3 1 − 3 (1 + 2.3α−1 )q b−1 . Thus 1 2b−1 q 3 1 − 3 (1 + 2.3α−1 ).q b−1 = 1 + (n+ − n− )q b−1 . This equation yields n+ − n− = − 3 (1 + 2.3α−1 ). However, since α > 3, 1 + 2.3α−1 is not divisible by 3, so the right side of above equation is not an integer while the left side is an integer. This contradiction shows that equation (3.8) cannot hold. Suppose that equation (3.9) holds. Then (3m−1 + 1)(3m − 3 + c − 2d) = 4c or equivalently, 32m−1 − 3 + (3m−1 − 3)c = 2(3m−1 + 1)d. Substitute c = 1 (q 2b−1 − q b−1 ) − d − 1, 2 we have 32m−1 − 3m−1 + 1 (3m−1 − 3).q b−1 .(q b − 1) = (3m − 1)d, as q b = 3αb = 3m , 2 2.3m−1 (3m −1)+(3m−1 −3).q b−1 (3m −1) = (3m −1).2d, hence 2d = 2.3m−1 +q b−1 (3m−1 −3) = 1 2b−1 q 3 + (2.3α−1 − 3).q α−1 . Combining this with 2d = 1 q 2b−1 + (n+ − n− )q b−1 , we have 3 n+ − n− = 2.3α−1 − 3 and n+ + n− = 3α−1 . Solving this system of equations, we get 1 n− = 2 (3 − 3α−1 ). As n− is a non-negative integer, 3 − 3α−1 0, it follows that 3α−1 3 or α 2. This contradicts to our assumption that α > 3. Thus (3.9) cannot hold. (b) Case r = 3. Let ζ be a root of x3 − x + 1 in F. Then ζ = F∗ , T (ζ 2 ) = 2 and kerT = 1, ζ F. Let ξ = ±1 and x1 = e1 +ξζ 2 f1 , x2 = e1 +ξ(ζ 2 +1)f1 , x3 = e1 +ξ(ζ 2 +2)f1 . Then γi = Q (xi ) = ξ(ζ 2 + i − 1) and Q(xi ) = T Q (xi ) = T (ξ(ζ 2 + i − 1)) = 2ξ. Hence xi , 1 i 2 3, are non-singular vectors of V and belong to the same Ω−orbit. Let Qi = 2 {γi , γiν , γiν }. We have γi = ξ(ζ 2 + i − 1), γiν = γi3 = ξ(ζ 2 + ζ + i), γiν = γi9 = ξ(ζ 2 + 2ζ + i). By (3), xi MΩ = { w | Q (w) ∈ Qi }, and so, by Lemma 2.11, 1 + ci + di = | xi MΩ | = 3 2b−1 (q 2 − q b−1 ). To determine di , argue as in case (a), w ∈ xi MΩ ∩ xi ⊥ if and only V if w ∈ V , Q (w) ∈ Qi and f (w, xi ) = ϕ ∈ kerT. Write w = ϕf (xi , xi )−1 xi + w0 , where w0 ∈ xi ⊥ , hence Q (w0 ) = Q (w) − γi−1 ϕ2 . Notice that ζ ≡ −1(mod (F∗ )2 ). V (i) First orbit x1 MΩ . As (x1 ⊥ )V = {e1 − ξζ 2 f1 , e2 , f2 , . . . }, Q (e1 − ξζ 2 f1 ) = −ξζ 2 and e1 − ξζ 2 f1 is a plus vector in (x1 ⊥ )V , it follows that w0 ∈ (x1 ⊥ )V is a plus vector if and only if Q (w0 ) ≡ −ξζ 2 ≡ −ξ(mod (F∗ )2 ). (i1 ) If Q (w) = γ1 = ξζ 2 , then Q (w0 ) = γi−1 (γi2 − ϕ2 ) = ζ 2 − ζ −2 ϕ2 . We need to determine the number of values of ϕ in kerT under which Q (w0 ) is congruence to ξ or 109 −ξ modulo (mod (F∗ )2 ). If ϕ ∈ {0, ±ζ}, then Q (w0 ) ∈ {ξζ 2 , ξζ 12 }. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±1, ±(ζ + 1), ±(ζ − 1)}. Then Q (w0 ) ∈ {ξζ 5 , ξζ 11 , ξζ}, hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). Thus there are 6(q 2b−2 + q b−1 ) + 3(q 2b−2 − q b−1 ) = 9q 2b−2 + 3q b−1 such vectors w0 . −1 4 3 (i2 ) If Q (w) = γ1 = ξ(ζ 2 + ζ + 1), then Q (w0 ) = γ1 (γ1 − ϕ2 ) = ζ 2 + ζ + 1 − ζ −2 ϕ2 . If ϕ ∈ {0, ±ζ, ±(ζ + 1), ±(ζ − 1)}, then Q (w0 ) ∈ {ξζ 6 , ξζ 10 , ξζ 4 , ξζ 16 }. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±1}, then Q (w0 ) ∈ {ξζ 23 }, hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). Thus there are 2(q 2b−1 + q b−1 ) + 7(q 2b−2 − q b−1 ) = 9q 2b−2 − 5q b−1 such vectors w0 . −1 10 9 (i3 ) If Q (w) = γ1 = ξ(ζ 2 + 2ζ + 1), then Q (w0 ) = γ1 (γ1 − ϕ2 ) = ζ 2 − +1 − ζ −2 ϕ2 . If ϕ ∈ {0, ±ζ, ±(ζ + 1), ±(ζ − 1)}, then Q (w0 ) ∈ {ξζ 18 , ξζ 4 , ξζ 2 , ξ}. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±1}, then Q (w0 ) ∈ {ξζ 15 }, hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). Thus, there are 2(q 2b−2 + q b−1 ) + 7(q 2b−2 − q b−1 ) = 9q 2b−2 − 5q b−1 such vectors w0 . Since q = 33 , 3 2d1 = 9q 2b−2 +3q b−1 +2.9q 2b−2 −5q b−1 = q 2b−1 −7q b−1 . Now, as 1+c1 +d1 = 2 (q 2b−1 −q b−1 ), c1 = q 2b−1 + 2q b−1 − 1 and c1 − 2d1 = 9q b−1 − 1 = 33b−1 − 1 = 3m−1 − 1 as m = 3b. Hence equation (3.8) holds. (ii) Second orbit x2 MΩ . As (x2 ⊥ )V = {e1 − ξ(ζ 2 + 1)f1 , e2 , f2 , . . . }, Q (e1 − ξ(ζ 2 + 1)f1 ) = −ξ(ζ 2 + 1) and e1 − ξζ 2 f1 is a plus vector in (x2 ⊥ )V , it follows that w0 ∈ (x1 ⊥ )V is a plus vector if and only if Q (w0 ) ≡ −ξ(ζ 2 + 1) ≡ −ξζ 21 ≡ ξ(mod (F∗ )2 ). −1 2 (ii1 ) Q (w) = γ2 = ξ(ζ 2 + 1). Then Q (w0 ) = γ2 (γ2 − ϕ2 ) = ζ 5 (1 − ζ − ϕ2 ). If ϕ ∈ {0, ±1, ±ζ}, then Q (w0 ) ∈ {ξζ 21 , ξζ 19 , ξζ 3 }. Hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). If ϕ ∈ {±(ζ + 1), ±(ζ − 1)}, then Q (w0 ) ∈ {ξζ 20 , ξζ 22 }, hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). Thus there are 5(q 2b−2 − q b−1 ) + 4(q 2b−2 − q b−1 ) = 9q 2b−2 − q b−1 such vectors w0 . −1 4 3 (ii2 ) If Q (w) = γ2 = ξ(ζ 2 + ζ − 1), then Q (w0 ) = γ1 (γ2 − ϕ2 ) = ζ 5 (ζ 2 + ζ + 1 − ϕ2 ). If ϕ ∈ {0, ±1, ±(ζ + 1)}, then Q (w0 ) ∈ {ξζ 11 , ξζ 15 , ξζ 19 }. Hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). If ϕ ∈ {±ζ}, then Q (w0 ) ∈ {ξζ 14 }. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±(ζ − 1)}, then Q (w0 ) = 0. Thus there are 5(q 2b−2 − q b−1 ) + 2(q 2b−2 + q b−1 ) + 2q 2b−2 = 9q 2b−2 − 3q b−1 110 such vectors w0 . −1 10 9 (ii3 ) Q (w) = γ1 = ξ(ζ 2 + 2ζ + 2). Then Q (w0 ) = γ1 (γ1 − ϕ2 ) = ζ 5 (ζ 2 − ϕ2 ). If ϕ ∈ {0, ±1, ±(ζ − 1)}, then Q (w0 ) ∈= {ξζ 7 , ξζ 17 , ξζ}. Hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). If ϕ ∈ {±(ζ + 1)}, then Q (w0 ) ∈ {ξζ 8 }, hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±ζ}, then Q (w0 ) = 0. Thus, there are 5(q 2b−2 − q b−1 ) + 2(q 2b−2 + q b−1 ) + 2q 2b−2 = 9q 2b−2 − 3q b−1 such vectors w0 . Therefore, 2d2 = 9q 2b−2 − q b−1 + 2.9q 2b−2 − 3q b−1 = q 2b−1 − 7q b−1 . As 1 + c2 + d2 = 3 (q 2b−1 − q b−1 ), c2 = q 2b−1 + 2q b−1 − 1 and c2 − 2d2 = 9q b−1 − 1 = 33b−1 − 1 = 2 3m−1 − 1. Hence equation (3.8) holds. (iii) Third orbit x3 MΩ . As (x3 ⊥ )V = {e1 − ξ(ζ 2 − 1)f1 , e2 , f2 , . . . }, Q (e1 − ξ(ζ 2 − 1)f1 ) = ξζ 25 and e1 − ξ(ζ 2 − 1)f1 is a plus vector in (x3 ⊥ )V , it follows that w0 ∈ (x3 ⊥ )V is a plus vector if and only if Q (w0 ) ≡ ξζ 25 ≡ −ξ(mod (F∗ )2 ). (iii1 ) If Q (w) = γ3 = ξζ 2 , then Q (w0 ) = γi−1 (γi2 −ϕ2 ) = ζ 2 −1+ζϕ2 . If ϕ ∈ {0, ±ζ, ±(ζ + 1), ±(ζ − 1)}, then Q (w0 ) ∈ {ξζ 12 , ξζ 6 , ξζ 16 , ξζ 24 }. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±1}, then Q (w0 ) = ξζ 11 , hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). Thus there are 7(q 2b−2 − q b−1 ) + 2(q 2b−2 + q b−1 ) = 9q 2b−2 − 5q b−1 such vectors w0 . −1 4 3 (iii2 ) If Q (w) = γ3 = ξ(ζ 2 + ζ), then Q (w0 ) = γ3 (γ3 − ϕ2 ) = ζ 2 + ζ + ζϕ2 . If ϕ ∈ {0, ±1, ±(ζ − 1)}, then Q (w0 ) ∈ {ξζ 10 , ξζ 4 , ξζ 8 }. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±ζ, ±(ζ + 1)}, then Q (w0 ) ∈ {ξζ 7 , −ξ}, hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). Thus there are 5(q 2b−1 − q b−1 ) + 4(q 2b−2 + q b−1 ) = 9q 2b−2 − q b−1 such vectors w0 . −1 10 9 (iii3 ) If Q (w) = γ3 = ξ(ζ 2 + 2ζ), then Q (w0 ) = γ1 (γ1 − ϕ2 ) = ζ 2 + ζ − ζϕ2 . If ϕ ∈ {0, ±1, ±ζ}, then Q (w0 ) ∈ {ξζ 4 , ξζ 2 , ξζ 12 , ξ}. Hence Q (w0 ) ≡ ξ(mod (F∗ )2 ). If ϕ ∈ {±(ζ + 1), ±(ζ − 1)}. Then Q (w0 ) ∈ {ξζ 3 , ξζ 5 }, hence Q (w0 ) ≡ −ξ(mod (F∗ )2 ). Thus there are 5(q 2b−2 − q b−1 ) + 4(q 2b−2 + q b−1 ) = 9q 2b−2 − q b−1 such vectors w0 . Hence 3 2d3 = 9q 2b−2 − 5q b−1 + 2.9q 2b−2 − q b−1 = q 2b−1 − 7q b−1 . As 1 + c3 + d3 = 2 (q 2b−1 − q b−1 ), we have c3 = q 2b−1 + 2q b−1 − 1 and c3 − 2d3 = 9q b−1 − 1 = 33b−1 − 1 = 3m−1 − 1 as m = 3b. Therefore equation (3.8) holds. 111 Finally as 3 i=1 1 9 | xi MΩ | = 2 (q 2b−1 − q b−1 ) = 2 (32m−1 − 3m−1 ) = |E+ (V )|, MΩ has ξ exactly there orbits on E+ (V ). ξ (c) Case r = 2. By part (2), if b is even then φ ∈ Ω(V, Q, F), where m = 2b. If b is odd then φ ∈ S \ Ω. Suppose that b is odd. As f (e1 − f1 , e1 − f1 ) = 1 ∈ (F∗ )2 , by Lemma 3.44, re1 −f1 ∈ S \ Ω. Let ψ = re1 −f1 φ . Then ψ ∈ I φ ψ ∈ MΩ . It follows that Ω ψ MΩ I φ ∩ Ω and hence = I ψ . Now, Q (wψ) = Q (wre1 −f1 φ ) = Q (wre1 −f1 )ν = Q (w)ν as re1 −f1 ∈ I . Deﬁne ψ to be ψ if b is odd and φ if b is even. Then for any z ∈ V , since Ω ψ follows that z MΩ = z Ω ψ MΩ I φ = I ψ , and z I = z Ω , it = { w | Q (w) ∈ {Q (z), Q (z)3 }}. Let ζ be a root of x2 − x − 1 in F. Then ζ = F∗ , ζ 4 = −1, (ζ + 1)2 = −1. Let η = ζ + 1, ξ = ±1. Let x1 = e1 − ξf1 , and x2 = e1 + ξζf1 . We have Q (x1 ) = −ξ ∈ F∗ , Q (x2 ) = ξζ, and hence x1 MΩ = x1 Ω and x2 MΩ = x2 Ω ψ = { w ∈ V | Q (w) ∈ {ξζ, (ξζ)3 }}. 1 By Lemma 2.11, | x1 MΩ | = 2 (q 2b−1 − q b−1 ) and | x2 MΩ | = 2 1 (q 2b−1 − q b−1 ). Now, as 2 1 1 | x1 MΩ | + | x2 MΩ | = 2 (3q 2b−1 − 3q b−1 ) = 2 3m−1 (3m − 1) = |E+ (V )|, MΩ has only two orbits on E+ (V ). Proposition 3.46 Assume G is nearly simple primitive rank 3 of type Ω− (3) and M is 2m + of type O2b (3α ), where α is a prime divisor of 2m, and 2m α = 2b 4. There is an M -orbit on E− (V ) such that equation (3.1) does not hold unless α = 3 or α = 2 and G = I. If ξ α = 3 then M has three orbits on E− (V ) and equation (3.1) holds for r = t and for all ξ M -orbits on E− (V ); if α = 2 and G = I then M has two orbits on E− (V ) so that 1G P ξ by Corollary 3.7, and hence these cases are in Table 1.1. Proof. Let β = {v1 , . . . , v2b } be an F basis of V such that fβ is either I2b or diag(µ, 1, . . . , 1) according as D = D(V ) is a square or non-square, where µ is a generators F∗ . Deﬁne    φβ ,Q (ν) if D = φ =  φ  β ,Q (ν)diag(µ, 1, . . . , 1) if D = 112    φ if D = and ψ =  φ2 if D =  1G M ; . We ﬁrst show the following: (1) Q (wφ ) = Q (w)ν for any w ∈ V ; (2) if α is odd then ψ ∈ Ω(V, Q, F) so that MΩ = Ω ψ . If α = 2 then MΩ (3) If α is odd then for any z ∈ V , z MΩ = { w | Q (w) ∈ {γ, γ 3 , . . . , γ 3 γ = Q (z). For any w ∈ V , write w = 2b i=1 α−1 I; }}, where λi vi . We have wφ = λ3 λv1 + 1 2b i=2 λ3 vi , where λ is 1 or µ i 2b i=2 according as D is a square or non-square. Now Q (wφ ) = (λ3 λ)2 Q (v1 )+ 1 −(λ2 λ)3 − 1 2b i=2 λ6 Q (vi ) = i λ6 = (−λ2 λ − 1 i 2b i=2 λ2 )3 = Q (w)ν , which gives (1). For (2), assume i α−1 ﬁrst that r is odd. Let βn = {ζ, ζ 3 , . . . , ζ 3 2 i } be a normal basis of F over F. For 2b, denote by βi the set βn ⊗ vi and β1 = βn ⊗ ηv1 , where η = 1 if D = . Observe that det((φ )βi ) = 1 for any i 2 and and η satisﬁes η 2 λ = −1, when D = det((φ )β1 ) is 1 if D is a square and it is −1 if D is a non-square. Thus det(φ ) = 1 or −1, whenever D is a square or non-square, accordingly. Hence if D = then ψ = φ ∈ S, otherwise, as ψ = φ2 , det(ψ ) = det(φ )2 = 1, we also have ψ ∈ S. Since o(ψ ) = α and [S : Ω] = 2, ψ 2 ∈ Ω. However, as ψ = ψ2 Ω, ψ ∈ Ω. It follows from the proof of Proposition 4.3.16 in [29] that [MΩ : Ω ] = α. Since ψ ∈ Ω and / ψ ∈ I φ ∩ Ω = MΩ , it follows that MΩ = Ω ψ . Secondly assume that α = 2. As 1 sgn(V ) = −, and 2 (32 − 1)b = 4b is even, by Proposition 2.6, D = that η 2 = −1. Let β1 = {ηv1 , ηµv1 } and βi = {vi , ηvi } for i and (φ )βi = diag(1, −1) so that det(φ ) = det((φ )β1 ) 2b i=2 . Take η  F such ∈  0 −1 2. Then (φ )β1 =   1 0 det((φ )βi ) = 1.(−1)2b−1 = −1. Thus det(φ ) = det(φ3 ) = −1. As [MΩ : Ω ] = α = 2, for any g ∈ MΩ \Ω , g ∈ I φ , hence g = g φi , where g ∈ I and 1 i 3 as o(φ ) = 2α = 4. By part (ii) of Lemma 3.44, any S. It follows that reﬂections in V are in S, and since I is generated by reﬂections, I 1 = det(g) = det(g )det(φi ) = (−1)i , which forces i even. Therefore g = g ψ j for some j = 0, 1. For any w ∈ V , by part (1), Q (wψ ) = Q (w)ν = Q (w), since o(ν) = α = 2, 2 113 which means that ψ ∈ I . Thus MΩ I . This proves (2). For (3), assume that α 3. By part (2), MΩ = Ω ψ . Let z ∈ V be any non-zero vector. Then z MΩ = z Ω ψ . Let g = g ψ i ∈ Ω ψ . We have Q (wg) = Q ((zg )ψ i ) = Q (zg )ν = Q (z)ν , by part (1) and the fact that g ∈ I . Now, as o(ν) = α is odd, ν 2 is also a generator for ν = Gal(F /F). This proves (3). Observe that when r is odd then for any non-singular vector z ∈ V , we have z Ω ψ = z MΩ = z MI . Thus it suﬃces to compute the parameters for MΩ in L. When α = 2 then by Proposition 2.6, D(Q) = 1 as 2 (3 − 1)m = 2b is even. It follows 2 2 ¨ from Proposition 2.8.2 in [29] that I = r , also Ω = S and |I : Ω| = 2. Therefore either ¨ G = Ω or G = I. (a) Case α 5. Let z ∈ V be one of the vectors v2 or v2 + v3 so that Q (z) = ξ ∈ F∗ . As Q(z) = T Q (z) = T (ξ) = αξ ∈ F∗ , z is a non-singular vector in V. Now, since Q (z) = ξ is invariant under ν, by part (3) above, z MΩ = { w | Q (w) = Q (z)}. 1 ⊥ By Lemma 2.11, | z MΩ | = 2 (q 2b−1 + q b−1 ), where q = 3α . As w ∈ z MΩ ∩ zV if and only if w ∈ V , Q (w) = Q (z) = ξ, f (w, z) = ϕ ∈ KerT, write w = ϕf (z, z)−1 z + w0 , ⊥ where w0 ∈ zV . Then Q (w0 ) = ξ(1 − ϕ2 ). Since T (±1) = 0, Q (w0 ) = 0 for any ϕ ∈ ⊥ kerT. Consider the classical orthogonal sub-geometry (zV , F , (Q )zV ) of (V , F , Q ) with ⊥ ⊥ ⊥ dimzV = 2(b − 1) + 1. Let z+ , z− be plus and minus type vectors in (zV , (Q )zV , F ), ⊥ respectively. Denote by n+ , n− the number of ϕ ∈ kerT such that Q (w0 ) = Q (z+ ) and q q Q (z0 ) = Q (z− ), accordingly. As |kerT| = 3α−1 = 3 , n+ + n− = . By Lemma 2.11 again, 3 1 1 1 2b−1 ⊥ 2b−2 b−1 2b−2 b−1 d = | z MΩ ∩ zV | = 2 n+ (q + q ) + 2 n− (q − q ) = 6q + 1 (n+ − n− ).q b−1 . 2 1 Thus 2d = 3 q 2b−1 + (n+ − n− )q b−1 . Now, suppose that equation (3.8) holds. Then 2d − c = 3m−1 + 1 = 1+c+d = 1 2b−1 (q 2 1 b q 3 + 1. As + q b−1 ), it follows that 2d = 1 2b−1 q 3 1 + 3 (1 + 2.3α−1 ).q b−1 . Thus 1 n+ − n− = 3 (1 + 2.3α−1 ). However since α > 1, 1 + 2.3α−1 is not divisible by 3, so that the right side of above equation is not an integer while the left side is an integer. This contradiction shows that equation (3.8) cannot hold. 114 Next suppose that equation (3.9) holds. Then (3m−1 − 1)(3m + 3 − c + 2d) = 4c or 1 equivalently, 32m−1 −3+2d(3m−1 −1)c = 2(3m−1 +3)c. Substitute c = 2 (q 2b−1 +q b−1 )−d−1, we have 32m−1 + 3m−1 + (3m + 1)d = 1 m−1 (3 2 + 3).q b−1 .(q b + 1), as q b = 3αb = 3m , 2.3m−1 (3m +1)+(3m +1).d = (3m−1 +3).q b−1 (3m +1), hence 2d = q b−1 (3m−1 +3)−2.3m−1 = 1 2b−1 q 3 1 + (3 − 2.3r−1 ).q b−1 . Combining this with 2d = 3 q 2b−1 + (n+ − n− )q b−1 , we have n+ − n− = 3 − 2.3α−1 and n+ + n− = 3α−1 . Solving this system of equations, we get 2n+ = 3 − 3α−1 . As n+ is non-negative, 3 − 3α−1 This contradicts to our assumption that α 0, it follows that 3α−1 3 or α 2. 5. Thus (3.9) cannot hold. (b) Case α = 3. Let ζ be a root of x3 − x + 1 in F. Then ζ = F∗ , and ζ 3 = ζ − 1. Let z be a non-singular vector in V , and also a non-singular vectors in V. By part (3), z MΩ = { w | Q (w) ∈ {Q (z), Q (z)3 , Q (z)9 }}. By Lemma 2.11, 1 + cz + dz = ⊥ | z MΩ | = 3 (q 2b−1 + q b−1 ). To determine dz , argue as in case (a), w ∈ z MΩ ∩ zV if 2 and only if w ∈ V , Q (w) ∈ {Q (z), Q (z)3 , Q (z)9 }} and f (w, xi ) = ϕ ∈ kerT. Write ⊥ w = ϕf (z, z)−1 z + w0 , where w0 ∈ zV , hence Q (w0 ) = Q (w) − Q (z)−1 ϕ2 . If equation (3.8) holds, then 2dz − cz = 3m−1 + 1, so that 2dz = q 2b−1 + 7q b−1 . If equation (3.9) holds then (3m−1 − 1)(3m + 3 − cz + 2dz ) = 4cz , and hence 2dz = q 2b−1 − 9q b−1 . Notice that ζ ≡ −1(mod (F∗ )2 ). As sgnV = −, by Proposition 2.6, D = 1 3 (3 2 if and only if − 1)b = 13b is odd. Therefore D is a square if and only if b is odd. We will show that 2dz = q 2b−1 + 7q b−1 , so that equation (3.8) holds. Deﬁne z− = v2 , z+ = v2 + v3 and for any ξ = ±,    z1 = ζzξ    z = ζ 4 z−ξ  2     z3 = ζ 6 zξ . As Q (zξ ) = ξ, we have Q (zi ) = ξ(ζ 2 + i − 1) for i = 1, · · · , 3. Let γi = Q (zi ) and Qi = {γi , γi3 = ξ(ζ 2 + ζ + i), γi9 = ξ(ζ 2 − ζ + i)}. Observe that −1 and λ, where λ is any 115 generator for F∗ , are non-square in F∗ . We can check that D v1 , zξ sgn v1 , zξ ⊥ ⊥ ≡ −ξ1 (mod(F∗ )2 ), ⊥ = ξ(−)b and Q (v1 ) ≡ (−1)b−1 (mod(F∗ )2 ). It follows that for w0 ∈ (zξ )V , w0 is a ξ(−)b vector if and only if Q (w0 ) ≡ (−1)b (mod(F∗ )2 ), or equivalently, w0 is a + ⊥ vector if and only if Q (w0 ) ≡ ξ (mod(F∗ )2 ), where w0 ∈ (zξ )V . −1 3 (i) First orbit z1 MΩ . We have Q (w0 ) = γ1 − γ1 ϕ2 , where j = 0, 1, 2, and ϕ ∈ j kerT = {0, ±1, ±ζ, ±(ζ + 1), ±(ζ − 1)}. If j = 0 then Q (w0 ) ≡ ξ(ζ 4 − ϕ2 ) (mod(F∗ )2 ), and there are 6 values of ϕ in kerT such that ζ 4 − ϕ2 ≡ 1 (mod(F∗ )2 ) and three values of ϕ ∈ kerT such that ζ 4 − ϕ2 ≡ −1 (mod(F∗ )2 ). Now if ζ 4 − ϕ2 ≡ 1 (mod(F∗ )2 ), then Q (w0 ) ≡ ξ (mod(F∗ )2 ), and if ζ 4 − ϕ2 ≡ −1 (mod(F∗ )2 ), then Q (w0 ) ≡ −ξ (mod(F∗ )2 ). ⊥ ⊥ Since z1 = zξ , there are 6(q 2b−2 + q b−1 ) + 3(q 2b−2 − q b−1 ) = 9q 2b−2 − 3q b−1 such vectors w0 in this case by Lemma 2.11 and argument above. If j = 1, then Q (w0 ) ≡ ξ(ζ 8 − ϕ2 ) (mod(F∗ )2 ). There are 7 values of ϕ ∈ kerT for which Q (w0 ) ≡ ξ(mod(F∗ )2 ) and 2 such values of ϕ with Q (w0 ) ≡ −ξ(mod(F∗ )2 ), hence there are 9q 2b−1 + 5q b−1 . If j = 2, then Q (w0 ) ≡ ξ(ζ 20 − ϕ2 ) (mod(F∗ )2 ). There are 7 values of ϕ ∈ kerT for which Q (w0 ) ≡ ξ(mod(F∗ )2 ) and 2 such values of ϕ with Q (w0 ) ≡ −ξ(mod(F∗ )2 ), hence there are 9q 2b−1 + 5q b−1 . Therefore, 2d = 27q 2b−2 + 7q b−1 = q 2b−1 + 7q b−1 . −1 3 ⊥ ⊥ (ii) Second orbit z2 MΩ . We have Q (w0 ) = γ2 − γ2 ϕ2 , and z2 = z−ξ . It follows ⊥ that w0 ∈ (z2 )V is a plus vector if and only if Q (w0 ) ≡ −ξ (mod(F∗ )2 ). Firstly, if j j = 0 then Q (w0 ) ≡ −ξ(ζ 16 − ϕ2 ) (mod(F∗ )2 ). There are 5 values of ϕ in kerT such that Q (w0 ) ≡ −ξ (mod(F∗ )2 ); and 4 values of such ϕ such that Q (w0 ) ≡ ξ (mod(F∗ )2 ). Thus there are 5(q 2b−2 + q b−1 ) + 4(q 2b−2 − q b−1 ) = 9q 2b−2 + q b−1 such vectors w0 in this case. Secondly if j = 1 then Q (w0 ) ≡ −ξ(ζ 6 − ϕ2 ) (mod(F∗ )2 ). There are 5 values of ϕ in kerT such that Q (w0 ) ≡ −ξ (mod(F∗ )2 ); 2 values of such ϕ such that Q (w0 ) ≡ ξ (mod(F∗ )2 ); and 2 values of ϕ ∈ kerT such that Q (w0 ) = 0. Thus there are 5(q 2b−2 + q b−1 ) + 2(q 2b−2 − q b−1 ) + 2q 2b−2 = 9q 2b−2 + 3q b−1 such vectors w0 in this case. Finally if j = 2, then Q (w0 ) ≡ −ξ(ζ 2 − ϕ2 ) (mod(F∗ )2 ). There are 5 values of ϕ in kerT such that Q (w0 ) ≡ 116 −ξ (mod(F∗ )2 ); 2 values of such ϕ such that Q (w0 ) ≡ ξ (mod(F∗ )2 ); and 2 values of ϕ ∈ kerT such that Q (w0 ) = 0. Thus there are 5(q 2b−2 + q b−1 ) + 2(q 2b−2 − q b−1 ) + 2q 2b−2 = 9q 2b−2 +3q b−1 such vectors w0 in this case. Therefore 2d = 9q 2b−2 +q b−1 +2(9q 2b−2 +3q b−1 ) = q 2b−1 + 7q b−1 . −1 3 ⊥ ⊥ (iii) Third orbit z3 MΩ . We have Q (w0 ) = γ3 − γ3 ϕ2 , and z3 = zξ . It follows that ⊥ w0 ∈ (z2 )V is a plus vector if and only if Q (w0 ) ≡ ξ (mod(F∗ )2 ). Firstly, if j = 0 then j Q (w0 ) ≡ ξ(ζ 24 − ϕ2 ) (mod(F∗ )2 ). There are 7 values of ϕ in kerT such that Q (w0 ) ≡ ξ (mod(F∗ )2 ); and 2 values of such ϕ such that Q (w0 ) ≡ −ξ (mod(F∗ )2 ). Thus there are 7(q 2a−2 + q a−1 ) + 2(q 2a−2 − q a−1 ) = 9q 2a−2 + 5q a−1 such vectors w0 in this case. Secondly, if j = 1 then Q (w0 ) ≡ ξ(ζ 22 − ϕ2 ) (mod(F∗ )2 ). There are 5 values of ϕ in kerT such that Q (w0 ) ≡ ξ (mod(F∗ )2 ); 4 values of such ϕ such that Q (w0 ) ≡ −ξ (mod(F∗ )2 ). Thus there are 5(q 2b−2 + q b−1 ) + 4(q 2b−2 − q b−1 ) = 9q 2b−2 + q b−1 such vectors w0 in this case. Finally if j = 2, then Q (w0 ) ≡ −ξ(ζ 16 − ϕ2 ) (mod(F∗ )2 ). There are 5 values of ϕ in kerT such that Q (w0 ) ≡ ξ (mod(F∗ )2 ); 4 values of such ϕ such that Q (w0 ) ≡ −ξ (mod(F∗ )2 ). Thus there are 5(q 2b−2 + q b−1 ) + 4(q 2b−2 − q b−1 ) = 9q 2b−2 + q b−1 such vectors w0 in this case. Therefore 2d = 9q 2b−2 + 5q b−1 + 2(9q 2b−2 + q b−1 ) = q 2b−1 + 7q b−1 . (c) Case r = 2 Assume ﬁrst that G = Ω. Then Ω have MΩ M G S and as in proof of (2), we I F∗ . Therefore z MΩ = z M = z Ω , for any non-singular point z in , and so V has a basis β with fβ = diag(ζ, 1, . . . , 1), where ζ 2 = ζ + 1. V . Since D = Let z1 = v2 + (ζ − 1)v3 and z2 = ηz1 , where η ∈ F∗ , η 2 = −1. Then Q (zi ) = (−1)i−1 ζ, and Q(zi ) = T Q (zi ) = T ((−1)i−1 ζ) = (−1)i = 0. Thus zi are non-singular vectors in both V and V. As v1 , zi sgn v1 , zi ⊥ ⊥ V 1 = (ζ − 1)v2 − v3 , v4 , . . . , v2a , D v1 , zi ⊥ = −ζ ≡ ζ (mod(F∗ )2 ), hence = (−) 2 (9−1)a−1 = (−)4a−1 = − and Q (v1 ) = −ζ ≡ ζ (mod(F∗ )2 ). It follows that for any w0 ∈ (zi⊥ )V , w0 is a − vector if and only if Q (w0 ) ≡ ζ (mod(F∗ )2 ), or equivalently, w0 is a + vector if and only if Q (w0 ) ∈ (F∗ )2 , where w0 ∈ (zi⊥ )V . By Lemma 2.11, 1 + ci + di = | zi MΩ | = 1 2a−1 (q 2 + q a−1 ). For any w ∈ zi MΩ ∩ zi⊥ , write w = 117 ϕf (zi , zi )−1 zi + w0 , where w0 ∈ zi ⊥ and ϕ ∈ kerT. Then f (w, zi ) = ϕ and Q (w0 ) = V Q (w) − Q (zi )−1 ϕ2 = (−1)i−1 ζ −1 (ζ 2 − ϕ2 ), as Q (w) = Q (zi ) = (−1)i−1 ζ. As kerT = {0, ±η}, if ϕ = 0 then Q (w0 ) = (−1)i−1 ζ ≡ ζ (mod(F∗ )2 ), and if ϕ = ±η, then Q (w0 ) = (−1)i−1 ζ −1 ( 2 −1) = (−1)i−1 ζ −1 ζ = (−1)i−1 ∈ (F∗ )2 . Therefore 2di = 2(q 2a−2 + q a−1 )+q 2a−2 −q a−1 = 3q 2a−2 +q a−1 , and so ci = 3q 2a−2 −1 = 32m−1 −1, as m = 2a, q = 32 . It follows that 2di −ci = q a−1 +1 = 3m−2 +1. Clearly (3.8) cannot hold as 3m−2 +1 = 3m−1 +1. If (3.9) holds then (3m−1 −1)(3m +3+2d−c) = 4ci . However, (3m−1 −1)(3m +3+2d−c) = (3m−1 − 1)(3m + 3 + 3m−2 + 1) = (3m−1 − 1)(3m + 3m−2 + 4) = 4(32m−1 − 1). Thus, (3.1) cannot hold in this case. Finally, assume that G = I. Then φ ∈ G and hence Ω φ Thus G has only two orbits on E− (V ), and so (3.1) holds. Proposition 3.47 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m of type O2a+1 (32 ), where m = 2a + 1. There is an M -orbit on Eε (V ) such that equation ξ (3.1) does not hold unless D(V ) = so that ε = −, or D = and G = I, so that ε = +. HG . In these cases M has two orbits on Eε (V ) so that 1G P ξ M is in Table 1.1. 1G by Corollary 3.7, and hence M Proof. Let β = {v1 , . . . , v2a+1 } be a basis of V over F with fβ = λI2a+1 , where λ = 1 if D = D(V ) = and λ = F∗ if D = . Deﬁne φ as in Proposition 3.20. Then Q (wφ ) = Q (w)ν . Let Wi = spanF vi . Then Wi are non-degenerate 2-spaces in V with D(Wi ) = −N (λ), by Lemma 3.44(i) . Also V = W1 ⊥ W2 ⊥ · · · ⊥ W2a+1 is a nondegenerate 2-space decomposition of V. Let ζ be a root of x2 − x − 1. Then ζ 2 = ζ + 1. (a) Case D = . By Proposition 2.6, we have ε = (−)2a = +. Also by Propositions ¨ ¨ 2.7.1 and 2.7.3 in [29], S = Ω and I = r . It follows that either G = Ω or G = I. Assume ﬁrst that G = S, so that Ω G S. It follows from Proposition 2.7 that D(Wi ) = 32 −1 for each i. Thus −N (λ) ∈ (F∗ )2 , and hence N (λ) ∈ (F∗ )2 . Therefore, λ 3−1 = λ4 ∈ (F∗ )2 , / so that λ = 1. Recall that (ξvi )φ = ξ 3 λvi . Let η ∈ F∗ be such that η 2 = −1. Let βi = {vi , ηvi }. Then (φ )βi = diag(1, −1), and hence det(φ ) = −1. According to Lemma 118 3.44(ii), rv ∈ S for any v ∈ V with f (v , v ) = 0, and hence I MΩ M S. Consequently, I . Let z1 = v1 + (ζ − 1)v2 , z2 = ηz1 . Since Q (zi ) = (−1)i−1 ζ and Q(zi ) = T Q (zi ) = T ((−1)i−1 ζ) = (−1)i−1 = 0, zi are non-singular vectors in V. As (zi⊥ )V = (ζ −1)v1 −v2 , v3 , . . . , v2a+1 , D((zi⊥ )V ) = det(diag(−ζ, 1, . . . , 1)) = −ζ ≡ and so by Proposition 2.6(iii), MΩ sgn(zi⊥ )V = (−1) 2a−1 a(9−1)/2−1 ( mod (F∗ )2 ), = (−)4a−1 = −. Now, as I , and dimF V = 2a + 1, zi MΩ = zi Ω , and so as zi are minus points, by Lemma 2.11, 1 + c + d = | zi Ω | = 1 (q 2a − q a ), where q = 32 . To compute parameter 2 d, for any w ∈ zi MΩ ∩ (zi⊥ )V , write w = ϕf (zi , zi )−1 zi + w0 , where w0 ∈ (zi⊥ )V , and ϕ ∈ kerT, hence Q (w0 ) = (−1)i−1 ζ −1 (ζ 2 − ϕ2 ). As T (±ζ) = 0, Q (w0 ) = 0 for any ϕ ∈ kerT. Apply Lemma 2.11 together with the fact that w0 ∈ (zi⊥ )V , sgn(zi⊥ )V = −, dim(zi⊥ )V = 2a, and |kerT | = 3, we have 2d = 3(q 2a−1 + q a−1 ). Hence c = 3q 2a−1 − 6q a−1 − 1 = (q a + 1)(3q a−1 − 1) and c − 2d = −q a − 1 = −3m−1 − 1. As D = , according to Proposition 2.6, ε = sgnV = (−)m(3−1)/2−1 = (−)m−1 = +. Clearly, equation (3.8) cannot hold as c − 2d = −3m−1 − 1 = 3m−1 − 1. Assume that equation (3.9) holds. Then (3m−1 + 1)(3m − 3 + c − 2d) = 4c. As 3m−1 = 32a = q a 3m = 32a+1 = 3q a , and c − 2d = −q a − 1, this equation becomes: 2(q a + 1)(q a − 2) = 4(q a + 1)(3q a−1 − 1). Dividing both sides by 2(q a + 1) yields, q a − 2 = 6q a−1 − 2, or equivalently q = 6, a contradiction. Therefore, equation (3.1) cannot hold. Assume that G = I. Then φ ∈ G, and hence G has only two orbits on E+ (V ), so that equation (3.1) holds. . Arguing as previous case  have D(Wi ) = we for all i, and so   0 1 λ = F∗ . Take βi = {vi , λvi }. Then (φ )βi =   , and hence det(φ ) = 1, so −1 0 φ ∈ S. Let v ∈ V be such that f (v , v ) = 1. By Lemma 3.44(ii), rv ∈ S \ Ω. Deﬁne    φ ψ =   rv φ (b) Case D = if φ ∈ Ω otherwise. 119 Then ψ ∈ Ω∩I φ = MΩ , and Q (wψ) = Q (w)ν , so that MΩ = Ω ψ . For i = 1, 2, put xi = η i (v1 + v2 ), yi = η i (v1 + λv2 ). Then Q (xi ) = (−1)i λ, Q (yi ) = (−1)i+1 and Q(xi ) = T ((−1)i λ) = (−1)i , Q(yi ) = T ((−1)i+1 ) = (−1)i , as T (λ) = 1, T (1) = −1, so that for ﬁxed i = 1, 2, {xi , yi } are non-singular vectors in V which belong to the same Ω-orbit. Ob⊥ serve that (x⊥ )V = v1 − v2 , v3 , . . . , v2a+1 , (yi )V = λv1 − v2 , v3 , . . . , v2a+1 , D((x⊥ )V ) = i i det(diag(λ, −λ, . . . , −λ)) = −λ2a = 2a−1 ⊥ , D((yi )V ) = det(diag(1, −λ, . . . , −λ)) = −λ2a−1 = 2a−1 ⊥ , and so sgn((x⊥ )V ) = (−)4a = +, sgn((yi )V ) = (−)4a−1 = −. Since Q (yi ) = (−1)i+1 i is invariant under ν, where ν = Gal(F /F), yi MΩ = yi Ω , and so by Lemma 2.11, | yi MΩ | = 1 (q 2a − q a ) as yi is a minus point in V . Next, as xi MΩ = xi Ω ψ 2 = { w ∈ V | Q (w) ∈ {Q (xi ), Q (xi )3 }} and xi is a plus point in V , | xi MΩ | = 2 1 (q 2a + 2 q a ) = q 2a + q a . As D = D(V ) = 1 , sgnV = (−)m = −, hence |E− (V )| = 2 3m−1 (3m + 1), 1 by Lemma 3.35. Now, since | xi MΩ | + | yi MΩ | = q 2a + q a + 2 (q 2a − q a ) = 1 q a (3q a + 1) = 2 1 m−1 m 3 (3 2 + 1) = |E− (V )|, it follows that MΩ has exactly two orbits on E− (V ). Proposition 3.48 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m of type GUm (3). Then M has only one orbit on Eε (V ) so that 1G P ξ and hence M is in Table 1.1. Proof. M has only one orbit on Eε (V ) by (3.5.2b) and (3.6.1c) in [34]. The tensor product subgroups C4 Proposition 3.49 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M 2m ε ε is of type On1 (3) ⊗ On2 (3), with (n1 , ε1 ) = (n2 ε2 ) and ni = 2mi 1 2 1G by Corollary 3.7, M 4, i = 1, 2. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold so that M is not in Tables ξ 1.1-1.3. Proof. Write Xi for X(Vi , fi ), where X ranges over the symbols I, S and Ω. Let vi ∈ Vi be non-singular vectors taken in some orthogonal bases of Vi , and x = v1 ⊗ v2 . Then x 120 is non-singular V. By (4.4.15) and Lemma 4.4.13(iii) in [29], S1 ⊗ S2 MΩ M MI = (I1 ⊗I2 ) z , where z 2 ∈ I1 ⊗I2 . Let Ni be the stabilizer in Si of vi . Then Ni ∼ SOni −1 (3) and = N1 ⊗N2 MΩ x . As |Ii : Ni | = 2.3mi −1 (3mi −εi ), A = | x MI | [(I1 ⊗I2 ) z : N1 ⊗N2 ] 2.3m1 +m2 , 8.3m1 +m2 −2 (3m1 −ε1 )(3m2 −ε2 ). Since (3m1 −ε1 )(3m2 −ε2 ) we deduce that A (3m1 +1)(3m2 +1) 16.32m1 +2m2 −2 < 32m1 +2m2 +1 . Now as dimV = 2m = 2m1 .2m2 , 32m1 +2m2 +1 . This is 2. As ni 4 for all i, m = 2m1 .m2 . By inequality (3.10), it suﬃces to show that 3m−1 equivalent to 2m1 m2 − 1 we have mi 2m1 + 2m2 + 1, or (m1 − 1)(m2 − 1) 2 for i = 1, 2. If mi 3 for some i, then clearly this inequality is correct. Thus we only need to consider the case when m1 = m2 = 2 and ε1 = ε2 . This implies that 1 n = 16. Without loss, assume that ε1 = +, so that ε2 = −. As 2 (3 − 1)m2 = 2 is even, by Proposition 2.6, D(V2 ) = , and hence by Proposition 6.3.4 in [29], MΩ is not maximal in L. In fact, MΩ < Sp2 (3) ⊗ Sp8 (3). Proposition 3.50 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m ε of type On1 (3) ⊗ On2 (3), with n2 = 2m2 + 1 odd and n1 = 2m1 4. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. ξ Proof. If n2 = 3 then MΩ is not maximal in L by Proposition 6.3.2[29]. Thus we can assume that n2 MΩ = Ω1 ⊗S2 5 or m2 2. As n1 = 2m2 4, m1 2. By Proposition 4.4.17[29], MI = I1 ⊗S2 . Let vi ∈ Vi be non-singular vectors and x = v1 ⊗v2 . Then x is a non-singular point in V and x MI = x MΩ = v1 Ω1 ⊗ v2 S2 = v1 Ω1 ⊗ v2 Ω2 . By Lemma 2.11, A = | x MI | |v1 Ω1 |.|v2 Ω2 | = 3m1 −1 (3m1 − ε).3m2 (3m2 + ξ), where (3m1 + 1).(3m2 + 1) 2.3m1 +m2 < 3m1 +m2 +1 , ξ = ρV2 (v2 ). As (3m1 − ε).(3m2 + ξ) we have A < 32m1 +2m2 . In view of inequalities (3.10) and (3.11), we need to show that 3m−2 32m1 +2m2 −1 , or equivalently m−2 1 2m1 +2m2 , where m = 2m1 m2 +m1 = 2 dimV. The last inequality is equivalent to (m1 −1)(2m2 −1) 3. This is true as mi 2, i = 1, 2. 121 Let (U, Fq , f ) be a symplectic geometry. Then f is skew-symmetric and bilinear on U. An ordered pair (u1 , u2 ) in U is called a hyperbolic pair if f (u1 , u2 ) = 1. A subspace of U generated by a hyperbolic pair (u1 , u2 ) is called a hyperbolic plane, with basis {u1 , u2 }. By Proposition 2.4.1 in [29], U is of even dimension, say 2 , and has a basis β = {e1 , f1 , . . . , e , f } such that f (ei , fi ) = δij and f (ei , ej ) = f (fi , fj ) = 0, such a basis is called a standard basis or symplectic basis for U. Assume that q is odd, and λ ∈ F∗ . Deﬁne δf ,β (λ) by ei δf ,β = λei and fi δf ,β = fi . (3.13) Denote by H(U ) the set of all distinct hyperbolic pairs in U, and deﬁne HP(U ) to be the set of all distinct hyperbolic planes in U. The following lemma will give us the cardinalities of these sets. Lemma 3.51 Let (U, F, f ) be a symplectic geometry and β = {e1 , f1 , . . . , e , f } be a standard basis for U. Let λ ∈ F∗ . Then (1) |H(U )| = q 2 −1 (q 2 − 1); q 2 −2 (q 2 − 1) ; (2) |HP(U )| = q2 − 1 (3) |{(u1 , u2 ) ∈ H(U ) | f (u1 , e1 ) = λ}| = q 4 −2 ; (4) |{ u1 , u2 ∈ HP(U ) | f (u1 , e1 ) = λ}| = q 4 −4 ; Proof. As dimU = 2 , there are q 2 − 1 non-zero vector u1 in U. Since dim(u⊥ ) = 2 − 1, 1 there are q 2 − q 2 −1 vectors u2 with f (u1 , u2 ) = 0. As |F∗ | = q − 1, there are 1 (q 2 q−1 − 1)(q 2 −q 2 −1 ) = q 2 −1 (q 2 −1) pairs (u1 , u2 ) with f (u1 , u2 ) = 1, which gives (1). By applying (1), in each hyperbolic plane u1 , u2 , there are q(q 2 − 1) hyperbolic pairs. This proves (2). For (3), assume that (u1 , u2 ) is a hyperbolic pair in U with f (u1 , e1 ) = λ = 0. We can write u1 = −λf1 + u0 , where u0 ∈ e⊥ . Since dim(e⊥ ) = 2 − 1, there are q 2 −1 choices 1 1 for u0 , hence there are the same number of choices for u1 . When u1 is chosen, clearly, there are q 2 −q 2 −1 q−1 = q 2 −1 possibilities for u2 such that f (u1 , u2 ) = 1. Hence (3) follows 122 by multiplying these two values. To prove (4), it suﬃces to show that if W = u1 , u2 is a hyperbolic plane with f (u1 , e1 ) = λ, then there are only q 2 hyperbolic pairs (v1 , v2 ) in W satisfying f (v1 , e1 ) = λ. Write v1 = ζ1 u1 + β1 u2 and v2 = ζ2 u1 + β2 u2 . It follows from f (v1 , v2 ) = 1 that ζ1 β2 − ζ2 β1 = 1. (3.14) Also, as f (u1 , e1 ) = λ, and v1 = ζ1 u1 + β1 u2 , we have f (v1 , e1 ) = ζ1 f (u1 , e1 ) + β1 f (u2 , e1 ) = λζ1 + β1 µ, where µ = f (u2 , e1 ). Now, since f (v1 , e1 ) = λ, λζ1 + β1 µ = λ. (3.15) If µ = 0, then ζ1 = 1 and β2 = 1 + ζ2 β1 . As ζ2 , β1 can be chosen arbitrarily in F, there are q 2 quadruples (ζ1 , β1 , ζ2 , β2 ) satisfying (3.14) and (3.15). Therefore there are q 2 such hyperbolic pairs (v1 , v2 ). Assume that µ = 0. If ζ1 = 0, then ζ2 β1 = −1, β1 = λµ−1 . Hence   ζ  1       β2  ζ  2       β 1 = 0 ∈ F = −µλ −1 (3.16) = λµ−1 Thus there q solutions to (3.16). If ζ1 = 0, then β1 = λµ−1 , ζ1 = 1 − (λ−1 µ)β1 and −1 β2 = ζ1 (1 + ζ2 β1 ). Thus equations (3.14), (3.15) become   β  1       ζ2  β  2       ζ 1 ∈ F \ {λµ−1 } ∈ F = (1 − λ−1 µβ1 )−1 (1 + ζ2 β1 ) = 1 − µλ−1 β1 . (3.17) Since β1 = λµ−1 and β2 ∈ F, there are (q − 1)q solutions to (3.17). Therefore, there are 123 q + q(q − 1) = q 2 such hyperbolic pairs (v1 , v2 ). The proof is now completed. Let (Vi , Fq , fi ), i = 1, 2 be classical symplectic geometries with dimVi = 2mi . Assume (V, f ) = (V1 , f1 ) ⊗ (V2 , f2 ) be a tensor decomposition of V with m1 < m2 . Let β1 = {a1 , b1 , . . . , am1 , bm1 }, β2 = {e1 , f1 , . . . , em2 , fm2 } be standard bases for (V1 , f1 ), (V2 , f2 ), respectively, and Ii = I(Vi , fi ). Then β = β1 ⊗ β2 is a basis for V = V1 ⊗ V2 . Let x = a1 ⊗ e1 + ξb1 ⊗ f1 ∈ V, with ξ = ±1. Lemma 3.52 Assume the notation above. Then (1) | x (I1 ⊗ I2 )| = 1 q 2m1 +2m2 −3 (q 2m1 (q−1)(q 2 −1) − 1)(q 2m2 − 1); (2) |{ v ∈ x (I1 ⊗ I2 ) | (v, x) = 0}| = q 4m1 +4m2 −6 . Proof. (1) We ﬁrst determine (I1 ⊗ I2 )x . For any g ∈ (I1 ⊗ I2 )x , there exist gi ∈ Ii such that g = g1 ⊗ g2 and xg = x. Let w1 = e1 g2 , w2 = f1 g2 , and    a1 g1 = ζ1 a1 + β1 b1 + u1  b g = ζ a +β b +u ,  1 1 2 1 2 1 2 where ui ∈ U1 = a1 , b1 ⊥ and ζi , βi ∈ F for i = 1, 2. Since gi ∈ Ii , we have f2 (w1 , w2 ) = f2 (e1 g2 , f1 g2 ) = f2 (e1 , f1 ) = 1, and similarly f1 (a1 g1 , b1 g1 ) = f1 (a1 , b1 ) = 1. Now as f1 (a1 g1 , b1 g1 ) = ζ1 β2 − ζ2 β1 + f1 (u1 , u2 ), it follows that ζ1 β2 − ζ2 β1 + f1 (u1 , u2 ) = 1 (3.18) We have xg = a1 g1 ⊗e1 g2 +ξb1 g1 ⊗f1 g2 = (ζ1 a1 +β1 b1 +u1 )⊗w1 +ξ(ζ2 a1 +β2 b1 +u2 )⊗w2 , or xg = a1 ⊗ (ζ1 w1 + ξζ2 w2 ) + b1 ⊗ (β1 w1 + ξβ2 w2 ) + u1 ⊗ w1 + u2 ⊗ (ξw2 ). We deduce from xg = x that a1 ⊗(ζ1 w1 +ξζ2 w2 )+b1 ⊗(β1 w1 +ξβ2 w2 )+u1 ⊗w1 +ξu2 ⊗w2 = a1 ⊗e1 +ξb1 ⊗f1 , and so a1 ⊗(ζ1 w1 +ξζ2 w2 −e1 )+b1 ⊗(β1 w1 +ξβ2 w2 −f1 )+u1 ⊗w1 +ξu2 ⊗w2 = 0. As {a1 , b1 } is linearly independent and ui ∈ a1 , b1 ⊥ , ζ1 w1 + ξζ2 w2 − e1 = 0 = β1 w1 + ξβ2 w2 − f1 , 124 and u1 ⊗ w1 + ξu2 ⊗ w2 = 0. Furthermore {w1 , w2 } is also linearly independent, hence u1 = 0 = u2 . In summary, we have   ζ w + ξζ w  1 1 2 2       β1 w1 + ξβ2 w2   u1       u 2 = e1 = f2 = 0 = 0. (3.19) As u1 = u2 = 0, it follows from (3.18) that ζ1 β2 − ζ2 β1 = 1. Using this together with the ﬁrst two equations in (3.19), we get    w1 = e1 g2 = β2 e1 − ξζ2 f1  w = f g = −β e + ζ f .  2 1 2 1 1 1 1 We conclude that g1 ∈ Sp( a1 , b1 ) × Sp(U1 ), g2 ∈ Sp( e1 , f1 ) × Sp(U2 ), where U1 = a1 , b1 ⊥ , U2 = e1 , f1 ⊥ . Let B1 = {a1 , b1 }, and B2 = {e1 , f1 }. Then for any ζi , βi ∈ F, i = 1, 2, with ζ1 β2 − ζ2 β1 = 1, set     (3.20) −ξζ2  ζ1 β1   β2 A1 =   and A2 =  , ζ2 β2 −ξβ1 ζ1 and deﬁne T to be a subgroup of I1 ⊗I2 consisting of all elements g1 ⊗g2 , where (gi )Bi = Ai as above and gi act trivially on Ui . Then T ∼ Sp2 (3) ∼ SL2 (3). Hence = = (I1 ⊗ I2 )x = (Sp(U1 ) ⊗ Sp(U2 )).T, where Sp(Ui ) ∼ Sp2mi −2 (3). Therefore = 1 q 2m1 +2m2 −3 (q 2m1 − 1)(q 2m2 − 1), −1 |x(I1 ⊗ I2 )| = |(I1 ⊗ I2 ) : (I1 ⊗ I2 )x | = q2 125 and hence (1) follows. (2) For g = g1 ⊗ g2 ∈ I1 ⊗ I2 , write u1 = a1 g1 , u2 = b1 g1 , w1 = e1 g2 , w2 = f1 g2 , then xg = u1 ⊗ w1 + ξu2 ⊗ w2 and (u1 , u2 ), (w1 , w2 ) are hyperbolic pairs in V1 , V2 respectively. As Sp(Vi ) act transitively on the set of all hyperbolic pairs H(Vi ) by Proposition 3.3 in [14], it follows that x(I1 ⊗ I2 ) consists of all elements of the form u1 ⊗ w1 + ξu2 ⊗ w2 , where (u1 , u2 ) ∈ H(V1 ) and (w1 , w2 ) ∈ H(V2 ). However when a couple of hyperbolic pairs ((u1 , u2 ), (w1 , w2 )) ∈ H(V1 ) × H(V2 ) are ﬁxed, there are q(q 2 − 1) pairs of hyperbolic pairs ((u1 A1 , u2 A1 ), (w1 A2 , w2 A2 )) ∈ H(V1 ) × H(V2 ) which give rise to the same element u1 ⊗ w1 + ξu2 ⊗ w2 , where Ai , i = 1, 2 are as in (3.20). Thus x(I1 ⊗ I2 ) = {u1 ⊗ w1 + ξu2 ⊗ w2 | (u1 , u2 ) ∈ HP ∗ (V1 ), (w1 , w2 ) ∈ H(V2 )} (3.21) where HP ∗ (V1 ) is deﬁne to be the set of of hyperbolic pairs which generates distinct hyperbolic planes in V1 . Let v = u1 ⊗ w1 + ξu2 ⊗ w2 be such that f (v, x) = λ = 0, where (u1 , u2 ) ∈ HP ∗ (V1 ) and (w1 , w2 ) ∈ H(V2 ). Then f1 (u1 , a1 )f2 (w1 , e1 ) + ξf1 (u1 , b1 )f2 (w1 , f1 ) + ξf1 (u2 , a1 )f2 (w2 , e1 ) + f1 (u2 , b1 )f2 (w2 , f1 ) = λ. Since λ is non-zero, one of the terms in the left side must be non-zero. Without loss, we assume that f1 (u1 , a1 )f2 (w1 , e1 ) = 0. It follows that f2 (w1 , e1 ) = f1 (u1 , a1 )−1 (λ−ξf1 (u1 , b1 )f2 (w1 , f1 )−ξf1 (u2 , a1 )f2 (w2 , e1 )−f1 (u2 , b1 )f2 (w2 , f1 )). By Lemma 3.51(4), there are q 4m1 −4 hyperbolic planes with one component ﬁxed in V1 , and by (3) of the same Lemma, there are q 4m2 −2 hyperbolic pairs with one component ﬁxed in V2 . Thus |{v ∈ x(I1 ⊗ I2 ) | (v, x) = 0}| = (q − 1).q 4m1 −4 .q 4m2 −2 = (q − 1)q 4m1 +4m2 −6 . Proposition 3.53 Assume G is nearly simple primitive rank 3 of type Ω+ (3) and M 2m is of type Sp2m1 (3) ⊗ Sp2m2 (3), with m1 < m2 . There is an M -orbit on Eε (V ) such that ξ 126 equation (3.1) does not hold unless m1 = 1, in which case M has only one orbit on E+ (V ) ξ so that 1G P 1G and hence M is in Table 1.1. M −1 Proof. Retain the notations before Lemma 3.52. Set δi = δβi ,fi (−1), and z = δ1 ⊗ δ2 . It follows from Lemma 4.4.5 and Proposition 4.4.12 in [29] that z ∈ S, z 2 ∈ I1 ⊗I2 , z ∈ I1 ⊗I2 and    I1 ⊗ I2 MΩ =   (I1 ⊗ I2 ) z if n ≡ 4 (mod 8) if n ≡ 0 (mod 8). Since x = a1 ⊗ e1 + ξb1 ⊗ f1 ∈ V, with ξ = ±1 and f = f1 ⊗ f2 , (x, x) = f (x, x) = ξf (a1 ⊗e1 , b1 ⊗f1 )+ξf (b1 ⊗f1 , a1 ⊗e1 ) = ξf1 (a1 , b1 )f2 (e1 , f1 )+ξf1 (b1 , a1 )f2 (f1 , e1 ) = 2ξ = 0, so that x is non-singular in V. As I1 ⊗I2 MΩ MI = (I1 ⊗I2 ) z , and z ﬁxes x, it follows that x(I1 ⊗ I2 ) = x(I1 ⊗ I2 ) z . Thus xMΩ = x(I1 ⊗ I2 ). Therefore by Lemma 3.52(1), | x MΩ | = 1 q 2m1 +2m2 −3 (q 2m1 −1)(q 2m2 −1), (q−1)(q 2 −1) and so 1+c+d = 1 2m1 +2m2 −3 2m1 3 (3 − 16 1)(32m2 − 1). We ﬁrst assume that m1 3. As 16 > 32 and 32mi − 1 < 32mi , we have 1 + c + d < 34m1 +4m2 −5 . We will show that 34m1 +4m2 −5 < 3m−1 so that 1 + c + d < 3m−1 and so in view of (3.10), equation (3.1) cannot hold. As m = 2m1 m2 , 34m1 +4m2 −5 < 3m−1 is equivalent to (m1 − 2)(m2 − 2) > 0. This is true because m2 > m1 need to consider cases 1 m1 3. Therefore we only 1 2. If m1 = 1, then m = 2m2 and 1+c+d = 2 3m−1 (3m −1). By Lemma 3.35, 1 + c + d = |E+ (V )|, hence MΩ has only one orbit on E+ (V ). If m1 = 2, then m = 4m2 and by Lemma 3.52(2), we have 1 + c = |{ v ∈ x (I1 ⊗ I2 ) | (v, x) = 0}| = 34m1 +4m2 −6 and so d = 2.34m2 +1 − 5.32m2 +1 , c − 2d = 10.32m2 +1 − 34m2 +1 − 1. Assume that (3.8) holds. Then c − 2d = 3m−1 − 1, and so 10.32m2 +1 − 34m2 +1 − 1 = 34m2 −1 − 1. It follows that 32m2 −2 = 1, hence m2 = 1, a contradiction to our assumption m2 > m1 = 2. Assume that (3.9) holds. Then (3m−1 + 1)(3m − 3 + c − 2d) = 4c or equivalently, 32m−1 − 3 + (3m−1 − 3)c = (2.3m−1 + 2)d. Since A := 1 + c + d = 5.32m2 +1 (32m2 − 1), we have c = A − 1 − d, and hence 32m−1 − 3 + (3m−1 − 3)(A − d − 1) = (2.3m−1 + 2)d, or 3m−1 (3m − 1) + 3A(3m−2 − 1) = (3m − 1)d. It follows that 3A(3m−2 − 1) is divisible by 127 3m − 1, where m = 4m2 . However as 3A(3m−2 − 1) = 1 (34 − 1).32m2 +2 (32m2 16 − 1)(3m−2 − 1) and m − 2 < m, 4 < m and 2m2 < m, by Zsigmondy’s Theorem 3m − 1 has a primitive prime divisor which does not divide 3A(3m−2 − 1). Thus (3.9) cannot hold. Therefore (3.1) cannot hold when m1 = 2. The Symplectic-type normalizers C6 Let α be a prime. Recall that an α-group R is extraspecial if Z(R) = Φ(R) = R ∼ Zα . = Also, R is of symplectic-type if every characteristic abelian subgroup of R is cyclic. Assume that α = 2. Up to isomorphism, there are only two extraspecial groups of order 23 , namely, Q8 and D8 . Then any extraspecial group R is either a central product of copies of D8 , or − 1 copies of D8 and one Q8 . The ﬁrst type is denoted by 21+2 and the latter one 21+2 . + − + Suppose that R ∼ 21+2 . Then CAut(R) (Z(R)) ∼ 22 .O2 (2), and R is of symplectic-type = + = with exponent 4. By Proposition 4.6.3[29], R has exactly one faithful absolutely irreducible representation ρ1 over an algebraically closed ﬁeld of characteristic p = 2 = α. Moreover, this representation has degree 2 and leaves invariant a non-degenerate quadratic form. Thus the members of C6 (Ξ) are groups MΞ = NΞ (R), where R ∼ 21+2 . By (4.6.1)[29], = + + M Ξ ∼ CAut(R) (Z(R)) ∼ 22 .O2 (2). = = Proposition 3.54 Assume G is nearly simple primitive rank 3 of type Ω+ (3) and M is 2m + of type 21+2 .O2 (2), with + 3. There is an M -orbit on Eε (V ) such that equation (3.1) ξ does not hold so that M is not in Tables 1.1-1.3. Proof. By Proposition 4.6.8[29], M Ω = MΩ ∼ 22 .Ω+ (2) and MI = 2 Let xξ be any non-singular vector in V. Since | xξ M | = |M : M that A = 1 + c + d 22 2+ + M Ξ ∼ 22 .O2 (2). = xξ | |MI |, it follows + 22 |O2 (2)| < 22 2+ +1 . In view of (3.10), it suﬃces to show that 8. If = 3, then 2m = 23 = 8. By [9], M Ω 7. Let R ∼ 21+2 . We now = + +1 < 3n/2−1 . Observe that this is true if is not maximal in P Ω+ (3). Therefore, we assume that 4 8 describe the faithful irreducible ρ1 of R. As in discussion previous Proposition 4.6.8[29], 128 for j = 1, . . . , , let (Vj , F, fj ) be a 2-dimensional orthogonal geometry with D(fj ) = Let βj = {vj,1 , vj,2 } be an orthonormal basis of Vj and set xj , yj ∈ I(Vj , F, fj ) satisfy     .  0 1 1 0  (xj )βj =   and (yj )βj =  . −1 0 0 −1 Then Rj := xj , yj ∼ D8 and = R ∼ 21+2 ∼ R1 ◦ · · · ◦ R = + = I(V1 ⊗ · · · ⊗ V , F, f1 ⊗ · · · ⊗ f ) ∼ O2 (3). = + Let R = R/Z(R). Then R is a 2 -dimensional vector space over F2 and R admits a nondegenerate quadratic form P such that P (x) = 0 or 1 according as |x| = 2 or 4. Also, fP (x, y) = [x, y] for any x, y ∈ R. Observe that if E is a subgroup of R, then E is a totally singular subspace of R if and only if E is an elementary abelian subgroup. It follows that (R, F2 , P ) is a classical orthogonal geometry of Witt index 0, so that I(R, F2 , P ) ∼ O2 (2). = + Identify R with its image Rρ1 in Ω(V ) and set Eξ = yξ , y2 , . . . , y , where ξ = ± and y+ = y1 , y− = x−1 y1 . As |yi | = 2 = |yξ | and {yξ , y2 , . . . , y } are pair-wise commuting so 1 that Eξ are elementary abelian subgroups of R, hence E ξ are maximal totally singular subspace of R. Let v+ = v1,1 and v− = v1,1 + v1,2 and xξ = vξ ⊗ v2,1 ⊗ · · · ⊗ v ,1 ∈ V, where V = V1 ⊗ · · · ⊗ V . Observe that vξ yξ = vξ and vj,1 yξ = vj,1 for any j true for all yj with j 2, hence xξ E ξ = xξ . (3.22) 2, and this is also Let A = Aut(R), K = CA (Z(R)) ∼ M Ξ and A∗ = Out(R). By Exercise 8.5, p.116 = in [1], we have Inn(R) = CA (R) and A∗ ∼ O2 (2). As Inn(Eξ ) centralizes Eξ , E2 ∼ = = + Inn(Eξ ) CK (Eξ ), and in (R, F2 , P ), E ξ is a maximal totally singular subspace, hence if α ∈ NA∗ (E ξ ) then α leaves E ξ invariant. For each α ∈ NA∗ (E ξ ), there exist an element β ∈ A so that β = α and Eξ β = Eξ . In fact, ﬁx a representative α ∈ A of 129 α, deﬁne xi β = xi α and yi β = yi , where yi α = z ji yi , yi ∈ Eξ , ji = 0, 1, i = 1, . . . , 2 and z = Z(R). Then β is the required element. Since NΩ+ (2) (E ξ ) ∼ 2( − )/2 .GL (2), = 2 and M Ω ∼ Inn(R).Ω+ (2), it follows that |M = 2 | xξ M | 32 −1 −1 xξ | |Inn(Eξ )|.|NΩ+ (2) (E ξ )|, and hence 2 |M : M xξ | 2 −1 i i=1 (2 + 1). Since 4 7, we have 2 −1 i i=1 (2 + 1) < . Thus (3.1) cannot hold. The tensor product subgroups C7 Let V1 be an α-dimensional vector space over F, with a non-degenerate bilinear form f1 . Let (Vi , fi ), i = 1, . . . , b be classical geometries which are similar to (V1 , f1 ). Then for each i, there exists a similarity ηi : (Vi , fi ) → (V1 , f1 ) satisfying fi (vηi , wηi ) = λi f1 (v, w) for all v, w ∈ V1 , where λi ∈ F∗ is independent of v, w. Let D be the tensor decomposition (V, f ) = (V1 ⊗ · · · ⊗ Vb , f1 ⊗ · · · ⊗ fb ). Then dimV = αb and V is spanned by the vectors v1 η1 ⊗ · · · ⊗ vb ηb with vi ∈ V1 . Deﬁne αi : Ξ1 → Γ L(Vi ) by (vηi )(gαi ) = (vg)ηi (g ∈ Ξ1 , v ∈ V1 ). Then αi is an isomorphism from Ξ1 to Ξi . We will write v1 ⊗ · · · ⊗ vb and g1 ⊗ · · · ⊗ gb instead of v1 η1 ⊗ · · · ⊗ vb ηb and g1 α1 ⊗ · · · ⊗ gb αb . Now let β ∈ Sb and deﬁne gβ by (v1 ⊗ · · · ⊗ vb )gβ = v1β −1 ⊗ · · · ⊗ vbβ −1 . Then Sb ∼ J := gβ | β ∈ Sb = be the tensor decomposition D in I. As described in §4.7, I. Let MI MI = (I1 ⊗ · · · ⊗ Ib ) zi,i+1 | i < b .J (3.23) −1 where zi,j = δi ⊗ δj and δi = δβi ,fi (µ), with µ = F∗ and δβi ,fi is deﬁned as in (2.4.3), (2.4.8) in [29]. Denote X = X1 ⊗ · · · ⊗ Xb for X ∈ {I, S, Ω}. Proposition 3.55 Assume G is nearly simple primitive rank 3 of type Ω+ (3) and M is 2m ε of type O2a (3) Sb , and a 3 if ε = + and a 2 if ε = −. There is an M -orbit on Eε (V ) ξ such that equation (3.1) does not hold so that M is not in Tables 1.1-1.3. Proof. Let V1 be a 2a-dimensional vector space over F, with a non-degenerate symmetric bilinear form f1 . Let (Vi , fi ), i = 1, . . . , b be classical geometries which are similar to 130 (V1 , f1 ) and (V, f ) = (V1 ⊗ · · · ⊗ Vb , f1 ⊗ · · · ⊗ fb ). Let ε = sgnV1 . We ﬁrst assume that b 3. Let β = {e1 , . . . , ea , f1 , . . . , fa } be a standard basis for V1 . For ξ = ±1, put xξ = e1 ⊗ · · · ⊗ e1 + ξf1 ⊗ · · · ⊗ f1 ∈ V. As f = f1 ⊗ · · · ⊗ fb and fi = λi f1 , it follows that f (x, x) = 2ξ b i=2 λi f1 (e1 , f1 )b−1 = 2ξ b i=2 λi = 0, so that xξ is a non-singular vector Ω, and hence S MΩ ⊥ in V. Now by Proposition 4.4.13(iii)(a) in [29], S Ci = CSi ( e1 , f1 ) ∼ SO( e1 , f1 = ⊥ MI . Let = ε and ). Since sgn e1 , f1 = +, we have sgn e1 , f1 ε so Ci ∼ SO2a−2 (3). Set N = C1 ⊗ · · · ⊗ Cb = MΩ . Then xξ N = xξ . Further observe (MΩ )xξ . Therefore that J and zi,i+1 , 1 |MΩ : (MΩ )xξ | i < b also ﬁx xξ . Thus N zi,i+1 |i < b .J ε ε |MI : N zi,i+1 |i < b .J| = |O2a (3) : SO2a−2 (3)|b , and hence | xξ MΩ | < ε ε |O2a (3) : SO2a−2 (3)|b = [2.32a−2 (3a − ε)(3a−1 + ε)]b . As (3a − ε)(3a−1 + ε) < 2.32a−1 and 4 < 32 , it follows that | xξ MΩ | < 3(4a−1)b . In view of (3.10), it suﬃces to show that 3(4a−1)b < 3m−1 , where n = 2m = (2a)b . This is equivalent to (4a−1)b correct as b 3 and a 2b−1 ab −1. This is 2. Hence equation (3.1) cannot hold. Now, assume that b = 2. Let vξ = e1 + ξf1 , w = e1 + f1 and xξ = vξ ⊗ w. Then f (xξ , xξ ) = f1 (vξ , vξ )f2 (w, w) = ξλ2 = 0, and so xξ is non-singular in V = V1 ⊗V2 . As above, we have S1 ⊗S2 MΩ . Let C1 = CS1 (vξ ) and C2 = CS2 (w). Since vξ , w are non-singular vectors in V1 and V2 respectively, we have Ci ∼ SO2a−1 (3). Set N = C1 ⊗C2 . Then xξ N = xξ . Clearly, as vξ I1 = vξ S1 and wI2 = wS2 , = we have xξ (I1 ⊗ I2 ) = xξ (S1 ⊗ S2 ). Thus |xξ MΩ | = |xξ (S1 ⊗ S2 ) z1,2 J| 2 z1,2 = 1 and J ∼ Z2 . Therefore | xξ MΩ | = 1 ε |SO2a (3) 2 |S1 : C1 |2 .22 as : SO2a−1 (3)|2 .4 = 2.32a−2 (3a − ε)2 . By (3.10), we need to prove that 2.32a−2 (3a − ε.1)2 < 3m−1 , where n = 2m = (2a)2 . As 2(3a − ε.1)2 4a 2(3a + 1)2 < 32a+1 , The above inequality is equivalent to 34a−1 2. Thus equation (3.1) cannot hold. 3m−1 or 2a2 . This is obviously true for any a Proposition 3.56 Assume G is nearly simple primitive rank 3 of type Ω+ (3) and M 2m is of type Sp2a (3) Sb with b even and a 2. There is an M -orbit on Eε (V ) such that ξ equation (3.1) does not hold so that M is not in Tables 1.1-1.3. 131 Proof. Let V1 be a 2a-dimensional vector space over F, with non-degenerate symplectic bilinear form f1 . Let (Vi , fi ), i = 1, . . . , b be classical geometries which are similar to (V1 , f1 ) and (V, f ) = (V1 ⊗ · · · ⊗ Vb , f1 ⊗ · · · ⊗ fb ). Then f = f1 ⊗ · · · ⊗ fb and fi = λi f1 , where λi ∈ F∗ . Let β = {e1 , . . . , ea , f1 , . . . , fa } be a standard basis for V1 and let xξ = e1 ⊗ · · · ⊗ e1 + ξf1 ⊗ · · · ⊗ f1 , where ξ = ±1. As I1 is quasi-simple, I Let Ci = CIi ( e1 , f1 ) ∼ Sp( e1 , f1 = Then N, J and zi,i+1 , 1 We have | xξ MI | (32a−1 .(32a − 1))b ⊥ Ω, and hence I MΩ MI . ). Hence Ci ∼ Sp2a−2 (3). Let N = C1 ⊗ · · · ⊗ Cb . = i < b .J (MΩ )xξ . i < b ﬁx vector xξ , so that N. zi,i+1 | 1 |MI : N. zi,i+1 | 1 i < b .J| = |Sp2a (3) : Sp2a−2 (3)|b = 4. Since m = 2b−1 ab and a 2, b 4, we 3(4a−1)b . Assume that b have 3m−1 > 3(4a−1)b , and so by (3.10), equation (3.1) cannot happen. Next assume that b = 2 and a 4. Then as (4a − 1)b = 8a − 2 < 2a2 − 1 = 2b−1 ab − 1, so that equation (3.1) a −1 3. Set z = z1,2 = δ1 ⊗ δ2 , g = cannot hold. Thus we can assume that b = 2 and 2 g(12) ∈ J and x = λ−1 e1 ⊗ e1 + ξf1 ⊗ f1 , where f2 = λ2 f1 . Then f (x, x) = 2ξ, hence x is 2 non-singular in V. We will show that xMΩ = x(I1 ⊗ I2 ). Write x = e1 ⊗ (λ−1 e1 ) + ξf1 ⊗ f1 . 2 Then f1 (e1 , f1 ) = 1 and f2 (λ−1 e1 , f1 ) = λ2 f1 (λ−1 e1 , f1 ) = 1 so that (e1 , f1 ), (λ−1 e1 , f1 ) are 2 2 2 hyperbolic pairs in (V1 , f1 ), (V2 , f2 ), respectively. For any g1 ⊗ g2 ∈ I1 ⊗ I2 , xg1 ⊗ g2 = e1 g1 ⊗ λ−1 e1 g2 + ξf1 g1 ⊗ f1 g2 . Clearly (e1 g1 , f1 g1 ) and (λ−1 e1 g2 , f1 g2 ) are hyperbolic pairs 2 2 in V1 and V2 . Argue as in Lemma 3.52, we have x(I1 ⊗ I2 ) = {u1 ⊗ λ−1 w1 + ξu2 ⊗ 2 w2 | (u1 , u2 ) ∈ HP ∗ (V1 ), (λ−1 w1 , w2 ) ∈ H(V2 )}, where HP ∗ (V1 ) is deﬁned to be the set 2 of of hyperbolic pairs which generates distinct hyperbolic planes in V1 and HP(V2 ) is the set of all hyperbolic pairs in V2 . Observe that if (u1 , u2 ) and (λ−1 w1 , w2 ) are hyperbolic 2 pairs in V1 , V2 , respectively, and v = u1 ⊗ λ−1 w1 + ξu2 ⊗ w2 , then vz = (−u1 δ1 ) ⊗ 2 −1 −1 −1 −1 (−λ−1 w1 δ2 )+ξu2 δ1 ⊗w2 δ2 . Since f1 (u1 δ1 , u2 δ1 ) = −f1 (u1 , u2 ) and f2 (λ−1 w1 δ2 , w2 δ2 ) = 2 2 −1 −1 −f2 (λ−1 w1 , w2 ), it follows that f1 (−u1 δ1 , u2 δ1 = 1 and f2 (−λ−1 w1 δ2 , w2 δ2 ) = 1. Hence 2 2 −1 −1 (−u1 δ1 , u2 δ1 ), (−λ−1 w1 δ2 , w2 δ2 ) are still hyperbolic pairs in V1 , V2 . Thus x(I1 ⊗I2 ) z = 2 x(I1 ⊗ I2 ). Next, we have vg = (u1 ⊗ λ−1 w1 )g + ξ(u2 ⊗ w2 )g = λ−1 w1 ⊗ u1 + ξw2 ⊗ u2 = 2 2 132 w1 ⊗ λ−1 u1 + ξw2 ⊗ u2 . Since f2 (λ−1 w1 , w2 ) = 1, λ2 = f2 (w1 , w2 ) = λ2 f1 (w1 , w2 ), hence 2 2 f1 (w1 , w2 ) = 1. Similarly f2 (λ−1 u1 , u2 ) = λ2 f1 (λ−1 u1 , u2 ) = f1 (u1 , u2 ) = 1. Thus vg ∈ 2 2 x(I1 ⊗ I2 ). Therefore x(I1 ⊗ I2 ) = xMI , and so, as I1 ⊗ I2 xMΩ = x(I1 ⊗I2 ). By Lemma 3.52, we have 1+c+d = MΩ MI , it follows that c = 38a−6 −1, d = 1 4a−3 2a 3 (3 −1)2 , 16 1 34a−3 ( 16 (32a − 1)2 − 34a−3 ) and hence c − 2d = 38a−5 − 1 − 1 .34a−3 (32a − 1)2 . Assume that 8 1 equation (3.8) holds. Then c−2d = 3m−1 −1. Hence 38a−5 −1− 8 .34a−3 (32a −1)2 = 32a 2 −1 −1. As 4a − 3 8a − 5 and 4a − 3 2a2 − 1, it follows that 8(34a−2 − 32(a−1) ) = (32a − 1)2 . We 4, then 2(a − 1)2 − (4a − 2) > 0, so a 3, then 4a − 2 > 2(a − 1)2 , hence 2 have 2(a − 1)2 − (4a − 2) = 2(a − 2)2 − 4, and so, if a that 8(34a−2 − 32(a−1) ) < 0 < (32a − 1)2 . Now if 2 2 2 2 2 8(34a−2 − 32(a−1) ) = 8.32(a−1) (38a−4−2a − 1) = (32a − 1)2 . It follows that 2(a − 1)2 = 0. However this cannot happen as a 2. Thus equation (3.8) cannot hold. Finally assume 1 4a−2 m−2 3 (3 16 that equation (3.9) holds. Then (3m − 1)d = 3m−1 (3m − 1) + − 1)(32a − 1)2 or equivalently, 16(3m − 1)(d − 3m−1 ) = 34a−2 (3m−2 − 1)(32a − 1)2 . Since m > m − 2 and m = 2a2 > 2a, this equality cannot hold by Zsigmondy’s Theorem. Thus equation (3.1) cannot hold. 3.5.3 Permutation Characters of Maximal subgroups in S Embedding of Symmetric and Alternating groups Proposition 3.57 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m almost simple of type An , with n 10, and V is the fully deleted permutation module for An in characteristic p = 3. There is an M -orbit on Eε (V ) such that equation (3.1) does ξ not hold unless n = 5, 6, 7, 12 or 18. If n = 5, 6, 7, then ε = − and M has at most 2 orbits on E− (V ) so that 1G P ξ 1G by Corollary 3.7; if n = 12, then ξ = M , ε = + and M has two , ε = −, M has orbits on E+ (V ), so that 1G P ξ 1G by Corollary 3.7; if n = 18 then ξ = M 4 orbits on E− (V ) and equation (3.1) holds for r = s and for any M -orbits on E− (V ). ξ ξ Thus these cases appear in Table 1.2. 133 Proof. Assume the construction and notation before Lemma 3.23. We have      2,   (ei , ej ) = −1,      0 if i = j if |i − j| = 1 if |i − j| 2. 2 −1 · · · 2 . . . 0 0 ··· ... ··· ··· 0 0    −1   . and fβ =  .  .   0  0   0 0  . .  . . , . .    2 −1  −1 2 so that D(fβ ) = n−ε3 (n). Let v = ε1 −ε2 , or ε1 +ε2 −ε3 −ε4 ∈ V. Let Q be the quadratic form associated to the natural bilinear form f on Fn , we have Q(v) = 0, hence v is a 3 non-singular vector in V with ξ = Q(v) = , , respectively. We see that n − 1 − ε3 (n) 1. is even if and only if n = 6k − 1, n = 6k or n = 6k + 1, where k (a) Parameters for v = ε1 − ε2 . We have ξ = Q(v) = 2, and by Lemma 3.23,    1 + c + d = 1 n(n − 1)  2   c = 2n − 4      d = 1 (n − 2)(n − 3). 2 As 2m = n − 1 − ε3 (n) n − 2, we have 3m−2 3(n−6)/2 . We see that if n 15 then 3(n−6)/2 > 1 n(n − 1), and hence 3m−2 > 1 + c + d. In view of (3.10) and (3.11), equation 2 (3.1) cannot hold. Thus we can assume that 5 n 14. However as n has the form 6k − 1, 6k or 6k + 1, it follows that n ∈ {5, 6, 7, 11, 12, 13}. (1) Case n = 5. Then 2m = 5 − 1 − ε3 (5) = 4, D = D(fβ ) = 5 ≡ −1 = , hence ε = sgnV = (−)m−1 = −. Thus A5 can embed in Ω− (3) ∼ A6 . Let z = ε1 +ε2 +ε3 +ε4 −ε5 . = 4 Then z ∈ V, Q(z) = Q(v) and | z M | = 5. As |E− (V )| = 1 3m−1 (3m − ε.1) = 15 = 10 + 5, 2 2 it follows that M has only 2 orbits on E− (V ). 2 (2) Case n = 6. Then 2m = 6 − 1 − ε3 (6) = 4, D = 5 ≡ −1 = , hence ε = sgnV = (−)m−1 = −. Thus A6 can embed in Ω− (3) ∼ A6 . Hence M has one orbit on E− (V ). = 4 2 134 (3) Case n = 7. Then 2m = 7−1−ε3 (7) = 6, and so m = 3. Also D = D(fβ ) = 7 ≡ 1 = , hence ε = sgnV = (−)m = −. Thus A7 can embed in Ω− (3) ∼ U4 (3). In this case, = 6 there are only two orbits with representatives v and z = ε1 + ε2 + ε3 + ε4 − ε5 , and 7! = 105. | z M| = 1!4!(7 − 5)! (4) Case n = 11. Then 2m = 11 − 1 − ε3 (11) = 10, D = 11 ≡ −1 = , hence ε = sgnV = 1 (−)m−1 = +. Thus A11 can embed in Ω+ (3). Now 3m−1 = 34 > 55 = 2 n(n − 1), in view 10 of (3.10), equation (3.1) cannot hold. (5) Case n = 12. Then 2m = 12 − 1 − ε3 (12) = 10, D = D(fβ ) = 11 ≡ −1 = , hence ε = sgnV = (−)m−1 = +. Thus A12 can embed in Ω+ (3). As 3m−1 = 34 = 81 > 66 = 10 1 n(n 2 − 1), in view of (3.10), equation (3.1) cannot hold. , hence (6) Case n = 13. Then 2m = 13 − 1 − ε3 (13) = 12, D = D(fβ ) = 13 ≡ 1 = ε = sgnV = (−)m = +. Thus A13 can embed in Ω+ (3). Now 3m−1 = 35 = 243 > 78 = 12 1 n(n 2 − 1), by (3.10), equation (3.1) cannot hold. (b) Parameters for v = ε1 + ε2 − ε3 − ε4 . We have ξ = Q(v) = 1, and by Lemma 3.23    1 + c + d = 1 n(n − 1)(n − 2)(n − 3)  8   c = 2n3 − 25n2 + 111n − 172      d = 2 + 4(n − 4)2 + 1 (n − 4)(n − 5)(n − 6)(n − 7). 8 If n 1 26, then 3(n−6)/2 > 8 n(n − 1)(n − 2)(n − 3). Thus 3m−2 3(n−6)/2 > 1 + c + d. Therefore, (3.1) cannot happen. So we assume that 5 n 25. As n = 6k − 1, 6k or 6k + 1, it follows that n ∈ {5, 6, 7, 11, 12, 13, 17, 18, 19, 23, 24, 25}. If n = 5, then A5 Ω− (3) ∼ A6 , and | w M | = = 4 1 5.4.3.2 8 = 15 = |E− (V )|. Thus there is only one orbit. 1 Ω− (3). Let 6 As in previous case, A6 has only one orbit on E− (V ). If n = 7, then A7 1 z = ε1 + ε2 + ε3 + ε4 + ε5 − ε6 − ε7 ∈ V. Then (z, z) = (v, v) and since there is no g ∈ S7 7! 1 = 21. Now, |E− (V )| = 2 3m−1 (3m + ε.1) = which maps z to −z, | z M | = |zM | = 1 5!2! 126 = 105 + 21 = | w M | + | z M |. Hence M has two orbits on E− (V ). Next suppose 1 135 that n = 6k − 1, 6k or 6k + 1, where 2 k 4. (1) Subcase n = 6k + 1. Let z = ε1 + · · · + ε6k−3 + ε6k−2 + ε6k−1 − ε6k − ε6k+1 . Then n! = 1 n(n − 1). If k 3 then n 19, z ∈ V, (z, z) = (v, v) and | z M | = |zM | = 2 (n − 2)!2! 1 hence 3m−2 3(n−6)/2 > 2 n(n − 1) = |zM |. If k = 2 then n = 13, so that ε = sgnV = +. In this case 2m = 12, and so 3m−1 = 35 > 78 = | z M |. In view of (3.10), (3.1) cannot hold. Ω+ (3). Let z = ε1 + · · · + ε5 − ε6 − · · · − ε10 ∈ V. 10 12! 1 1 = 8316. Also | v M | = 8 n(n − We have (z, z) = (v, v) and | z M | = 2 |zM | = 2.(5!)2 .2! 1 1)(n − 2)(n − 3) = 1485. As |E+ (V )| = 2 3m−1 (3m − 1) = 1 .34 (35 − 1) = 9801. Observe 1 2 (2) Subcase n = 12. Then A12 9801 = 8316 + 1485, hence E+ (V ) = z M ∪ v M. Thus M has only two orbits on E+ (V ). 1 1 (3) Subcase n = 18. Then m = 8, 3m−1 + 1 = 2188, and D = 18 − ε3 (18) = 17 ≡ −1 = so that ε = −, hence A18 , Ω− (3). In this case, M has 4 orbits with representatives 16 v1 = ε1 + ε2 − ε3 − ε4 , v2 = ε1 + · · · + ε5 − ε6 − ε7 , v3 = ε1 + · · · + ε8 − ε9 − · · · − ε16 and v4 = ε1 +· · ·+ε5 −ε6 −· · ·−ε10 . Using GAP, the parameters ci , di , i = 1, · · · , 4 of vi M are as follow: (d1 , c1 ) = (3789, 5390), (d2 , c2 ) = (223497, 444806), (d3 , c3 ) = (328914, 655640) and (d4 , c4 ) = (1838565, 3674942). We see that 2di − ci = 2188 = 3m−1 + 1, for all i, so that equation (3.8) holds with ε = −, r = t and m = 8. (4) Subcase n = 24. Then c = 15740, d = 16137, c − 2d = −16534, m = 11 and D = that ε = +. Then equation (3.1) cannot hold. (5) Subcase n = 11. Then A11 Ω+ (3). However as the partition λ = (11, 1) is a 10 , so JS-partition and m(λ) = λ, by Theorem 2.33, D(11,1) ↓A11 ∼ D(10,1) is irreducible, so = that A11 A12 Ω+ (3). Notice that by using GAP, M has 4 orbits on E+ (V ) with 10 1 representatives v1 = ε1 +· · · ε8 −ε9 −ε10 , v2 = ε1 +ε2 −ε3 −ε4 , v3 = ε1 +· · ·+ε5 −ε6 −· · ·−ε10 and v4 = ε1 + · · · + ε5 − ε6 − ε7 and equation (3.1) holds with r = s. (6) Subcase n = 17. Then m = 8, D = , and so ε = (−)m−1 = −, hence A17 Ω− (3). 16 We have d = 2823, c = 4816 and 2d − c = 1330 = 3m−1 + 1 and 4c is not divisible by 136 3m−1 + 1 = 37 + 12188. Thus equation (3.9) and (3.8) cannot hold. (7) Subcase n = 23. Then m = 11, D = , and so ε = (−)m−1 = +, hence A23 Ω+ (3). 22 1 In this case 3m−1 = 310 > 8 23.22.21.20 = | v M |, by (3.10), equation (3.1) cannot hold. Proposition 3.58 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m almost simple of type An with n 12, and V is not the fully deleted permutation module for An in characteristic 3. There is an M -orbit on Eε (V ) such that equation (3.1) does ξ not hold so that M is not in Tables 1.1-1.3. Proof. By Lemma 2.32, we have dim(V ) or m 1 (n2 4 1 (n2 2 − 5n + 2). Hence 2m 3 n2 −5n−6 4 1 (n2 2 − 5n + 2), − 5n + 2). However when n 12, 3m−2 > n! = |Aut(An )|, which contradicts (3.10) and (3.11). Thus equation (3.1) cannot hold. Proposition 3.59 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M 2m is almost simple of type An with 5 n 11, and V is not the fully deleted permutation module for An in characteristic 3. There is an M -orbit on Eε (V ) such that equation (3.1) ξ does not hold so that M is not in the Tables 1.1-1.3. Proof. By (3.10) and (3.11), we have dimV 2log3 (n!) + 4. As m 2, dimV 4, and so by [26], we only need to consider the following case (An , dimV ) = (A11 , 34). It follows from the Appendix in [24] that the symmetric group S11 has exactly two inequivalent irreducible representations of degree 34 in characteristic 3 with corresponding partitions λ = (9, 2) and m(λ) = (5, 4, 12 ). By Theorem 2.31, Dλ ↓ A11 = Dm(λ) ↓ A11 is irreducible. Thus we can assume that V = Dλ . By Theorem 2.25, dimS λ = 44 and by Lemma 2.27, S λ = Dλ + D(10,1) , dimDλ = 34. By Theorem 2.24, S λ has a basis consisting of standard polytabloids. If t is a standard λ-tableaux, then t is one of 1 2 j1 · · · j7 1 3 2 j9 137 j1 · · · j7 or , j8 j9 where 3 j1 < j2 < · · · < j7 11, 3 j8 < j9 11, and jk are all pairwise distinct. Let t1 = 1 2 5 6 7 8 9 10 11 3 4 t2 = 1 2 4 6 7 8 9 10 3 5 11 and t3 = 1 2 3 4 7 8 9 10 5 6 11 . Set ei = eti . To simplify the notation, we write just the second row for a tabloid. With this notation, we have e1 = 34−14−32+12, e2 = 35−15−32+12, and e3 = 56−16−52+12. Let x1 = e1 + S λ ∩ S λ⊥ and x2 = e1 + e2 + e3 + S λ ∩ S λ⊥ . Then xi are non-singular in V and S{7,8,9,10,11} Mxi . Thus | xi M | 11! 5! |S11 : S{7,8,9,10,11} | = 11! . 5! As 2m = 34, m = 17. Clearly, 3m−2 = 315 > | xi M |, hence equation (3.1) cannot hold. Cross-characteristic embedding of ﬁnite groups of Lie type Proposition 3.60 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m almost simple of type S, where S is a ﬁnite simple Chevalley group in cross-characteristic. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold unless (S, L) = ξ + + (L3 (4), Ω− (3)), (O8 (2), O8 (3)), in which cases M has at most two orbits on Eε (V ) so that 6 1G P 1G by Corollary 3.7, and so M is in Table 1.2 or (L, S) = (Ω+ (3), 2.F4 (2)) is in 52 M Table 1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eε (V ). ξ Recall the hypothesis in section 3.3, V is an irreducible F3 S-module of even dimension with Frobenius-Schur indicator + and realizes over F3 . Moreover (3, q) = 1. Case SL2 (q). Assume ﬁrst that q ≡ 1(mod 4). It follows from Table 2.7(b) that either dimV = 1 (q 2 1 (q 2 + 1) or dimV q − 1. However the ﬁrst case cannot happen as q − 1. If q > 29 then + 1) ≡ 1(mod 2), hence dimV is odd. Thus 2m = dimV 3(q−5)/2 > q 4 138 3(q−5)/2 > q 4 so that 3m−2 |Aut(S)|. By (3.10) and (3.11), equation (3.1) cannot hold. Thus 5 q 29, (q, 3) = 1, q ≡ 1(mod 4) and q is a prime power, hence q ∈ {5, 13, 17, 25, 29}. If q = 29 then |Aut(S)| = 29(292 − 1) < 3(29−1)/2 , and so 3m−2 > 1 + c + d. Similarly if q = 52 then |Aut(S)| = 2.25(252 − 1) < 3(25−1)/2 , so that 3m−2 > 1 + c + d. Therefore q = 5, 13 or 17. Since L2 (5) ∼ A5 , we can assume that = q = 13 or q = 17. By [26], L2 (13) has three irreducible representations in characteristic 3 with Frobenius -Schur indicator + and their degrees are 12. However they cannot realize over F3 . Finally by [26] again, we need to consider two cases (S, dimV ) = (L2 (17), 16) or (L2 (17), 18). If (S, dimV ) = (L2 (17), 16) then L2 (17) P Ω− (3). For each extension of S, 16 there exist two non-singular points of diﬀerent type with (c, d) as follow: if M = L2 (17).2 then (c, d) = (32, 120), (722, 501); if M = L2 (17) then (c, d) = (32, 120), (1643, 804). If (S, dimV ) = (L2 (17), 18) then L2 (17) P Ω+ (3). For each extension of S, there exist two 18 non-singular points of diﬀerent type with (c, d) as follow: if M = L2 (17) then (c, d) = (1635, 812), (1611, 836); if M = L2 (17).2 then (c, d) = (3185, 1711), (3280, 1615). We can check that equation (3.1) cannot hold in any of these cases. Assume that q ≡ 3(mod 4). As q is odd, dimV = q − 1 or dimV = q + 1 by Table 2.7(c). As above, we have q 29, and so q ∈ {7, 11, 19, 23}. If q = 23 then |Aut(S)| = 23(232 − 1) < 3(23−1)/2 , and so 3m−2 > 1 + c + d. Therefore q = 7, 11 or 19. By [26], we need to consider the following cases: (L2 (7), 6), (L2 (11), 10). By [9], Ωε (3) has no 6 maximal subgroups with socle isomorphic to L2 (7) ∼ L3 (2) or its covers. If (S, dimV ) = = (L2 (11), 10) then L2 (11) P Ω+ (3). For each extension of S, there exist two non-singular 10 points of diﬀerent type with (c, d) as follow: if M = L2 (17) then (c, d) = (20, 45), (116, 48); if M = L2 (17).2 then (c, d) = (3185, 1711), (443, 216). We can check that equation (3.1) cannot hold in any of these cases. Assume that q ≡ 0(mod 2). As q is even, q ± 1 are odd, by Table 2.7(d), dimV = q and 3 q +1. As L2 (4) ∼ A5 , we need to consider case (L2 (16), 16). If (S, dimV ) = (L2 (16), 16) = then L2 (16) P Ω− (3) and Out(L2 (16)) ∼ Z4 . For each extension of S, there exist two = 16 139 non-singular points of diﬀerent type with (c, d) as follow: if M = L2 (16) then (c, d) = (1355, 648), (30, 105); if M = L2 (16).2 then (c, d) = (30, 105), (620, 399); if M = L2 (16).4 then (c, d) = (30, 105), (5378, 2781). We can check that equation (3.1) cannot hold in any of these cases. Case Ln (q), n 3. Assume (n, q) = (3, 2), (3, 4), (4, 2). By Table 2.6, 2m 2 (q n − 2q + |Aut(Ln (q))| 1)/(q−1). Now, if (n, q) = (3, 2), (4, 2), (5, 2), (3, 4), (3, 5), then 3m−2 > q n so that (3.1) cannot hold. By the isomorphisms L2 (7) ∼ L3 (2), and L4 (2) ∼ A8 we only = = need to consider the following cases: L5 (2), L3 (4), L3 (5). If S = L3 (4), then by [26], either dimV = 6 or dimV 36. If the latter case holds then 3m−2 316 > 49 |Aut(L3 (4))|, hence (3.1) cannot hold. By [9], L3 (4) embeds in Ω− (3) and L3 (4) has only one orbit 6 on Eε (V ). If S = L3 (5), then by [26], dimV = 2m 30. As 3m−2 313 > 2.58 |Aut(L3 (5))| = 2.53 (52 − 1)(53 − 1), hence (3.1) cannot hold. If S = L5 (2), then by [26], either dimV = 30 or dimV 124. If the latter case holds then 3m−2 360 > 225 = |Aut(L5 (2))|, hence (3.1) cannot hold. Case P Sp2n (q), q odd. By Table 2.6, 2m then 3m−2 3(q n −9)/4 1 n (q 2 − 1). If (n, q) = (2, 5), (2, 7), (3, 5), > q 2n 2 +n+1 |Aut(S)|, so that (3.1) cannot hold. Thus, we need to consider cases S4 (5), S4 (7), S6 (5). If S = S6 (5), then by Theorem 2.1 in [16], either 1 dimV = 2 (q n ± 1) = 62, 63 or dimV (q n − 1)(q n − q)/(2q + 2) = 1240. By [19], if 1240, hence 3m−2 > |Aut(S)|. If S = S4 (7) 40 dimV = 62, then ind(V ) = −. Thus dimV then dimV 126 by [19] and so 3m−2 > |Aut(S)|. Finally if S = S4 (5) then dimV 318 > 511 |Aut(S)|, so that (3.1) cannot happen. by [26]. Then 3m−2 Case P Sp2n (q), q even. As Sp4 (2) ∼ A6 , we can assume that (n, q) = (2, 2). By = Table 2.6, 2m 3m−2 > q 2n 2 +n+1 (q n − 1)(q n − q)/(2q + 2). If (n, q) = (2, 2), (3, 2), (4, 2), (2, 4) then |Aut(S)|, so that equation (3.1) cannot hold. Thus, we need to consider cases: S4 (4), S6 (2), S8 (2). By [26] and [19], we need to consider the following cases: (S4 (4), 18), (S4 (4), 34), (S4 (4), 50), (S6 (2), 14), (S6 (2), 34), (S8 (2), 50). If S is one of S8 (2) 140 or S4 (4) and dimV = 50, then m = 25, |Aut(S8 (2))| = 216 .(22 −1)(24 −1)(26 −1)(28 −1) < 216+20 = 236 , |Aut(S4 (4))| = 2.44 (42 − 1)(44 − 1) < 2.410 = 221 . Now, as 323 > 236 > 221 , we have 3m−2 = 323 > |Aut(S8 (2))| > |Aut(S4 (4))|. Similarly if S = S6 (2) or S4 (4) and dimV = 34, then m = 17, |Aut(S6 (2))| = 29 (22 − 1)(24 − 1)(26 − 1) < 221 . We have 315 > 221 so that 3m−2 > |Aut(S6 (2))| and 3m−2 > |Aut(S4 (4))|. Thus (3.1) cannot hold in any of these cases. Case Un (q), (q, 3) = 1, n 3, n odd. By Table 2.6, 2m (q n − q)/(q + 1). If (n, q) = (3, 2), (5, 2), (7, 2), (3, 4), (3, 5) then 3m−2 > |Aut(S)|, so that equation (3.1) cannot hold. Since U3 (2) ∼ 32 .Q8 is not simple, we need to consider cases: U5 (2), U7 (2), U3 (4), U3 (5). = If S = U3 (4), then by [26], dimV = 64. But then 3m−2 > |Aut(S)|. If S = U3 (5), then by [26], either dimV = 28 or dimV 340 > 59 84. If the latter case holds then 3m−2 |Aut(U3 (5))|, hence equation (3.1) cannot hold. If S = U5 (2) then by [26], |Aut(S)|. If S = U7 (2), then by [19], dimV > 250. dimV = 44. As 3m−2 = 320 > 225 Then 3m−2 > |Aut(S)|. Case Un (q), n 3, even. Assume (n, q) = (4, 2). By Table 2.6, 2m (q n − 1)/(q + 1). If (n, q) = (6, 2) then 3m−2 > |Aut(S)|, so that equation (3.1) cannot hold. As U4 (2) ∼ S4 (3) ∼ O5 (3), we need to consider cases U6 (2). By [26], dimV = = 3m−2 = 326 > 236 |Aut(S)|. 4. Assume q > 3. By Table 2.6, 2m 2 −n+1 56. But then Case P Ω+ (q), n 2n As |Aut(S)| 3q 2n (q n −1)(q n−1 +q)/(q 2 −1)−2. < q 2n 2 −n+2 , m−2 (q n − 1)(q n−1 + q)/(2q 2 − 2) − 3 and 4, q 4. It follows (q n − 1)(q n−1 + q)/(2q 2 − 2) − 3 > (2n2 − n + 2)log3 (q) for any n that 3m−2 > |Aut(S)|. Assume that q = 2 and n (q n − 1)(q n−1 − 1)/(q 2 − 1). Then |Aut(S)| 5. By Table 2.6 again, dimV = 2m 22n 2 −n+1 and m − 2 (q n − 1)(q n−1 − 1)/(2q 2 − 2) − 2. We have (q n − 1)(q n−1 − 1)/(2q 2 − 2) − 2 > (2n2 − n + 1)log3 (2), so that 3m−2 > |Aut(S)|. Thus we are left with S = P Ω+ (2). By [9], either dimV = 8, 28 8 or dimV 48. Suppose that the latter case holds. As |Aut(S)| 3.229 and m 24, we 141 have 3m−2 322 > 3.229 |Aut(P Ω+ (2))|. 8 P Ω+ (3). In this case, there exist two 28 If (S, dimV ) = (P Ω+ (2), 28) then P Ω+ (2) 8 8 non-singular points ui , i = 1, 2, of diﬀerent type with orbit sizes 3780, 45360 respectively. Clearly | ui M | < 312 = 3m−2 . By (3.10) and (3.11), equation (3.1) cannot hold. + + If dimV = 8 then by [9], S = O8 (2) is a maximal subgroup of L = O8 (3), and there are 3 classes of S in L. By [13] again, the permutation characters 1L = 1a + 260cde + S 9450a + 18200e, 1a + 260bdf + 9450a + 18200d, 1a + 260aef + 9450a + 18200c, respectively, while 1L = 1a + 260a + 819a, 1a + 260d + 819a, 1a + 260b + 819b, 1a + 260e + 819b, 1a + M 260c + 819c, 1a + 260f + 819c. Thus M has at most two orbits on Eε (V ). Case P Ω− (q), n 2n 4. Assume that (n, q) = (4, 2), (4, 4), (5, 2). By Table 2.6, 2m (q n + 1)q 2n 2 −n+1 (q n + 1)(q n−1 − q)/(q 2 − 1) − 1. As |Aut(S)| < q 2n 2 −n+2 , m−2 (q n + 1)(q n−1 − q)/(2q 2 − 2) − 3 and (q n + 1)(q n−1 − q)/(2q 2 − 2) − 3 > (2n2 − n + 2)log3 (q) for any (n, q) as above. It follows that 3m−2 > |Aut(S)| for (n, q) as above. If S = P Ω− (4) then dimV 8 dimV 1026, hence 3m−2 3511 > 430 > |Aut(S)|. If S = P Ω− (2) then 10 186 by [19], and hence 3m−2 > |Aut(S)|. Finally if S = P Ω− (2) then either 8 50 by [26]. Clearly if dimV 50 then 3m−2 > |Aut(S)|. If dimV = 34 or dimV (S, dimV ) = (P Ω− (2), 34) then P Ω− (2) 8 8 P Ω− (3). In this case, there exist two non34 singular points ui , i = 1, 2, of diﬀerent type which are the eigenvector of an element g of order 17. The normalizer N in S of the subgroup generated by g is of order 68, and N ﬁxes ui , i = 1, 2. We have | ui M | (3.11), equation (3.1) cannot hold. Case P Ω2n+1 (q), n As |Aut(S)| q 2n 2 +n+1 |Aut(S)|/|N | < 315 = 3m−2 , in view of (3.10) and 3, q > 3, q odd. By Table 2.6, 2m , m−2 (q 2n − 1)/(q 2 − 1) − 2. (q 2n − 1)/(2q 2 − 2) − 3 and (q 2n − 1)/(2q 2 − 2) − 3 > (2n2 + n + 2)log3 (q) for any (n, q) as above. It follows that 3m−2 > |Aut(S)|. Case E6 (q). By Table 2.6, 2m q(q 4 + 1)(q 6 + q 3 + 1) − 2. As |Aut(S)| q 79 , m − 2 q(q 4 + 1)(q 6 + q 3 + 1)/2 − 3 > 79.log3 (q) for any q 142 2. It follows that 3m−2 > |Aut(S)|. Case E7 (q). By Table 2.6, 2m e(S), where e(S) = qΦ7 (q)Φ12 (q)Φ14 (q) − 3 = q 134 , m − 2 q(q 6 + · · · + q + 1)(q 6 − q 5 + q 4 − q 3 + q 2 − q + 1) − 3. As |Aut(S)| 1 (e(S) 2 − 4) > 134.log3 (q) for any q 2. It follows that 3m−2 > |Aut(S)|. Case E8 (q). By Table 2.6, 2m e(S), where e(S) q 27 (q 2 − 1). As |Aut(S)| q 249 , e(S) − 4 > 249.log3 (q) for any q 2. It follows that 3m−2 > |Aut(S)|. m−2 2 Case F4 (q), q odd. As q is odd and (q, 3) = 1, it follows that q 5. We have 2m q q 4 (q 6 − 1) and |Aut(S)| 5, hence 3m−2 > |Aut(S)|. Case F4 (q), q even. Assume that q > 2. Then q 1 e(S) = 2 q 7 (q 3 − 1)(q − 1). Now, as m − 2 1 e(S) 2 q 53 . Since m − 2 1 4 6 q (q 2 − 1) − 2 > 53.log3 (q) for any 4. Thus by Table 2.6, 2m − 2 > 53log3 (q) for any q > 2, and so 3m−2 > |Aut(S)| if q 4. If q = 2 then by [19], either dimV = 52 or dimV > 250. If the latter case holds then clearly, 3m−2 > |Aut(S)|. Thus we only need to consider case dimV = 52 and S ∼ 2.F4 (2). = Case G2 (q). As G2 (2) ∼ U3 (3) · 2 and (q, 3) = 1, we can assume that q = ﬁrst that q > 4. Then by Table 2.6, 2m m−2 1 q(q 2 2 4. Suppose q 15 , and 4. q(q 2 − 1). Since |Aut(G2 (q))| − 1) − 2 > 15log3 (q) for any q > 4, we have 3m−2 > |Aut(S)| when q 64 and hence 3m−2 330 > 415 |Aut(S)|. q 79 , and m − 2 If S = G2 (4) then by [26], dimV Case 2 E6 (q). By Table 2.6, 2m 1 9 2 q (q 2 q 9 (q 2 − 1). Since |Aut(S)| − 1) − 2 > 79log3 (q), 3m−2 > |Aut(S)|. q 3 (q 2 − 1). Since |Aut(S)| 3.q 29 , and m − 2 > Case 3 D4 (q). By Table 2.6, 2m 1 3 2 q (q − 1) − 3 2 > 29log3 (q) for any q 3, 3m−2 > |Aut(S)| whenever q > 2. If S = 3 D4 (2) 32 4 > 3.22 9 > |Aut(S)|. q 4 . q/2(q − 1). Since |Aut(S)| q 27 , then dimV 52 by [26], and hence 3m−2 Case 2 F4 (q), q = 22a+1 . By Table 2.6, 2m and m − 2 1 4 q . 2 q/2(q − 1) − 2 > 27log3 (q) for any q 8, 3m−2 > |Aut(S)|. We are left with case S = 2 F4 (2) . By [19], dimV 124, so that 3m−2 > |Aut(S)|. 8. Suppose Case Sz(q), q = 22a+1 . As Sz(2) is not simple, we can assume that q 143 ﬁrst that q > 8. By Table 2.6, 2m 1 2 q/2(q − 1). Since |Aut(S)| q 6 , and m − 2 40 by [19]. q/2(q − 1) − 2 > 6log3 (q), 3m−2 > |Aut(S)|. If S = Sz(8) then dimV 318 > 86 = 218 |Aut(S)|. Then 3m−2 Deﬁning characteristic embedding of ﬁnite groups of Lie type Proposition 3.61 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m almost simple of type S, where S is simply connected of type A or 2 A over F3f . There is an M -orbit on Eε (V ) such that equation (3.1) does not hold so that M is not in Tables ξ 1.1-1.3 . Proof. Assume (3.1) holds for some r ∈ {s, t} and for any points in Eε (V ). Let β = {e1 , · · · , e +1 } be a basis for N, the natural module of S over F3f . Case f = 1. We can assume that 2. By Theorem 2.35 and 2.36, there exists a 3-restricted dominant weight λ ∈ X3 such that V ∼ L(λ). By Corollary 3.36, and the fact = that 1 + c + d that 2m |Aut(S)| 3( +1) , we have 3m−2 18, then 1 3 8 2 3( +1) , and hence m − 2 2 ( + 1)2 , so 2 2 + 4 + 6. If 2 2 + 4 + 6, and so by Theorem 5.1 in [38], λ is one of the following 3-restricted dominant weights {λ1 , λ , λ2 , λ −1 , 2λ1 , 2λ , λ1 + λ }. Since L(λ) is self-dual and 18, by Proposition 2.38, λ = λ1 + λ . If < 18, then by Theorem 4.4, Appendix A6 through A21 in [38], either λ = λ1 + λ or λ = λ( +1)/2 , for = 3, 5, 7. As dimL(λ1 +λ ) = only if 2 +2 −ε3 ( +1), it follows that dimL(λ1 +λ ) = 2m is even if and = 6k − 2, 6k − 1, 6k for some positive integer k. Using the constructions prior to Proposition 3.28, L(λ1 + λ ) ∼ V := V1 /(V1 ∩ V2 ). Let U be the subgroup of S consisting = of all matrices of the form diag(I2 , A), where A ∈ SLε−1 (3). Then U ∼ SLε−1 (3). For = ξ = ±1, let xξ = E1,2 + ξE2,1 + V1 ∩ V2 , when ε = +, and x+ = E1,2 + E2,1 + V1 ∩ V2 , x− = µE1,2 + µE2,1 + V1 ∩ V2 when ε = −. Then xξ ∈ V and Q(xξ ) = 0. It follows that xξ is a non-singular point in V, of plus or minus type depending on ξ and . As V1 ∩ V2 is 144 ﬁxed under natural action of U, and clearly, U centralizes xξ , it follows that U the stabilizer of xξ in S. We have 3m−2 |M : M xξ | S xξ , |Aut(S) : U | = [Aut(Lε+1 (3)) : 8, then 2 SLε−1 (3)] < 34 +2 . We deduce that 2m < 8 + 8. If that 2m = 2 + 2 − 1 > 8 + 8, so + 2 − ε3 ( + 1) 2 + 2 − 1 > 8 + 8, which is a contradiction. Thus we has the form 6k − 2, 6k − 1, 6k, must be one of assume that 2 7. However since the following values {4, 5, 6}. Computation shows that with n = + 1 : 32n−3 (3n − εn )(3n−1 − εn−1 ) (6 − 2ε)(3 − ξ) 1 + cxξ + d xξ = dξ = 32n−4 (3 − ξ) (1 + ξε)(3n−2 − εn−2 )32n−4 − 2 3−ε 32n−4 ((3n − εn )(3n−1 − εn−1 ) + 2(3n−2 − εn−2 )(3n−3 − εn−3 ) − (32 − ε2 )(3 − ε)) + (3 − ξ)(6 − 2ε) 32n−3 (3n − εn )(3n−1 − εn−1 ) 32n−4 (3 − ξ) 32n−4 (1 + ξε)(3n−2 − εn−2 ) − + −1 (6 − 2ε)(3 − ξ) 2 3−ε 32n−4 ((3n − εn )(3n−1 − εn−1 ) + 2(3n−2 − εn−2 )(3n−3 − εn−3 ) − (9 − ε2 )(3 − ε) − (6 − 2ε)(3 − ξ) cξ = We can check that equation (3.1) cannot hold in any of these cases. If V ∼ = = 3, λ = λ2 then dimL(λ2 ) = 6, and L4 (3) ∼ P Ω+ (3). If = 6 3 = 5, λ = λ3 , then N and dimV = 20. However, in this case, the Frobenius-Schur indicator of V is −1 so that S does not leave invariant a non-degenerate symmetric form on V. Finally if = 7, λ = λ4 , then V ∼ = 4 N and dimV = 70. Then V has a basis consisting of + 1. Fix an isomorphism +1 +1 ei1 ∧ ei2 ∧ ei3 ∧ ei4 , where 1 i1 < i2 < i3 < i4 N to F3 . The bilinear form on V is the projection of w1 ∧ w2 to N for any w1 , w2 ∈ V. Let w = e1 ∧ e2 ∧ e3 ∧ e4 + ξe5 ∧ e6 ∧ e7 ∧ e8 , ξ = ±1. Then w is a non-singular vector in V and (SL( e1 , · · · , e4 ) × SL( e5 , · · · , e8 )).2 ∼ (SL4 (3) × SL4 (3)).2 ﬁxes w . Thus = |Aut(L8 (3)) : (SL4 (3) × SL4 (3)).2| 316 (35 + 1)(36 − 1)(37 + 1)(38 − 1)/(8.26.80) < (334 + 1)/2. Hence 1 + c + d < (3m−1 + 1)/2 so that equation (3.1) cannot hold. Case f > 1. We ﬁrst consider case = 1. Since SL2 (q) ∼ SU2 (q) ∼ Sp2 (q), we = = 145 can assume that ε = +. Moreover, as L2 (9) ∼ A6 , we also assume that f = any 3-restricted dominant weight then λ = cλ1 , where 1 c 3. If λ is 2, dimL(cλ1 ) = c + 1 and L(cλ1 ) is self-dual. By Proposition 2.41, dimV = (dimΨ)f , for some irreducible kS-module Ψ. Clearly Ψ = L(λ) for some λ ∈ X + , hence dimV have |Aut(L2 (3f ))| 34f and m − 2 2f −1 − 2. If f (dimΨ)f 2f . We 6, then 2f −1 − 2 > 4f and hence m − 2 > 4f so that 3m−2 > 34f . By (3.10) and (3.11), equation (3.1) cannot hold. Thus, we can assume that 3 dimΨ > 2 then dimΨ also m − 2 f 5. As dimV is even, dimΨ is also even. If 4f , so that m − 2 22f −1 − 2, and 4. Then dimV = 2m 22f −1 − 2 > 4f. Thus equation (3.1) cannot hold. Therefore dimΨ = 2, and hence Ψ = L(λ1 ). Since Ψ is invariant under a non-degenerate alternating form, it follows that V is invariant under a product of f alternating forms, hence it ﬁxes a nondegenerate symmetric form if and only if f is even. Therefore, we conclude that f = 4 and S = SL2 (24 ). Now, we consider case 2. It follows from the case f = 1 that if Ψ is an irreducible k S-module which leaves invariant a non-degenerate symmetric form then either dimΨ 2 + 2 − ε3 ( + 1) 2 + 2 − 1 or = 3, 5 and dimΨ 6, 20, respectively. By Proposition 2.41, dimV = (dimΨ)f , for some irreducible k S-module Ψ. Hence dimV ( 2 + 2 − 1)f if = 3, 5 and dimV 6f , 20f if = 3, 5, respectively. Using (3.10) and (3.11), we need to consider the case S = SLε (9) and dimW = 36. However as 4 Lε (9) ∼ P Ωε (9), and L2 (34 ) ∼ Ω− (9). The results follow by Proposition 3.63. = = 4 4 6 We next consider the following embedding of classical groups by using twisted tensor products: Cln (q α ) → Clnα (q). These embedding can arise as follows, for example, take S = P Sp2 (32 ). Let (N, F32 , f1 ) be a classical symplectic geometry. For any a ∈ F32 , denote by a the image under the involutory ﬁeld automorphism of F32 . Let f2 be the nondegenerate symplectic form on N (1) . Then S preserves a non-degenerate quadratic form Q on N ⊗N (1) which satisﬁes Q(v⊗w) = 0 for v, w ∈ N and Q(v+w) = Q(v)+Q(w)+f (v, w), where f is the product form f1 f2 . Let β0 = {v1 , v2 , · · · , v2 −1 , v2 } be a basis for N such that 146 f1 (v2i−1 , v2i ) = 1, f1 (v2i−1 , v2j−1 ) = 0 = f1 (v2i , v2j ), 1 i, j . Let µ be a generator for i 2 ,1 j<k 2 }. F∗2 , and let β = {vi ⊗ vi , vj ⊗ vk + vk ⊗ vj , µvj ⊗ vk + µvk ⊗ vj , 1 3 By Proposition 2.4 and 2.5 in [41], S ﬁxes the non-degenerate symmetric form f on V = spanF3 (β), so that S we have: Proposition 3.62 Suppose that S ∼ P Sp2 (32 ). Embed S into P Ωε )2 (3) via V, where = (2 ε = (−) . Assume that not hold. Proof. For ξ ∈ {±1}, let zξ = v1 ⊗ v1 + ξv2 ⊗ v2 ∈ V. As Q(zξ ) = Q(v1 ⊗ v1 ) + Q(ξv2 ⊗ v2 ) + f (v1 ⊗ v1 , ξv2 ⊗ v2 ) = ξ = 0, zξ are non-singular in V. Let U = {v1 , v2 } N and W2 4. There is an M -orbit on Eε (V ) such that equation (3.1) does ξ P Ωε )2 (3) where ε = (−) . With these results and notations, (2 be the F3 span of {v1 ⊗ v1 , v2 ⊗ v2 , v1 ⊗ v2 + v2 ⊗ v1 , µv1 ⊗ v2 + µv2 ⊗ v1 }. Then P Sp(U ) ∼ = P Sp2 (9) embeds into P Ω− (3) as argument above, and by comparing the order, we have 4 an isomorphism. It follows that the stabilizer in P Sp(U ) of the point zξ is isomorphic to Ω3 (3). Clearly P Sp(U ⊥ ) ∼ P Sp2 −2 (9) ﬁxes zξ , hence Ω3 (3) × P Sp2 −2 (3) = Therefore [Aut(P Sp2 (9)) : (Ω3 (3)×P Sp2 −2 (9))] = 34 −3 (34 −1) m−2 = 2 2 S zξ . 38 −3 . As 2m = (2 )2 , − 2. Observe that m − 2 − (8 − 3) = 2 2 − 2 − 8 + 3 = 2 ( − 4) + 1 > 0 as 4. Thus 3m−2 > 38 −3 , and so equation (3.1) cannot hold by (3.10) and (3.11). Next we exhibit the embedding of P Ω± (32 ) into P Ωε )2 (3). Let λ be a root of x2 − x − 1 2 (2 in F3 . Then λ is a generator for F∗ . Let µ = λ2 . We have µ2 = −1 and µ3 = −µ. 9 Let (V, F9 , Q) be a classical orthogonal geometry of type ξ. First assume ξ = +. As 1 2 (9 − 1) = 2 is even, by Proposition 2.6(i), D(Q) = , hence N has a basis β such that fβ is I2 . By Proposition 2.7 in [41], N ⊗ N (1) has a basis β consisting of = vi ⊗ vi ui aij = vi ⊗ vj + vj ⊗ vi , i < j bij = µvi ⊗ vj − µvj ⊗ vi , i < j 147 where 1 i, j 2 . Moreover ε = (−) and all the values of the product form on this module lie in F3 , so that S = Ω+ (32 ) preserves a non-degenerate symmetric form 2 on W = spanF3 (β), hence S embeds into P Ωε )2 (3). Next, assume that ξ = −. As (2 1 2 (9 − 1) = 2 is even, by Proposition 2.6,(ii), D(Q) is a non-square, and hence by (iv), N has a basis β such that fβ is diag(λ, 1, · · · , 1). By Proposition 2.7 in [41] again, chose a basis β N ⊗ N (1) which consists = µλ−1 v1 ⊗ v1 , u1 ui = vi ⊗ vi , i > 1 a1j = xvi ⊗ yvj + vj ⊗ vi , 1 < j, aij = vi ⊗ vj + vj ⊗ vi , i < j b1j where 1 P Ω− (32 ) 2 i, j = yvi ⊗ vj − xvj ⊗ vi , 1 < j, bij = µvi ⊗ vj − µvj ⊗ vi , i < j 2 , and x2 + y 2 = λ−1 , also ε = (−) . As above, we have an embedding P Ωε )2 (3) via the module V = spanF3 (β). We will use the usual notation (2 (., .) for the product form on N ⊗ N (1) . Proposition 3.63 Let S = P Ωξ (32 ). Embed S into P Ωε )2 (3) via V, where ε = (−) . 2 (2 Assume that 3. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold. ξ Proof. Case ξ = +. Let x = v1 ⊗ v1 . Then (x, x) = (v1 ⊗ v1 , v1 ⊗ v1 ) = (v1 , v1 )(v1 , v1 ) = 1. Hence x is non-singular in V. Let H be the stabilizer in S of v1 . It follows from Proposition 4.1.6 in [29] that H ∼ −1 × Ω2( −1)+1 (9). Therefore | v1 S| = |S : H| = = 1 −1 9 (9 2 1 − 1) = 2 32 −2 (32 − 1). Observe that if g ﬁxes v1 then g ν also ﬁxes v1 . Thus g ﬁxes x = v1 ⊗ v1 if g ∈ H. It follows that H in S, this forces H = S x . We have A = |M : M x | H| 3.8.9 −1 (9 − 1) < 34 +1 since |Out(S)| 3, m − 2 − 4 − 1 = 2 2 S x . However, as H is maximal |Aut(S) : H| = |Out(S)||S : 2 48. As 2m = (2 )2 , m − 2 = 2 − 2. Now, since − 4 − 3 = 2 ( − 2) − 3 > 0, and hence 3m−2 > 34 +1 > 1 + c + d, so that equation (3.1) cannot hold in view of (3.10) and (3.11). Next, let y = v1 ⊗ v1 + v2 ⊗ v2 ∈ V. We have (y, y) = 2 = 1 = (x, x). Thus y is a nonsingular point in V which belongs to diﬀerent S-orbits to x .S. Let V1 = v1 , v2 , and 148 V2 = V1⊥ . Then sgnVi = + and V = V1 ⊥ V2 . Setting Ii = I(Vi ). We have I1 = g1 , g2 , where  λ 0   0 −1 g1 =   and g2 =  . −1 0 λ −1 0    It is easy to check that yg1 = λ4 v1 ⊗ v1 + λ−4 v2 ⊗ v2 = −y and yg2 = v2 ⊗ v2 + v1 ⊗ v1 = y. Hence MΩ := (I1 × I2 ) ∩ Ω(V ) = S y . By Proposition 4.1.6 in [29] that MΩ ∼ (Ω+ (9) × = 2 Ω+ −2 (9)).22 . Therefore, | y S| = |S : MΩ | = 2 1 2 −2 9 (9 16 − 1)(9 −1 + 1). If = 4 or = 4 but the triality τ of Ω+ (9) does not involved in G, then by Lemma 2.7.3 in [29], 8 Out(S) ∼ D8 × Z2 which is generated by r , r , δf ,β0 (λ) and φβ (ν), where β0 is a standard = basis of V. As three of these generators ﬁx y , |M y | we have |M : M y | We have A (8 − 6) = 2 3 2 −2 9 (9 8 2 3.16 |S 8 23 |MΩ |. Thus, in any cases, 3 2 −2 9 (9 8 : MΩ | Therefore A | y M| − 1)(9 −1 + 1). − 1)(9 −1 + 1) 2 3 2 −2 2 −1 9 .9 4 94 −3 = 38 −6 . Also m − 2 − 4, then −2−8 +6 = 2 − 8 + 4 = 2 ( − 4) + 4. Thus if clearly m − 2 > 8 − 6, so that 3m−2 > 38 −6 (3.10). Assume that = 3. Then A 1 4 3 9 (9 8 A, and equation (3.1) cannot hold by − 1)(92 + 1). Now, by (3.10) again, it is enough to check that 1 (3m−1 + 3) > 1 94 (93 − 1)(92 + 1) =: A0 . As m = 2.32 = 18, 2 8 1 m−1 (3 +3)−A0 2 1 = 8 (318 +317 +12)− 1 (318 +316 −312 −38 ) = 1 (317 −314 +312 +38 +12) > 0. 8 8 Case ξ = −. Let x = u1 and y = u2 . Then (x, x) = µ2 = −1 and (y, y) = 1, so that x, y are non-singular vectors in V. As in previous case, we can check that the stabilizers of x and y are isomorphic to −1 × Ω2 −1 (9), so that | x S| = | y S| = 1 9 −1 (9 − 1). 2 The result follows as above. Let K be a ﬁeld and N be a ﬁnite dimensional vector space over MΩ with a quadratic form Q and associated bilinear form (., .). Let T 0 (N ) = MΩ , T 1 (N ) = N, T 2 (N ) = N ⊗ N, · · · , T k (N ) = N ⊗ · · · ⊗ N , and deﬁne T (N ) = T 0 (N )⊕T 1 (N )⊕· · ·⊕T k (N )⊕· · · . The bilinear map T r (N ) × T s (N ) → T r+s (N ), (x, y) → x ⊗ y, x ∈ T r (N ), y ∈ T s (N ) deﬁnes an operation on T (N ). Then T (N ) has a structure of an algebra and call Tensor algebra. Let 149 k I be an ideal of T(N) generated by all elements v⊗v−Q(v).1 for v ∈ N. The factor algebra C(N ) = C(N, Q) = T (N )/I is called the Cliﬀord algebra of (N, Q). Identifying N with elements of degree 1 in C(N ), we have v 2 = Q(v) and uv + vu = (u, v) for any u, v ∈ N. Denote by T+ (N ) (resp. T− (N )) the sum of all T k (N ) for k even (resp. k odd). Then T (N ) = T+ (N )⊕T− (N ). Also I = I∩T+ (N )⊕I∩T− (N ). Let C+ (N ) = T+ (N )/(I∩T+ (N )) and C− (N ) = T− (N )/(I ∩ T− (N )). Then, C(N ) = C+ (N ) ⊕ C− (N ). The linear map J : C(N ) → C(N ) deﬁned by J(u) = u if u ∈ C+ (N ) and J(u) = −u if u ∈ C− (N ) is called the main involution. Also, there exists a unique anti-involution α : C(N ) → C(N ) such that for any v1 , v2 , · · · , vt ∈ N, (v1 v2 . . . vt )α = vt vt−1 . . . v1 ∈ C(N ). We next describe a basis for C(N ) with respect to some ﬁxed basis of N. Let β = {v1 , · · · , vn } be a basis for N over MΩ . Then for any increasing sequence 0 < i1 < i2 < · · · < ir n, the elements vi1 · · · vir together with 1 form a basis for C(N ), hence dimC(N ) = 2n . Also C+ (N ), (resp. C− (N )) are subspace of C(N ) spanned by vi1 · · · vir with r even (resp. r odd). We now assume that dimN = n = 2 is even and (N, K, Q) is an orthogonal geometry of plus type. Let β = {e1 , · · · e , f1 , · · · , f } be a standard basis for N. Denote by E and F the subspaces of N generated by {ei }i=1 and{fi }i=1 , respectively. Then N = E ⊕ F and E, F are maximal totally singular subspace of N. Let C(E) be the sub-algebra of C(N ) generated by E and set C+ (E) = C(E) ∩ C+ (N ) and C− (E) = C(E) ∩ C− (N ). As E is totally singular, C(E) can be identiﬁed with the exterior algebra of E. Put f = f1 · · · f , then C(N )f = C(E)f. We can deﬁne a representation ρ : C(N ) → End(C(E)) by the condition vuf = (ρ(v)u)f for v ∈ C(N ) and for all u ∈ C(E).[6] Proposition 3.64 (Proposition 5.4.9[29]) The spin representation embeds (i)   +  Ω (q)  2   q even q odd, q odd, 150 ≡ 0, 3 (mod 4) ≡ 1, 2 (mod 4). B (q) in Ω+ (q)  2     Sp2 (q) (ii)   +  Ω −1 (q)  2        If , q even q odd, q odd, ≡ 0 (mod 4) ≡ 2 (mod 4). D (q) in Ω+ −1 (q) 2 Sp2 −1 (q) is odd then the representation is not self-dual. The representation cannot be realized over a proper subﬁeld of Fq . (iii)   SU −1 (q)  2     +   Ω −1 (q 2 ) 2  Ω+ (q 2 )   2 −1      Sp −1 (q 2 ) 2 odd , q even q odd, q odd, ≡ 0 (mod 4) ≡ 2 (mod 4). 2 D (q) in The representation cannot be realized over a proper subﬁeld of Fq2 . Proposition 3.65 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M 2m is almost simple of type S, where S is simply connected of type B over F3f . There is an M − orbit on Eε (V ) such that equation (3.1) does not hold unless (L, S, λ) = (O10 (3), O5 (3), 2λ1 ), ξ in which case M has 5 orbits on E− (V ) and the equation holds for all M -orbits on E− (V ) ξ + + with r = t; or (L, S, λ) = (O8 (3), O7 (3), λ1 ) or (O16 (3), O9 (3), λ1 ), in which cases M has only one orbit on E+ (V ) so that 1G P ξ Table 1.2. 1G by Corollary 3.7, hence these cases appear in M Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eε (V ). ξ Case f = 1. We claim that if λ is a 3-restricted dominant weight with dimL(λ) is even and greater than dimN = 2 + 1, then λ must be one of the following weights: (i) λ = λ −1 , even, 4, and dimL(λ) = 2 2 + ; (ii) λ = 2λ , = 6k, 6k + 1 or 6k + 2, for some non-negative integer k, and dimL(λ) = 2 2 + 3 ,2 2 + 3 − 1, 2 2 + 3 , respectively; = 2 and dimL(λ) = 10. (iii) λ = λ1 , 2 8, and dimL(λ) = 2 or λ = 2λ1 , 151 Using (3.10) and (3.11), we have 3m−2 1) 32 2+ |Aut(S)| = |Aut(Ω2 +1 (3))| = 3 5 then 3 3 2 2i i=1 (3 − . Hence dimL(λ) = 2m 4 2 +2 +4. Notice that if (5 − 1)2 − 5 = 11 > 0, and so −4 2 −2 −4 5 2 − 4 2 − 2 − 4 = ( − 1)2 − 5 then dimL(λ) For 2 4 2 +2 +3 < 3 > 4 2 + 2 + 4. If > 11, , and hence by Theorem 5.1, in [38], λ is either λ −1 or 2λ . 11, by Theorem 4.4 in [38] and the upper bound for dimension of L(λ) above, again, λ is one of the weights above or case (iii) holds. It remains to get the restriction on in cases (i) and (ii). From the reference above, we also have dimL(λ −1 ) = (2 + 1) 2 and dimL(2λ ) = 2 only if + 3 − ε3 (2 + 1). Now case (i) holds as dimL(λ −1 ) is even if and 2 is even. For case (ii), we can check that + 3 − ε3 (2 + 1) is even exactly when has the forms given in (ii). We now consider case (i). As in Proposition 3.29, let v = e1 ∧x+ξx∧f1 = (e1 −ξf1 )∧x, where ξ = ±1. Since Q(e1 ∧ x) = 0 = Q(x ∧ f1 ), we have Q(v) = (e1 ∧ x, x ∧ f1 ) = ξ. Hence v is non-singular. Let N1 be the subspace of N generated by {e1 − ξf1 , x}. As N1 is ⊥ non-degenerate, N = N1 ⊥ N1 . Denote by H the centralizer of N1 in Ω(N ) ∼ Ω2 +1 (3). = It follows that H ∼ Ω2 −1 (3), and H ﬁxes v. Using 3.10) and (3.11) we have 3m−2 = 1+c+d 2 2 |Aut(Ω2 +1 (3)) : Ω2 −1 (3)| = 2 · 32 −1 (32 − 1) 4, 2 34 . Hence m − 2 < 4 , so that 0. + = 2m < 8 + 4. However as + − (8 + 4) = 2 (2 − 4) + ( − 4) Thus m − 2 4 + 2, a contradiction. For case (ii), let v = e1 ⊗ ei + ξf1 ⊗ f1 , where ξ ∈ {±1}. Then Q(v) = ξ, hence v is non-singular in L(2λ ). Let N1 = e1 , f1 . Then N1 is a non-degenerate subspace of N. As in case (i), let H be the centralizer of N1 in Ω(N ), as H ∼ Ω2 −1 (3), we have 3m−2 = |Aut(Ω2 +1 (3)) : Ω2 −1 (3)| < 34 , hence 2m < 8 + 4. 4, 2m = 2 2 +3 −1 2·4 +3 −1 = As 2m = dimL(2λ ) = 2 2 +3 −ε3 (2 +1), and 8 + (3 − 1) 8 + 11 > 8 + 4. This gives a contradiction to 2m < 8 + 4. = 2 then S = Ω5 (3) For (iii), if the last case holds, that is λ = 2λ1 and Ωε (3). We have m = 5 and by [13], we see that ε = − and there are 5 orbits of 10 non-singular points of each type with orbit sizes 4320, 2592, 2160, 540, 270 and (c, d) = 152 (2852, 1467), (1700, 891), (1412, 747), (332, 207), (152, 117), respectively. We see that c − 2d = −3m−1 − 1. Thus equation (3.8) holds so that equation (3.1) holds with r = t. For the spin representation λ = λ1 , by Proposition 3.64, S ﬁxes a non-degenerate symmetric form on L(λ1 ) if and only if an embedding Ω7 (3) = 4 then Ω9 (3) ≡ 0, 3 (mod 4). Thus = 3, 4, 7, 8. If = 3 then we have Ω+ (3), and by [9] p.140, O7 (3) has only one orbit on E+ (V ). If 8 Ω+ (3). Moreover, by 4.6.3(a) in [34], Ω9 (3) has only one orbit of each 16 = 7, then B7 (3) Ω+7 (3). However 2 type of non-singular points on its spin module. If this is not a maximal embedding as B7 (3) D8 (3) Ω27 (3) by Proposition 3.64. In fact, the spin module for B7 is the restriction of the spin module of D8 , in which B7 is the stabilizer of a non-singular point in the natural module for D8 . Finally, B8 (3) = 8. Then Ω+8 (3). Let V be the spin module and v+ , v− be the maximal and minimal vec2 tors correspondingly. Let P be the parabolic subgroup of S which stabilizes v+ . Then P is of type A −1 and P = U.L1 , where U is the unipotent radical and L1 is the Levi factor of P. Observe that L1 centralizes v+ and since P op = U op L1 which ﬁxes the point v− , it follows that L1 ﬁxes z = v+ + ξv− with ξ = ±1. Hence L1 ∼ SL (3) = Thus 1 + c + d |Aut(B8 (3)) : SL (3)| 8.336 (32 + 1) · · · (38 + 1) 16.371 Sz. 374 . Since 2m = 28 , m − 2 = 27 − 2 = 126 > 74, hence 3m−2 > 374 hold. 1 + c + d. Thus (3.1) cannot Case f > 1. Let Ψ be an even dimensional, self-dual irreducible kS-module. It follows from case f = 1 that dimΨ 2 2 2 2 if 2 6, and 2 2 6 and dimΨ 2 2 + otherwise. Note that 2 2 + if and only if + 3 − ε3 (2 + 1) + for all 2. By Propositions 2.41 and 2.39, dimV = (dimΨ)f , for some self-dual irreducible k S-module Ψ. As dimV = 2m is even, it follows that Ψ is an even dimensional, self-dual irreducible k S-module. As above, we have 3m−2 2m < 2f (2 (2 2 2 |Aut(Ω2 +1 (3f ))| < f ·3f (2 (2 2 2+ ) 3f (2 2+ +1) . Hence + + 1) + 4. If 2 7, then 2m + )f . Then by induction, we have 6. Then 2m 2f . Hence + )f > 2f (2 + + 1) + 4, a contradiction. Thus 2 153 2f < 2f (2 2 + + 1) + 4. This inequality holds except when ( , f ) = (2, 2), (2, 3), (3, 2). 10, and hence dimV Case Ω5 (9). We have S ∼ Sp4 (9). If dimΨ > 4, then dimΨ = 102 = 100. If this is the case then clearly 3m−2 > |Aut(S)|. Thus dimΨ = 4 and V is the twisted tensor product N ⊗ N (1) . By [13], there exists an element of order 5 with nonsingular eigenvectors of both types and has orbit size 99630 with (c, d) = (65528, 34101) and we can check that equation (3.1) cannot hold. Case Ω7 (9). If dimΨ > 8, then dimΨ 21, and hence dimV 212 = 441. Clearly 3m−2 > |Aut(S)| in this case. Thus dimΨ = 8 and V is the twisted tensor product N ⊗ N (1) . However, since Ω7 (9) Ω+ (9) 8 Ω+ (3), NG (S) is not maximal in G. 64 10, Case Ω5 (27). We have S = P Ω5 (33 ) ∼ P Sp4 (33 ). If dimΨ > 4, then dimΨ = and hence dimV 102 = 100. But we still have 3m−2 > |Aut(S)|. Thus dimΨ = 4 and dimV = 43 = 64. However, this case cannot occur as S preserves no non-degenerate symmetric form on V. Proposition 3.66 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M 2m is almost simple of type S, where S is simply connected of type C over F3f , with 3. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold or = 5, 6, 7, f = 1, ξ and λ = λ −1 or Table 1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eε (V ). ξ Case f = 1. Let λ ∈ X3 be a 3-restricted dominant weight such that L(λ) ∼ V. We = ﬁrst show that if dimV > 2 , then λ must be one of the following weights: (i) λ = λ −1 , dimL(λ) = 2 (ii) λ = 2λ , 2 = 3, f = 2, λ = (1 + 3i )λ , 1 i 2 and these cases appear in the − − 1 − ε3 ( ); 2 even and dimL(λ) = 2 + ; (iii) (λ, , dimL(λ)) = (λ1 , 3, 14), (λ2 , 4, 40), (λ3 , 5, 110). As |Aut(P Sp2 (3))| = f ·3 and hence 2m 4 2 2 i=1 (3 2i −1) 32 2 2+ , by (3.10) and (3.11), 3m−2 3 32 3 2+ , + 2 + 4. If 5 then 4 154 +2 +4 < , and so dimV < . Now, if > 11, then dimL(λ) = 2m < 3 , and hence by Theorem 5.1, in [38], λ is either λ −1 or 2λ . For 2 11, by Theorem 4.4 in [38] and the upper bound for dimension of L(λ) above, again, λ is one of the weights in (i), (ii) or (λ, , dimL(λ)) are as in (iii). This proves our claims. We have dimL(2λ ) = (2 + 1), L(2λ ) ∼ S 2 (V ), and dimL(λ −1 ) = 2 = 2 − −1− εp ( ), L(λ −1 ) ∼ w⊥ /( w ∩ w⊥ ), where w = e1 ∧ f1 + · · · + e ∧ f . In these cases, S leaves = invariant a quadratic form Q induced from the symplectic form on N. First, suppose that λ = 2λ . Let v = e1 ⊗e1 +ξf1 ⊗f1 ∈ V. Since dimL(2λ ) = (2 +1) is even, must be even. Let H be the centralizer in S of the subspace generated by {e1 , f1 }. Then H ∼ Sp2 −2 (3). = 2 2.3 (32 − 1) m−2 < By (3.10) and (3.11), we have 3 |Aut(P Sp2 (3)) : P Sp2 −2 (3)| = 3( −1)2 34 . Hence 2m < 8 +4. As 2m = (2 +1), it follows that (2 +1) < 8 +4. As 3 and is even, we conclude that 4. Then (2 +1) 4(2 +1) = 8 +4, a contradiction. Next, consider case λ = λ −1 . As dimL(λ −1 ) = 2 2 − −1−εp ( ) is even, = 6k −1, 6k or 6k +1. Now L(λ −1 ) has a basis consisting of ei ∧ ej , fi ∧ fj , 1 i<j , ei ∧ fj , 1 i=j , and ei ∧ fi − ei+1 ∧ fi+1 , i = 1, · · · , − 1 − εp ( ). Let zξ = e1 ∧ e2 + ξf1 ∧ f2 , where ξ = ±. Then zξ is non-singular and Q(zξ ) = −ξ. Let N1 = e1 , e2 , f1 , f2 be a subspace of N. Then ∧2 N1 has a basis {e1 ∧ e2 , f1 ∧ f2 , e1 ∧ f2 , e2 ∧ f1 , e1 ∧ f1 − e2 ∧ f2 }. Observe that zξ is ⊥ of type ξ. Since N1 is non-degenerate, N = N1 ⊥ N1 . Let H, K be the centralizer in S of ⊥ N1 , N1 , respectively. Then H ∼ Sp2( −2) (3), K ∼ Sp4 (3) ∼ Spin5 (3), K zξ ∼ Spinξ (3) = = = = 4 and H centralizes zξ , so that E := Spinξ (3) × Sp2 −4 (3) 4 S zξ , hence 1 + c + d |Aut(S) : E| = |Aut(P Sp2 (3)) : (Ωξ (3) × P Sp2 −4 (3))| < 38 −7 . Hence 2m < 16 − 10. 4 Since 2m = 2 2 − − 1 − εp ( ) 2 2 − 17 + 8 < 0. If contradiction. Thus 2 2 − − 2, we have 2 2 − − 2 < 16 − 10, or equivalent 0, a 8, then 2 2 − 17 + 8 = 2 2 − 16 − ( − 8) = (2 − 1)( − 8) < 8. Since = 6k − 1, 6k, 6k + 1, it follows that = 5, 6, 7. If ( , λ) are as in (iii), then by Appendix A.2 in [38], ind(L(λ)) = −, so that S does not ﬁx any non-degenerate quadratic form on L(λ). 155 Case f > 1. Let Ψ be an even dimensional, self-dual irreducible k S-module. It follows from case f = 1 that either dimΨ = 2 or dimΨ 2 2 − − 1 − ε3 ( ) 2 2 − − 2. By Propositions 2.41 and 2.39, dimV = (dimΨ)f , for some such self-dual irreducible k Smodule Ψ. By (3.10) and (3.11), we have 3m−2 2m < 2f (2 2 |Aut(P Sp2 (3f ))| < 3f (2 2+ +1) . Hence + + 1) + 4. We assume ﬁrst that dimΨ = 2 . Then 2m = (2 )f and so (2 )f < 2f (2 2 + + 1) + 4. By induction, we can see that this holds only when f = 2. Now when f = 2, by Proposition 3.62, equation (3.1) cannot hold either. Finally, assume that dimΨ 2 2 − − 2. Then we have (2 2 − − 2)f < 2f (2 2 + + 1) + 4. Using induction, this inequality cannot occur. Thus equation (3.1) cannot hold. Proposition 3.67 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m almost simple of type S, where S is simply connected of type D or 2 D over F3f . There is an M -orbit on Eε (V ) such that equation (3.1) does not hold so that M is not in Tables ξ 1.1-1.3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eε (V ). ξ Case f = 1. Let (N, F, Q) be a classical orthogonal geometry of type ε with S = Ω(N ). Let β = {e1 , · · · , f } be a standard basis for N. Let λ ∈ X3 be a 3-restricted dominant weight such that L(λ) ∼ V. We will show that if dimV > 2 , then λ must be one of the = following weights: (i) λ = λ −1 , dimL(λ) = 2 (ii) λ = 2λ , 2 − , l even; 2 even and dimL(λ) = 2 10; + − 1 − ε3 ( ); (iii) λ = λ1 , λ2 , 4 (iv) = 4, λ = λ1 + λ2 , λ1 + λ4 , λ2 + λ4 , dimL(λ) = 56. 32 4, 3 2− As |Aut(P Ωε (3))| 2 4 2 − 2 + 8. Since +2 , by (3.10) and (3.11), 3m−2 32 2− +2 , and hence 2m 3 − (4 2 − 2 + 8) = ( 2 + 2)( − 4) 0, and so dimV = 2m . By Theorem 5.1, in [38], if > 11, then λ is either λ −1 or 2λ . For 4 156 11, by Appendix A.41 in [38] and the upper bound for dimension of L(λ), λ is one of the weights above or λ appears in (iii) and (iv). Firstly assume λ = λ −1 . Then V ∼ L(λ) ∼ ∧2 N and dimV = 2 = = As dimV = 2m is even, 2 − = (2 − 1). must be even. For ξ = ±, let zξ = (e1 − ξf1 ) ∧ (e2 + f2 ) ∈ V. Then zξ is non-singular in V. Let U = e1 − ξf1 , e2 + f2 . We can check that D(U ) = −ξ and sgn(U ) = ξ. It follows that sgn(U ⊥ ) = ξε. Let H = Ω(U ⊥ ). Then H ∼ Ωεξ−2 (3) and = 2 H ﬁxes zξ , so that H m − 2 = 1 (2 2 m−2 2 S zξ . Thus | zξ M | |Aut(S) : H| 34 . Since 2m = 2 2 − , − − 4). By (3.10) and (3.11) again, 3m−2 34 . This is equivalent to 8. This inequality is 1 4 , and so 2 (2 2 − −4) 4 or equivalently ( −4)(2 −1) satisﬁed only when = 4. Suppose that = 4. Then m = 14. Assume that ε = +. As Ω(U ), Ω(U ⊥ ) and re2 +f2 , re2 −f2 ﬁx zξ , we have |M : M zξ | H = P Ω(U ) ⊗ P Ω(U ⊥ ) ∼ P Ωξ (3) ⊗ P Ωξ (3) = 2 6 |M : M zξ | 1 7 2 3 (3 +1)(3+ξ)(33 +ξ) 2 1 |Out(S)|.|S 4 : H| 6|S : H|, where S zξ and |Out(S)| 24 as f = 1. Thus < 313 = 3m−1 . Assume that ε = −. As 1 (q−1) = 4 2 , and so by Proposition 2.8.2 in [29], S zξ 1 14 3 . 8 is even, it follows from Proposition 2.6 that D(Q) = |Out(S)| = 4. Therefore |M : M zξ | 4.|S : H| where H ∼ (Ωξ (3) ⊗ Ω−ξ (3)).22 = 2 6 1 6 4 3 (3 +1)(3−ξ)(33 +ξ) 16 which is the stabilizer in S of U. Therefore |M : M zξ | 1 1 However 1 (3m−1 + 3) − 8 314 = 1 (313 + 3) − 1 314 = 1 (4.313 + 12 − 314 ) = 8 (313 + 12) > 0 2 2 8 8 and hence |M : M zξ | < 1 (3m−1 + 3) which contradicts to (3.11). 2 Secondly assume that λ = 2λ . Then dimL(λ) = 2 is even, this implies that k 1. It follows that 2 + − 1 − ε3 ( ). Since dimL(λ) must have one of the following forms 6k − 1, 6k, 6k + 1, where 5, and so |Out(S)| = 2df 8. As in Proposition 3.31, take zξ = e1 ⊗ e1 + ξf1 ⊗ f1 + w ∩ w⊥ . Let H be the centralizer in S of e1 , f1 . Then H ﬁxes zξ and H ∼ P Ωε −2 (3). Thus |M : M zξ | = 2 2 2 + −1−ε3 ( ) 2 2 + −2, we have m−2 |Aut(S) : H| 34 −1 . Since 2m = 4 −1. 1 (2 2 + 2 −6). Thus 1 (2 2 + −6) 2 5. But this is equivalent to ( − 4)(2 + 1) Next assume that λ = λ1 , and 4 0, which is absurd as 10 so that V ∼ L(λ1 ) is the spin module. From = 157 Proposition 3.64, we have ε = + and ≡ 0(mod 4). Thus = 4, 8. However, if = 4, then dimL(λ1 ) = 23 = 8, and so this case cannot happen. Thus = 8 and D8 (3) Ω+7 (3). 2 Argue as in Proposition 3.65, with zξ = v+ + ξv− , we have SL8 (3) ﬁxes zξ so that |M : M zξ | 359 < 363 = 3m−1 as m = 26 = 64. This contradicts to (3.10). Finally assume that case (iv) holds. Without lost, we can take λ = λ1 + λ4 . Then V = L(λ) ∼ ∧3 N, where N is the natural module for S, and 2m = 56 so that m = 28. = Let zξ = e1 ∧ e2 ∧ e3 + ξf1 ∧ f2 ∧ f3 ∈ V. Then zξ is non-singular. Let K be the stabilizer of a totally singular decomposition e1 , e2 , e3 ⊕ f1 , f2 , f3 in P Ω( e1 , ·, f3 ). Then K ∼ P GL3 (3).2 and K ﬁxes zξ . Hence |M : M zξ | = |Out(S)||S : K| Case f 323 < 326 = 3m−2 . Thus (3.1) cannot hold. 2. Let Ψ be an even dimensional, self-dual irreducible k S-module. It 2 2 |Aut(P Ωε (3)) : K| 8 follows from case f = 1 that dimΨ = 2 or dimΨ when 4 − when 8 or dimΨ = 2 −1 7. By Propositions 2.41 and 2.39, dimV = (dimΨ)f , for some self-dual irreducible k S-module Ψ of even degree. If dimΨ = 2 then 2m = (2 )f . By (3.10) and (3.11), we have 3m−2 1 + f (2 2 |Aut(P Ωε (3f ))| < 3.3f (2 2 2− +1) or equivalently 2f −1 f −2 < − + 1). This happens only when f = 2. The result follows from Proposition 8, then dimV (dimΨ)f (2 2 3.63. Assume that dimΨ > 2 . If − )f . As above, 1 we have 2 (2 2 − )f − 2 < 1 + f (2 2 − + 1). Clearly this inequality cannot happen. Thus 4 7. Then dimV (dimΨ)f 2f ( −1) . Hence 2f ( −1)−1 − 2 < 1 + f (2 2 − + 1). This holds only when = 4 and f = 2. In this case, S = D4 (32 ) and Ψ is the spin representation of S. Since the spin representation of S can be obtained from the natural representation of S via triality, (3.1) cannot hold by Proposition 3.63. The proof is now completed. Proposition 3.68 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M 2m is almost simple of type S, where S is of exceptional type above and deﬁne over F3e with e 1. There is an M -orbit on Eε (V ) such that equation (3.1) does not hold unless ξ 158 (L, S, λ) = (P Ωε (3), F4 (3), λ1 ). In this case M is in the Table 1.3. 52 Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eε (V ). ξ (a) Case G2 . By (3.10) and (3.11), 3m−2 |Aut(G2 (3e ))| 315e . Thus dimV = 2m 30e + 4. Assume ﬁrst that e = 1. Then dimV = 2m dimL(λ) > 500, a contradiction. Assume that e 34. From Appendix A.49 in [38], 2. By Propositions 2.41 and 2.39, 2m = (dimΨ)e , for some self-dual irreducible k S-module Ψ of even degree. It follows that 2m 500e , and hence 30e + 4 500e , a contradiction. |Aut(F4 (3e ))| 353e . So 2m 106e + 4. (b) Case F4 . By (3.10) and (3.11), 3m−2 Assume that e = 1. Then dimV 110. It follows from Appendix A.50 in [38], λ = λ1 , L(λ) = L(F4 ), the simple Lie algebra of type F4 over F3 , and dimL(λ) = 52. In this case, we have an embedding F4 (3) 52e Ωε (3). Assume that e 52 2. 379e . Thus 2m 158e+4. 572, a 2. We have 2m 52e , so that 106e + 4. But this cannot happen for any e (c) Case ε E6 . By (3.10) and (3.11), 3m−2 |Aut(ε E6 (3e ))| Assume that e = 1. Then dimV contradiction. Assume that e 162. From Appendix A.51 in [38], dimL(λ) 2. Then clearly 572e > 158e + 4. |Aut(E7 (3e ))| 3134e . Thus 2m (d) Case E7 . By (3.10) and (3.11), 3m−2 268e+4. If e = 1 then by Appendix A.52 and A.2 in [38], dimL(λ) that e 2. Clearly 1330e > 268e + 3. |Aut(E8 (3e ))| 1330, a contraction. Assume (e) Case E8 . By (3.10) and (3.11), 3m−2 3249e . Thus 2m 498e+4. Assume that e = 1. From Appendix A.53 in [38], λ = λ8 , dimL(λ8 ) = 248 and V = L(λ8 ) = L(E8 ). Let Φ denote the root system of E8 and Π the fundamental system of roots of Φ. Let α0 be the highest root of Φ. Then the element eα0 in L(E8 ) is a maximal vector with high weight λ8 . Let ξ = ±1 and z = eα0 +ξe−α0 ∈ L(E8 ). Then z is non-singular in V and clearly E7 is contained in the stabilizer of z . Hence 1+c+d |E8 (3) : E7 (3)| < 3120 . Since dimV = 248, we have m = 124, so that 3m−2 = 3122 > 3120 . Therefore equation (3.1) cannot hold. Assume that e 2. Clearly 248e > 498e + 3 for any e 159 2. (f ) Case 3 D4 . By (3.10) and (3.11), 3m−2 If e = 1 then dimV |Aut(3 D4 (3e ))| 330e . Thus 2m 60e+4. 64. From Appendix A.53 in [38], λ = λ4 and dimL(λ4 ) = 8, or λ = λ3 , dimL(λ3 ) = 28 or λ = λ1 + λ2 , dimL(λ1 + λ2 ) = 56. Observe that λ1 , λ1 + λ2 are not invariant under the triality of D4 , but λ3 is invariant. Firstly, assume that λ = λ1 . Since the splitting ﬁeld for 3 D4 (3) is F33 , we have embedding 3 D4 (3) Ω+ (33 ) 8 Ω+ (3) 24 where the latter is the over-ﬁeld subgroup embedding. Thus this cannot happen. For the remaining cases, the same argument will lead to a contradiction. Finally, λ = λ3 . Then L(λ3 ) can be realized over F3 , hence we can take V ∼ L(λ3 ). If N is the natural module = + for O8 (33 ) then V ∼ ∧2 N. However this still does not gives rise to maximal embedding = as 3 D4 (3) Ω+ (33 ) 8 Ω+ (3). If e 28 2. 2 then by Proposition 5.4.8 in [29], dimV 24e . Clearly 24e > 60e + 4 for any e (g) Case 2 G2 . By (3.10) and (3.11), 3m−2 32e + 20. By Proposition 2.41, dimV can check that this cannot hold. Embedding of Sporadic groups |Aut(2 G2 (32e+1 ))| 38(2e+1) . Thus 2m 1. We 72e+1 , and so 72e+1 32e + 20, where e Proposition 3.69 Assume G is nearly simple primitive rank 3 of type Ωε (3) and M is 2m almost simple of type S, where S is a simple sporadic group. There is an M -orbit on Eε (V ) ξ such that equation (3.1) does not hold unless (S, L) = (2.Co1 , Ω+ (3)), in which case M 24 has only 2 orbits on E+ (V ) for only one type of points and so 1G P ξ In this case M is in Table 1.2. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eε (V ). ξ By (3.10) and (3.11), we have |Aut(S)| 3m−2 , hence 2m 2log3 (|Aut(S)|) + 4 = g3 (S). 1G by Corollary 3.7. M Since V is an absolutely irreducible F3 S-module with ind(V ) = + and dimV is even, by Lemma 3.33 and [19], we only need to consider the following cases: (S, dimV ) = (M11 , 10), (M12 , 10), (M23 , 22), (M24 , 22), (Co3 , 22), (Co1 , 24). Observe ﬁrst that the ﬁrst four cases 160 do not give rise to maximal embedding since Mn An P Ωε n−1−ε3 (n) (3), where the last embedding is obtained via the fully deleted permutation module for An . Therefore we need to consider the last two cases. Assume that S = Co1 and dimV = 24. It follows that m = 12, and sgnV = + by [13]. Observe that Co2 is a maximal subgroup of Co1 and it is also a stabilizer of a non-singular point, say x. Then |xS| = 98280 < 3m−1 = 177147 and (3.1) cannot hold in view of (3.10). For other type of points, inside Co2 , the maximal subgroup HS : 2 of Co2 ﬁxes a non-singular point y, by checking the orders, we can see that Sy = HS : 2 and so |yS| = 46872483840. Next choose an 2A-element g in Co1 . + Then the normalizer of g in Co1 is H := 21+8 . O8 (2). Choose a non-singular point z of the + same type as that of y in the eigenspace of g of dimension 8. We have |Hz | = 165150720. We now look for an involution h in Co2 not contained in H which ﬁxes z. Then H, h is exactly the stabilizer of z in Co1 , and |zS| = 165150720. In fact the stabilizer of z in Co1 is isomorphic to 211 : M23 which is a subgroup of index 24 in the maximal subgroup 211 : M24 of Co1 . Clearly these are two distinct S-orbits of the same type. As 1 |yS| + |zS| = |Eε (V )| = 2 311 (312 − 1), S has only two orbits on Eε (V ), hence (3.1) holds by Corollary 3.7. Finally assume that S = Co3 and dimV = 22. We have m = 11 and by [13], we have sgnV = +. We also see that M cL : 2 is the stabilizer of a non-singular point x with orbit size |xS| = 276 < 3m−1 . For other type of non-singular points, we can ﬁnd a non-singular point y with |yS| = 23054625, and we see that (3.1) cannot hold. 161 Appendix A Classification of nearly simple primitive rank 3 groups Let G be a primitive rank 3 group of ﬁnite degree n. Then one of the following holds: (i) T × T G T0 Z2 , where T0 is a 2−transitive group of degree n0 , the socle T of T0 is simple and n = n2 ; 0 (ii) G is an aﬃne group, i.e. G has a regular elementary abelian normal subgroup and n is a prime power; (iii) L := soc(G) is simple. Combining results of Kantor and Liebler ([28]), Liebeck and Saxl ([37]) and Bannai ([3]), we get the list of all nearly simple primitive rank three groups: Theorem A.1 (1) Let L be one of the groups Sp2m−2 (q), Ω± (q), Ω2m−1 (q) or SUm (q) 2m for m 3 and q a prime power. Let L G with G/Z(L) Aut(L/Z(L)). Assume that G acts as a primitive rank 3 permutation group on the set E of cosets of a subgroup P of G. Then at least one of the following holds up to conjugation under Aut(L/Z(L)). (i) E is an L-orbit of singular (or isotropic) points; (ii) E is an L-orbit of maximal totally singular (or isotropic) subspaces and L is one of the following groups: Sp4 (q), SU4 (q), SU5 (q), Ω− (q), Ω+ (q) or Ω+ (q); 6 8 10 (iii) E is any L-orbit of non-singular points and L = SUm (2), Ω± (2), Ω± (3) or Ω2m−1 (3); 2m 2m (iv) E is either orbit of nonsingular hyper-planes of L = Ω2m−1 (4) or Ω2m−1 (8), where G = Ω2m−1 (8) · 3 in the latter case; (v) L = SU3 (3), L ∩ P = P SL3 (2); (vi) L = SU3 (5), L ∩ P = 3 · A7 ; (vii) L = SU4 (3), L ∩ P = 4 · P SL3 (4); (viii) L = Sp6 (2), P = G2 (2); (ix) L = Ω7 (3), L ∩ P = G2 (3); (x) L = SU6 (2), L ∩ P = 3 · P SU4 (3) · 2; (2) Let L = P SLn (q) G Aut(L). Assume that G acts as a primitive rank 3 permutation group on the set E of cosets of a subgroup P of G. Then one of the following 162 n 351 416 2016 130816 n0 Table A.1: L is an exceptional group. L L∩P k; l {fs ; ft } G2 (3) U3 (3) · 2 126, 224 168; 182 G2 (4) J2 100, 315 65; 350 G2 (4) U3 (4) · 2 975, 1040 650; 1365 G2 (8) SU3 (8) · 2 32319, 98496 18468; 112347 E6 (q) D5 -parabolic k0 , l0 f0 , g0 comment two classes two classes G = G2 (8) · 3 two classes Table A.2: G ∼ Am = m n = [Am : P ] |P | Structure of P Decomposition of (1P )Am m 5 1 m(m − 1) (m − 2)! H(Σ2 ) ∩ Am [m] + [m − 1, 1] + [m − 2, 2] 2 6 15 24 H(Π2 ) ∩ A6 [6] + [4, 2] + [32 ] 8 35 576 H(Π4 ) ∩ A8 [8] + [6, 2] + [42 ] 9 120 1512 P Γ L2 (8) [9] + [42 , 1] + [5, 14 ]1 10 1260 1440 H(Π5 ) ∩ A10 [10] + [8, 2] + [6, 4] holds up to conjugation under Aut(L). (i) E is the set of lines for L, n 4; (ii) L = P SL2 (4) ∼ P SL2 (5), |E| = 5 , = 2 L = P SL2 (9) ∼ A6 , |E| = 6 , = 2 L = P SL4 (2) ∼ A8 , |E| = 8 , or = 2 G = P Γ L2 (8), |E| = 9 ; 2 (iii) L = P SL3 (4), L ∩ P = A6 ; (iv) L = P SL4 (3), L ∩ P = P Sp4 (3). (3) Let G be a nearly simple primitive rank 3 group of exceptional Lie type, that is, L/Z(L) G/Z(L) Aut(L/Z(L)), L is an exceptional group of Lie type and G acts as a primitive rank 3 permutation group on the set E of cosets of a subgroup P of G. Then L, L ∩ P are given in Table A.1. n0 = (q 12 − 1)(q 9 − 1)/(q 4 − 1)(q − 1); k0 = q(q 8 − 1)(q 3 + 1)/(q − 1), l0 = q 8 (q 5 − 1)(q 4 + 1)/(q − 1); f0 = q(q 9 − 1)(q 4 + 1)/(q 3 − 1), g0 = q 2 (q 6 + 1)(q 5 − 1)(q 4 + 1)/(q − 1). (4) L = Am , m 5. Let G is a nearly simple primitive rank 3 group of Alternating type Am , m 5 with point stabilizer P. Then m and P are given in Table A.2 and A.3. We denote by H(Σr ) (resp. H(Σr ) ∩ Am ) the subgroup of Sm (resp. Am ) which consists of the elements ﬁxing some given subset Σr with r < [m/2]. We also denote by H(Πu ) (resp. H(Πu ) ∩ Am ) the subgroup of Sm (resp. Am ) which consists of elements permuting a ﬁxed nontrivial complete set of blocks πi with |πi | = u. (5) The list of nearly simple primitive rank 3 groups of sporadic type is in Table A.4. 163 Table A.3: G ∼ Sm = m n = [Sm : P ] |P | Structure of P 1 m 5 2 m(m − 1) 2(m − 2)! H(Σ2 ) 6 15 48 H(Π2 ) 8 35 1152 H(Π4 ) 10 1260 2880 H(Π5 ) Decomposition of (1P )Sm [m] + [m − 1, 1] + [m − 2, 2] [6] + [4, 2] + [23 ] [8] + [6, 2] + [42 ] [10] + [8, 2] + [6, 4] L M11 M12 M22 M22 M23 M23 M24 M24 J2 HS M cL Suz Co2 Ru F i22 F i22 F i23 F i23 F i24 Table A.4: L P ∩L n M9 .2 55 M10 .2 66 24 · A6 77 A7 176 M21 · 2 253 4 2 · A7 253 M22 · 2 276 M12 · 2 1288 P SU3 (3) 100 M22 100 P SU4 (3) 275 G2 (4) 1782 P SU6 (2) · 2 2300 2 F4 (2) 4060 2 · P SU6 (2) 3510 Ω7 (3) 14080 2 · F i22 31671 P Ω+ (3) · S3 137632 8 F i23 306936 is a sporadic simple group k; l {fs ; ft } 18; 36 10; 44 20; 45 11; 54 16; 60 21; 55 70; 105 21; 154 42; 210 22; 230 112; 140 22; 230 44; 231 23; 252 495; 792 252; 1035 36; 63 36; 63 22; 77 22; 77 112; 162 22; 252 416; 1365 780; 1001 891; 1408 275; 2024 1755; 2304 783; 3276 693; 2816 492; 3080 3159; 10920 429; 13650 3510; 28160 782; 30888 28431; 109200 30888; 106743 31671; 275264 57477; 249458 comment two classes two classes two classes 164 Appendix B Nearly simple groups of type Ωε (2), SUm(2) and sporadic 2m We also obtain similar results when G is nearly simple primitive rank 3 of type L, where L is either Ωε (2), m 3, or SUm (2), m 4 and G acts on the L-orbit E(V ) of non-singular 2m points in the natural module V for L. The proof of Theorem B.1 is omitted. Theorem B.1 Let L be one of the following groups Ωε (2), m 4 or SUm (2), m 4, 2m and G be nearly simple primitive rank 3 of type L. Let P be the stabilizer of a non-singular point in V. Let M be any maximal subgroup of G. Then 1G 1G unless the pairs (L, M ) M P appear in Tables B.1-B.3. For reference purposes, we include here the similar results for almost simple primitive rank 3 groups of type S, where S is a simple sporadic group. The proof relies on [13]. Theorem B.2 Let L be a simple sporadic group and G be almost simple primitive rank 3 of type L. Let P be a maximal subgroup of G such that G acts as a primitive rank 3 permutation groups on the set of cosets of P. Let M be any maximal subgroup of G. Then 1G 1G unless the triples (G, P, M ) appear in Tables B.4-B.7. M P As all the permutation characters of maximal subgroups of groups in Theorem B.2 are stored in [13], except F i22 .2, HS.2 and F i24 .2, we can verify the Tables in Theorem B.2 easily. For the remaining groups, we will use the package ‘atlasrep’ to get the maximal subgroups and then apply equation (3.1) to determine the character containment. The following GAP code will produce a list of maximal subgroups together with their permutation characters of a group with identiﬁer ”name”. gap>permchars:=function(name) >local output, id, tbl, maxtab, chi, max, i; > tbl :=CharacterTable(name);output := [ ]; > if HasMaxes(tbl)=false then Print(‘fail’); return fail; else max :=Maxes(tbl); ﬁ; > for i in [1..Size(max)] do maxtab :=CharacterTable(max[i]); > id :=TrivialCharacter(maxtab); chi :=InducedClassFunction(maxtab, id, tbl); > output[i] := [max[i],PermCharInfo(tbl, chi).ATLAS [1]]; od; > return output; > end; 165 Table B.1: M ∈ C type of M conditions Pα 1 α m − O2a S2 ε=+ − O2 S4 ε=+ + α O2b (2 ) b 2, α = 2, 3 − O2b (23 ) b 2 − 2 O2b (2 ) b 2 GUm (2) SUn (2) GU1 (2) × GUn−1 (2) Pα 1 α n/2 GU1 (2) S4 GUa (2) S2 GUa (23 ) Spn (2) L ε Ω2m (2) Remarks orbits 2 m = 2a 2 m=4 2 m = bα α − 1 m = 3b 2 m = 2b 2 1 2 2 2 2 1 n=4 n = 2a n = 3a n even Table B.2: M ∈ S socle of M modules Remarks orbits + Ωn−1−ε2 (n) (2) An , n = 7, 9, 16 λ = (n − 1, 1) 2 − Ω10 (2) A12 λ = (n − 1, 1) 2 Ω− (2) L2 (7) ∼ L3 (2) r=s 3 = 8 − Ω14 (2) G2 (3) 2 − Ω14 (2) U3 (3) r=s 6 Ω+ (2) Lε (2) λ3 2 20 6 − Ω14 (2) P Sp6 (2) λ2 r=s 3 Ω+ (2) P Sp6 (2) Spin module 1 8 + Ω16 (2) P Sp8 (2) Spin module 1 + Ω22 (2) Co2 Leech lattice 2 Ω+ (2) Co1 Leech lattice 1 24 SU6 (2) M22 2 SU9 (2) J3 2 L 166 Table B.3: Exceptions L socle of M modules ε Ω2m (2) P Sp2 (2) λ −1 Ω+ (2) P Sp2 (2) Spin module 2 Ω+ (2) P Ω+ (2) Spin module 32 12 − Ω26 (2) F4 (2) adjoint module Ω− (2) E6 (2) adjoint module 78 + Ω56 (2) E7 (2) minimal module Ω− (2) Ru 28 SU9 (2) 34 .Sp4 (3) SU40 (2) S8 (3) Weil module SUn (2) Chev(2f ) Remarks = 5, 6, 7 = 5, 6 M ∈ C6 (G) G M11 M12 M12 Table B.4: Sporadic Groups P M 1G Remarks orbits P 2 3 : Q8 .2 1a + 10a + 44a A6 . 2 2 L2 (11) 1 M10 : 2 1a + 11a + 54a M11 1a + 11a 2 M11 1a + 11b 1 M10 : 2 1a + 11b + 54a 2 L2 (11) 1 M9 : S3 2 2 × S5 3 2 4 : D12 3 A4 × S3 3 M10 : 2 1a + 11b + 54a M11 1a + 11a 1 M11 1a + 11b 2 M10 : 2 1a + 11a + 54a 2 L2 (11) 1 M9 : S3 2 2 × S5 3 2 4 : D12 3 A4 × S3 3 167 G M22 P 2 : A6 4 M22 A7 M22 .2 24 : S6 M23 L3 (4) : 22 M23 24 : A7 M24 M22 : 2 M24 M12 : 2 HS M22 HS.2 M22 : 2 Table B.5: Sporadic Groups (continue) M 1G Remarks P 1a + 21a + 55a L3 (4) 1a + 21a A7 1a + 21a + 154a 1a + 21a + 154a L3 (4) 1a + 21a 4 2 : A6 1a + 21a + 55a 1a + 21a + 55a L3 (4).22 1a + 21a L2 (11).2 1a + 22a + 230a M22 1a + 22a 23 : 11 1a + 22a + 230a M22 1a + 22a 23 : 11 1a + 23a + 252a M23 1a + 23a M12 : 2 1a + 252a + 1035a 26 : 3. S6 26 : (L3 (2) × S3 ) L2 (23) L2 (7) 1a + 252a + 1035a M23 1a + 23a M22 : 2 1a + 23a + 252a 4 2 : A8 L3 (4) : S3 L2 (23) 1a + 22a + 77a U3 (5) : 2 1a + 175a, 2 classes S8 4. 24 : S5 2 × A6 . 22 5 : 4 × A5 1a + 22a + 77a S8 : 2 1+6 2+ : S5 (2 × A6 · 22 ).2 51+2 : [25 ] + 5 : 4 × S5 168 orbits 2 2 2 2 2 2 2 1 2 1 2 2 2 2 1 4 1 2 2 2 4 1 2 2 2 1 2 2 2 1 1 Table B.6: Sporadic Groups (continue) G P M 1G Remarks orbits P J2 U3 (3) 1a + 36a + 63a . 3 P GL2 (9) 2 21+4 : A5 2 − A4 × A5 2 A5 × D10 1 2 5 : D12 2 J2 .2 U3 (3) : 2 1a + 36a + 63a 3. A6 · 22 2 1+4 2− : S5 2 (A4 × A5 ) : 2 2 (A5 × D10 ) : 2 1 2 5 : (4 × S3 ) 1 M cL U4 (3) 1a + 22a + 252a 31+4 : 2S5 2 + . 2 A8 2 51+2 : 3 : 8 2 + c M L.2 U4 (3).23 1a + 22a + 252a U3 (5) : 2 2 31+4 : 4S5 2 + . 2 S8 2 51+2 : (24 : 2) 2 + Suz G2 (4) 1a + 780a + 1001a . 32 U4 (3) : 23 2 U5 (2) 1 1+6 2− · U4 (2) 2 5 3 : M11 1 M12 : 2 2 2+4 2 3 : 2(A4 × 2 ).2 2 Suz : 2 G2 (4) : 2 1a + 780a + 1001a 32 . U 4(3).(22 )133 2 U5 (2) : 2 1 21+6 · U4 (2).2 2 − 5 3 : (M11 × 2) 1 M12 : 2 × 2 2 32+4 : 2(S4 × D8 ) 2 169 G Co2 Ru P U6 (2) : 2 2 F4 (2) F i22 2 . U6 (2) F i22 : 2 2 . U6 (2) F i22 O7 (3) F i23 2 . F i22 F i23 + O8 (3) : S3 Table B.7: Sporadic Groups (continue) M 1G Remarks orbits P 1a + 275a + 2024a M cL 2 1a + 783a + 3276a 6 (2 : (U3 (3)) : 2 3 (22 × Sz(8)) : 3 2 3+8 2 : L3 (2) 3 U3 (5) : 2 3 A8 4 L2 (29) 1 1a + 429a + 3080a O7 (3) 2 classes 2 + O8 (2) : S3 2 2 F4 (2) 1 1a + 429a + 3080a G2 (3) : 2 2 + O8 (2) : S3 × 2 2 2 F4 (2) 1 1a + 429a + 13650a 2 . U6 (2) 2 + O8 (2) : S3 × 2 2 210 M22 2 2 F4 (2) 2 1a + 782a + 30888a + O8 (3) : S3 2 1+8 1+6 1+2.2S4 3+ .2− .3+ 2 3 7 3 .[3 ].(2 × L3 (3)) 2 1a + 30888a + 106743a 2 . F i22 2 2. 2 U6 (2).2 3 211 . M23 2 170 List of References [1] M. Aschbacher, Finite Group Theory, 2 ed., Cambridge Studies in Advanced Mathematics, vol. 10, Cambridge University Press, 2000. [2] M. Aschbacher, R. Guralnick, and K. 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Rank 3 Permutation Characters and Maximal Subgroups-PhD Thesis by Hung Phi Tong-Viet | Papers by Hung Phi

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