Rank 3 permutation characters and maximal subgroups more

Permutation characters, Classical Groups, Maximal Subgroups, and Compact Riemann surfaces

RANK 3 PERMUTATION CHARACTERS AND MAXIMAL SUBGROUPS HUNG P. TONG-VIET Abstract. Let G be a transitive permutation group acting on a ﬁnite set E and let P be the stabilizer in G of a point in E. We say that 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 (or permutation H module) over C of G on the cosets G/H. Let H and K be subgroups of G. G G if 1G − 1G is a character of G. Also a ﬁnite group G is We say 1H 1K K H called nearly simple primitive rank 3 on E 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 cosets of some subgroup of L. In this paper we classify all maximal subgroups M of a nearly simple primitive rank 3 group G of type L = Ω2m+1 (3), m 3, acting on an L-orbit E of non-singular points of the natural module for L such that 1G 1G , where P is the stabilizer of a P M non-singular point in E. This result has an application to the study of minimal genera of algebraic curves which admit group actions. 1. Introduction Let G be a transitive permutation group acting on a ﬁnite set E and let P be the stabilizer of a point in E. 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 H cosets G/H over C. We also use the same notation 1G for the permutation module. H Let H, K be subgroups of G. Consider the permutation characters 1G and 1G , we H K say 1G 1G if 1G − 1G is a character of G. In terms of permutation modules, H K K H 1G if and only if 1G is isomorphic to a submodule of 1G . A ﬁnite group L 1G K H H K 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 or almost simple with socle L if L G Aut(L) for some ﬁnite simple group L. It follows from deﬁnitions that if G is nearly simple of type L then G/Z(L) is almost simple of type L/Z(L). Assume that m 3 is an integer. Let L be one of the following quasi-simple groups Ω2m+1 (3), Ωε (3), Ωε (2) or SUm (2). Let G be a 2m 2m 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 and let P be the stabilizer of a non-singular point in E(V ). Then G is a primitive rank 3 group on E(V ). In this situation, we say that G is a nearly simple primitive rank 3 group of type L. In this paper, we shall classify all maximal subgroups M of a nearly simple primitive rank 3 group G of Date: April 26, 2011. 1991 Mathematics Subject Classiﬁcation. Primary 20B15; secondary 20C20. Key words and phrases. permutation characters, nearly simple groups, maximal subgroups. 1 2 HUNG P. TONG-VIET Table 1. M ∈ C L type of M conditions orbits ε Ω2m+1 (3) O1 (3) ⊥ O2m (3) Pα 1 α m 2 O1 (3) S2m+1 (2m + 1, ξ, r) = (5, ±, t) 2 (7, +, t) 2 Oα (33 ) 2m + 1 = 3α 3 Table 2. M ∈ S L socle of M Ω7 (3) A9 Ω7 (3) P Sp6 (2) Ω13 (3) P Sp6 (3) Ω7 (3) G2 (3) Ω25 (3) F4 (3) modules orbits λ = (8, 1) 2 2 λ2 2 λ1 , λ 2 1 λ4 2 Ref 5.19 5.20 5.23 5.25 Ref 5.3 5.6 5.10 5.13 Table 3. Possible exceptions L socle of M Ω41 (3) S8 (3) E6 (3) Ω77 (3) Ω133 (3) E7 (3) modules λ1 adjoint module adjoint module Ref 5.23 5.25 type Ω2m+1 (3) such that 1G 1G . The remaining types of nearly simple primitive M P rank 3 groups of type L have been considered in [33]. Theorem 1.1. Let L = Ω2m+1 (3), with m 3, and let G be a nearly simple primitive rank 3 group of type L. Let P be the stabilizer of a non-singular point in 1G unless the pairs (L, M ) V. Let M be any maximal subgroup of G. Then 1G M P appear in Tables 1-3. Remark 1.2. For all pairs (L, M ) in Tables 1 and 2, 1G is not contained in 1G . P M The pairs in Table 3 are the cases that we have not determined whether or not there is a containment. 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. Let X be a compact Riemann surface with G ∼ Aut(X). Let be a prime such that ( , |G|) = 1. Let JX = be the Jacobian of X and let JX [ ] be the -torsion points of JX . It is well known that JX [ ] is a characteristic module and the genus of the quotient curve X/P can be computed using the relation 2g(X/P ) = dim(JX [ ]P ), where P G and JX [ ]P is the ﬁxed point space of JX [ ] under P (see [9, Lemma 8.1]). Now combining the latter fact with the Frobenius reciprocity, we deduce that if P, M are proper subgroups of G such that 1G 1G , then g(X/P ) g(X/M ). For a proof of this P M result see [9, Corollary 8.2]. Theorem 1.1 with the discussion above now yield the following. RANK 3 PERMUTATION CHARACTERS 3 Corollary 1.3. Assume the assumption and notation of Theorem 1.1. Let X be a compact Riemann surface with G Aut(X), and inertia groups g1 , . . . , gr over X/G. Then g(X/P ) g(X/M ) for any maximal subgroup M of G which does not appear in Tables 1-3. If G is doubly transitive on a non-empty set E with point stabilizer P, then either 1G 1G or G = P M for any maximal subgroup M of G. The maximal P M factorization of almost simple groups was classiﬁed completely by M. Liebeck, C. Praeger and J. Saxl in [25], so for almost simple groups, we can tell exactly for which M we have the containment 1G 1G . In case of rank 3, M. Aschbacher, P M R. Guralnick and K. Magaard [1] 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 [9]. 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 [10], 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 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 belongs to a collection C(G) of geometric subgroups 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) have been determined by P. Kleidman and M. Liebeck in [21] 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 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 1G . From this P M assumption, we will get an 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 [10]. As in the almost simple doubly transitive case, we can get a list of maximal subgroups M such that 1G 1G . In Table 1, we list all the cases when M ∈ C(G), P M Table 2 contains all the cases when M ∈ S(C), and in the last table, we list the cases that we have not determined whether or not there is a containment. Notice that we only have a ﬁnite number of exceptions in Table 3. Also there is a ﬁnite number of cases in Table 2. 4 HUNG P. TONG-VIET For the notation 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 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 ‘5.3’ means that the case is dealt with in Proposition 5.3. 2. Preliminaries We adopt the constructions and notation of [21]. Fix a ﬁnite ﬁeld Fq , q a prime power. Let V be an Fq vector space of dimension n. The general linear group GL(V ) of V over Fq is a group of all non-singular Fq -linear transformations of V. The special linear group of V over Fq , SL(V ), is the group of all elements of GL(V ) with determinant 1. The projective linear groups P GL(V ) and P SL(V ) are obtained by factoring out the scalar matrices in the corresponding linear groups. For any subgroup X of GL(V ), we write P X or X for the corresponding projective group X/X ∩ F∗ . q A map g from V to itself is called an Fq -semilinear transformation of V if there is a ﬁeld automorphism σ(g) ∈ Aut(Fq ) such that for all v, w ∈ V and λ ∈ Fq , (2.1) (v + w)g = vg + wg and (λv)g = λσ(g) (vg) The general semilinear group of V over Fq , Γ L(V ) consists of all non-singular Fq -semilinear transformations of V. As F∗ Γ L(V ), we can factor out the scalars q to get the projective general semilinear group P Γ L(V ). Let κ be a left linear or a quadratic form on V. Observe κ is a map from V k to Fq where k = 1, 2. Deﬁne I(V, Fq , κ) = {g ∈ GL(V ) | κ(vg) = κ(v) for all v ∈ V k }; S(V, Fq , κ) = I(V, Fq , κ) ∩ SL(V ); Ξ(V, Fq , κ) = {g ∈ Γ L(V ) | κ(vg) = τ (g)κ(v)σ(g) for all v ∈ V k } where τ (g) ∈ F∗ , σ(g) ∈ Aut(Fq ), and q Λ(V, Fq , κ) = {g ∈ Ξ(V, Fq , κ) | σ(g) = 1}. Deﬁne A= and certain subgroup of index 2 in S in case O; S otherwise; where ι is an inverse-transpose automorphism of GL(V, Fq ). We get a sequence of groups: Ω S I Λ Ξ A. Note that in [21], Ξ(V, Fq , κ), and Λ(V, Fq , κ) are denoted by Γ (V, Fq , κ), and ∆(V, Fq , κ), respectively. For more details, see [21, Chapter 2]. Let (V, Fq , Q) be a classical orthogonal geometry with q odd, and dimV = n. Let ε = sgn(Q) be the sign of the quadratic form Q. Note that ε = ◦ when n is odd, otherwise, ε is either + or −. A square or a non-square in F∗ will be denoted q by or , respectively. If W is a non-degenerate subspace of V then we write sgn(W ) = sgn(QW ), where QW is the restriction of Q to W. 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 Q(v) = Q(x). When n is even, for any γ ∈ Fq , we denote by S ε (n, γ) the number of all vectors v ∈ V with Q(v) = γ. For x ∈ V \ {0}, a one-space with representative x will be called a point in V and denoted by x . We now deﬁne a Ω= Ξι Ξ in case L with n otherwise; 3; RANK 3 PERMUTATION CHARACTERS 5 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 dimV is even, then ρ(x) = Q(x). If dimV is odd, then ρ(x) = sgn(x⊥ ). Assume that dimV is odd. Let x be a nonsingular vector in V. We say x is a plus vector if ρ(x) = +; and x is a minus vector if ρ(x) = −. We also say that x is of plus or minus type according to whether its representative x is a plus or a 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 points of type ξ in V, ξ where ε = sgn(Q). When ε = ◦, we will write Eξ (V ) instead of E◦ (V ). ξ Lemma 2.1. Let (V, Fq , Q) be a classical orthogonal geometry with dimV odd. Two non-singular vectors x, y (two non-singular points x , y ) have the same type if and only if Q(x) ≡ Q(y) (mod (F∗ )2 ). Hence for any non-singular vector z, we q have Eρ(z) (V ) = { v ⊆ V | Q(v) ≡ Q(z) (mod (F∗ )2 )}. In particular if q = 3, then q Eρ(z) (V ) = { v ⊆ V | Q(v) = Q(z)}. Proof. Assume that Q(x) ≡ Q(y) (mod (F∗ )2 ). By [21, Proposition 2.5.4(ii)], x q 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 [21, Proposition 2.5.4(i)], there exists an isometry between x⊥ and y ⊥ . This isometry can extend to an isometry g of V such that (x⊥ )g = y ⊥ . Since (x⊥ )⊥ = x , and (y ⊥ )⊥ = y , ( x )g = y . Thus xg = µy for some µ ∈ F∗ . Therefore Q(x) = Q(xg) = Q(µy) = q µ2 Q(y). The other statements are obvious. The following lemma will be used to compute the Higman rank 3 parameters for orthogonal groups. Lemma 2.2. Let (V, Fq , Q) be a classical orthogonal geometry with q odd and ε = sgn(Q). q 2k−1 + ε(q k − q k−1 ) if γ = 0, (1) if dimV = 2k and γ ∈ Fq , then S ε (2k, γ) = q 2k−1 − εq k−1 if γ = 0; (2) if dimV = 2k + 1 and x ∈ V − {0} then S(2k + 1, x) = q 2k + ρ(x)q k . Proof. Statement (1) follows from [11, Proposition 9.10]. For (2), let γ = Q(x) and ξ = ρ(x). Assume ﬁrst that x is a non-singular vector. Then V = x ⊥ x⊥ , sgn(x⊥ ) = ξ and dim(x⊥ ) = 2k. For any vector v ∈ V with Q(v) = γ, write v = ϕx+v0 , where ϕ ∈ Fq and v0 ∈ x⊥ . We have 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 . Assume that x is a singular vector. Observe that V always contains a non-singular vector y. Let η = ρ(y) and µ = Q(y). We have V = y ⊥ y ⊥ , sgn(y ⊥ ) = η, Q(y) = µ ∈ F∗ and dim(y ⊥ ) = 2k. q For any v ∈ V with Q(v) = 0, write v = ϕy + v0 , where ϕ ∈ Fq 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 . The proof is complete. By the classiﬁcation of the ﬁnite simple groups, every non-abelian ﬁnite simple group is either an alternating group An , n 5, a ﬁnite simple group of Lie type, or 6 HUNG P. TONG-VIET Table 4. Upper bounds for the size of Aut(L). L f (L) 2 Ln (q), n 3 qn 2 q 2n +n+1 P Sp2n (q) 2 Un (q), n 3 qn 2 P Ω+ (q), n = 4 q 2n −n+1 2n P Ω+ (q) 3q 29 8 − 2n2 −2n+3 P Ω2n (q) q 2 q 2n +n+1 Ω2n+1 (q) E6 (q) q 79 E7 (q) q 134 L f (L) E8 (q) q 249 F4 (q) q 53 2 E6 (q) q 79 G2 (q) q 15 3 D4 (q) 3q 29 2 F4 (q) q 27 Sz(q) q6 2 q8 G2 (q) − 2n2 −2n+2 P Ω2n (q) 2q one of the 26 sporadic simple groups. The next lemma gives an upper bound for the full automorphism groups of the non-abelian ﬁnite simple groups of Lie type. The proof will be omitted. Lemma 2.3. If L is a ﬁnite simple group of Lie type, then |Aut(L)| f (L) is given in Table 4. f (L), where Let G be a ﬁnite group and P, M be subgroups of G. Denote by M \G/P, a set of representatives for the 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 2.4. 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 |. P M Proof. (i) Suppose that M \G/P = {x1 , . . . , xk }. If i, j ∈ {1, . . . , k} with i = j, then P xi M ∩ P xj M = ∅. Clearly P xi M are distinct orbits of M on G/P. As G = ∪k P xi M, {P xi M }k is a complete set of orbits of M on G/P. i=1 i=1 (ii) By [17, Corollary 5.5], the number of orbits of M on G/P is the inner product 1 (1M , (1G )M ) = |M | m∈M 1G (m), and so (1M , (1G )M ) = |M \G/P | by (i). Now by P P P the Frobenius reciprocity [17, Lemma 5.2], we have (1G , 1G ) = (1M , (1G )M ), so M P P that (1G , 1G ) = |M \G/P | as required. M P Note that (ii) could follow easily from the Mackey’s formula. For the representation theory of symmetric groups and the Mullineux conjecture, we refer to [19], [7] or [8]. Let λ = (λ1 , λ2 , . . . , λk ) n be a partition of n, and let p be a prime. We denote by [λ] the Young diagram of the partition λ, which consists of n nodes placed in decreasing rows. The rim of [λ] is its south-east border. 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 (see [7] or [8]). 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 prim 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, RANK 3 PERMUTATION CHARACTERS 7 M (λ) = h1 r1 h2 r2 ... ... hk . rk Now the p-regular partition m(λ) of n is deﬁned via M (m(λ)) = where εi = 0, 1, if p | hi if p hi h1 s1 h2 s2 ... ... hk , sk and si = hi − ri + εi . Note that the partition m(λ) can be reconstructed from the Mullineux symbol M (m(λ)). The following notation 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 . . . ak F rp (λ) =  b1 b2 . . . bk  ε1 ε2 . . . ε k where ai = hi − ri and bi = ri − εi . If λ has p-modular Frobenius symbol F rp (λ) as constructed above then the Mullineux map m is deﬁned by b1 F rp (m(λ)) = a1 ε1  b2 a2 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 the Mullineux map if and only if the ﬁrst two rows of F rp (λ) are the same (see [7]). 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) (see [8]), where Dλ denote the irreducible Sn -module in characteristic p, corresponding to the p-regular partition λ. Let λ be a 3-regular partition of n with λ1 n − 2. We will show that λ is rarely a ﬁxed point of the Mullineux map. Lemma 2.5. Assume that p = 3, n 5 and λ is a p-regular partition of n. Suppose that λ1 n − 2. Then m(λ) = λ if and only if 5 n 6 and λ = (n − 2, 12 ). Proof. As λ1 n−2, λ1 = 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 from [22, Lemma 1.8]. Assume that λ = (n − 2, 12 ). We ﬁrst compute the Mullineux symbol M (λ) of λ. We have 5 3 ... 3 a M (λ) = , 3 1 ... 1 1 where 3, and so 1, occurs t times with t = n−2 ] −1, and 0 a = n−2−3(t+1) 3 Hence   2 2 ... 2 a − 1 0 . F r3 (λ) = 2 1 . . . 1 1 0 ... 0 1 2. 8 HUNG P. TONG-VIET If t 1, or equivalently n 8, then clearly, the ﬁrst two rows of F r3 (λ) cannot be equal, so that λ = m(λ). Thus 5 n 7 and t = 0. Then   2 a−1 0 . F r3 (λ) = 2 1 1 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. We next prove a gap result between the minimal module and the second minimal module for alternating groups in characteristic 3. For k 1, denote by Rn (k) the set of irreducible Sn -modules D such that D ∼ Dλ or D ∼ Dm(λ) for some p-regular = = partition λ n, with λ1 n − k. Lemma 2.6. Let F be a splitting ﬁeld for An of characteristic p = 3. Suppose that n 12 and V is an irreducible FAn -module with dimV > n. Then dimV (n2 − 5n + 2)/2. Proof. It follows from [7, Theorem 2.1] that V = Dλ ↓An with m(λ) = λ or λ V = D± , where m(λ) = λ. Let U = Dλ for the partition λ obtained above. By [29, Proposition 2.2], either dimU > (n − 2)(n − 3), or U ∈ Rn (2). Observe that 1 dimV = dimU if m(λ) = λ and if m(λ) = λ then dimV = 2 dimU. Thus if dimU > 1 1 1 2 (n − 2)(n − 3), then clearly, dimV 2 dimU > 2 (n − 2)(n − 3) > 2 (n − 5n + 2). Therefore we can assume U ∈ Rn (2). Hence 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 [19, Theorem 6] that µ is neither (n) nor (n−1, 1), and so µ = (n−2, 2), or (n−2, 12 ). By Lemma 2.5, µ is not ﬁxed under the Mullineux map. Also, as the Mullineux map is an involutory map, m(m(µ)) = µ = m(µ). We conclude that Dµ and Dm(µ) are irreducible upon restriction to An . Since dimDm(µ) = dimDµ ⊗sgnn = dimDµ , we have dimV = dimDλ = dimDµ . Finally, the result follows from [18, Theorem 24.1, 24.15]. 3. 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 |∆(x) ∩ ∆(y)| = |Γ (x) ∩ Γ (y)| = Suppose that k l. µ if y ∈ Γ (x) λ if y ∈ ∆(x) λ1 µ1 if y ∈ Γ (x) if y ∈ ∆(x). Lemma 3.1. ([13, Lemma 5, 7]). Let G act primitively rank 3 on E. Then (i) |E| = k + l + 1; (ii) µl = k(k − 1 − λ); (iii) D = (λ − µ)2 + 4(k − µ) is a square; (iv) λ1 = l − k + µ − 1; RANK 3 PERMUTATION CHARACTERS 9 (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 → y∈∆(x) y and x → y∈Γ (x) y, for x ∈ E, and extending linearly. We have that 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. ([13, Lemma 6]). The eigenvalues diﬀerent from k of ∆ are: √ √ λ−µ− D λ−µ+ D and t = . s= 2 2 Lemma 3.3. ([1, 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. ([13]). We have 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 fs = 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 and β : VM → V by β(yg) = Σm∈M xmg for g ∈ G. 1 Then α and β are CG maps and θ = |P ||P ∩M | β ◦ α ∈ EndCG (V ). The map θ can be written in terms of the linear maps ∆ and Γ as follows: Lemma 3.5. ([1, 2.1]). We have θ=I+ d∆ cΓ + . k l Proof. For any g ∈ G, we have 1 θ(xg) = β(α(xg)) |P ||P ∩ M | 1 = β(ypg) |P ||P ∩ M | p∈P 1 = xmpg. |P ||P ∩ M | p∈P,m∈M Let O be one of the three orbits of P on E and vO = u∈O 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 | p∈P xmp = |O| vO . 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 θ(xg) = (vx + + )(x)g = (I + + )(xg). k l k l This proves the lemma. For r = s, t, let πr be the projection of V on Vr . 10 HUNG P. TONG-VIET Lemma 3.6. ([1, 2.2, 2.3]). Let r = s or t, then (1) θ ◦ πr = 0 if and only if Vr ker(θ); (2) If θ ◦ πr = 0 then α : Vr → VM is an injective CG map; (3) For r = s, t : θ ◦ πr = 0 if and only if dr (r + 1)c = . k l Proof. Statement (1) is clear as πr (V ) = Vr . By Lemma 3.4, Vs is not CGisomorphic to Vt , so that θ : Vr → Vr . As Vr is an 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: (3.1) 1+ dr (r + 1)c d∆ cΓ + ) ◦ πr = (1 + − )πr . k l k l dr (r + 1)c Thus θ ◦ πr = 0 if and only if 1 + − = 0. This ﬁnishes the proof. k l 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 P P 1 is the trivial character, and χs , χt are irreducible characters of G, aﬀorded by the irreducible modules Vs , Vt , with degrees fs , ft , respectively. From deﬁnition, 1G if 1G − 1G is a character of G. This is equivalent to (χr , 1G ) > 0 for 1G M P M M P any r ∈ {s, t}. By Lemma 3.6, for r ∈ {s, t}, (χr , 1G ) = 0 if and only if equation M (3.1) holds for any M -orbits in E. Note that the parameters c and d depend on x, or equivalently, on the conjugate of P. When we pick a diﬀerent conjugate 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 1G . Otherwise we need to change to CG map from Vr to VM and hence 1G M P diﬀerent conjugate of P or diﬀerent point in E. See Proposition 5.13 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. θ ◦ πr = (I + 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 let M be any subgroup of G. If M has at most two orbits on E, then 1G 1G . P M Proof. By way of contradiction, suppose that 1G 1G . Then φ = 1G − 1G is a P M M P G character of G. Since φ and 1P are characters of G, we have that (1G , 1G ) = (1G , φ + 1G ) = (1G , 1G ) + (1G , φ) = 3 + (1G , φ) P M P P P P P P By Lemma 2.4, (1G , 1G ) is the P M E with G/P, M has (1G , 1G ) P M 3, number of orbits of M on G/P. Now, by identifying 3 orbits on E, a contradiction. 4. Main Hypothesis and Notations From now on, we assume the following set up. Let L = Ω2m+1 (3) with m 2. Let V be the natural module for L over F = F3 . Denote by Eξ (V ) an L-orbit of non-singular points of type ξ in V. Let G be a nearly simple primitive rank 3 group of type L acting on Eξ (V ). Observe that L G I(V ), where I(V ) is the full RANK 3 PERMUTATION CHARACTERS 11 isometric group of V. Denote by P the stabilizer of a point in Eξ (V ). The letter M will be reserved for the maximal subgroup of G. If M ∈ C(G) and X is a group satisfying Ω(V ) X Ξ(V ), then there exists a subgroup H ∈ C(Ξ) such that H ∩ G = M and the subgroup H ∩ X ∈ C(X) is called the X-associate of M and 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 irreducible 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 ) = +. 5. Character containment for nearly simple groups of type L 5.1. Higman rank 3 parameters for L. We now assume the hypotheses and notation in Section 4 with L = Ω2m+1 (3), m 2. There are two types of nonsingular 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) = Eξ (V ) ∩ x⊥ and Γ (x) = Eξ (V ) ∩ (V − x⊥ − { x }). Then P has exactly three orbits { x }, ∆(x) and Γ (x) on the set Eξ (V ). Recall the Higman rank 3 parameters deﬁned in Section 3. For ξ = ±, we write ξ1 to denote +1 or −1 when ξ = + or ξ = −, respectively. Lemma 5.1. Let ξ ∈ {±} and x ∈ Eξ (V ). Then 1 (i) |Eξ (V )| = 2 3m (3m + ξ1); 1 m−1 m (3 − ξ1); (ii) k = 2 3 (iii) l = (3m − ξ)(3m−1 + ξ1); (iv) µ = λ = 1 3m−1 (3m−1 − ξ1); 2 √ (v) D = 2 · 3m−1 ; (vi) s = 3m−1 ; (vii) t = −3m−1 . Proof. Let γ = Q(x). We have ρ(x) = sgn(x⊥ ) = ξ. By Lemma 2.1, we obtain V Eξ (V ) = { v ⊆ V | Q(v) = γ}. Observe that each point v has 2 representatives v 1 and −v, by Lemma 2.2, |Eξ (V )| = 2 S(2m + 1, x) = 1 (32m + ξ3m ) = 1 3m (3m + ξ1), 2 2 which gives (i). From deﬁnition we get k = |∆(x)| = |Eξ (V ) ∩ x⊥ | = |{ v ⊆ x⊥ | Q(v) = Q(x)}|. 1 1 By Lemma 2.2 again, k = 2 S ξ (2m, γ) = 2 S ξ (2m, γ) = 1 (q 2m−1 − ξq m−1 ). The 2 parameter l can be computed from 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). We have 1 1 λ = |∆(x) ∩ ∆(y)| = |{v ∈ x⊥ ∩ y ⊥ | Q(v) = Q(x)}| = S(2m − 1, z), 2 2 where z ∈ W := x⊥ ∩ y ⊥ = x, y ⊥ with Q(z) = Q(x). We need to determine ⊥ ⊥ ρW (z) = sgn(zW ). Since x⊥ = y ⊥ (y ⊥ ∩ x⊥ ) = y ⊥ W and W = zW ⊥ z , ⊥ ⊥ ⊥ we deduce that x = y, z ⊥ zW , where dim(zW ) = 2m − 2, dimW = 2m − 1 and dim y, z = 2. As D y, z = det(diag(−γ, −γ)) = , by [21, Proposition 2.5.13], ⊥ sgn y, z = (−)1 = −. It follows from [21, Proposition 2.5.10] that sgn(zW ) = 12 HUNG P. TONG-VIET −sgn(x⊥ ) = −ξ. Thus by Lemma 2.2, λ = 1 (32m−2 − ξ3m−1 ), which gives (iv). 2 The remaining parameters follow from Lemmas 3.1 and 3.2. Corollary 5.2. 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 (5.1) (5.2) (iii) If m (5.3) 2 then 1+c+d c − 2d = ξ3m − 1. (ξ3m−1 + 1)(ξ3m − 1 + c − 2d) = 2c (ii) If (ξ, r) = (+, t) or (−, s) then equation (3.1) has the form 3m + 1 3m−1 . 2 Proof. From deﬁnitions we have c 0 and d 0. Let A = 1 + c + d. Multiply both sides of equation (3.1) by l, we have (r + 1)c = l + drl/k. Subtracting drl/k from both sides, drl = l. (5.4) (r + 1)c − k (1) If ξ = +, r = s, then by Lemma 5.1, we have r = 3m−1 , l = (3m −1)(3m−1 +1), and l/k = 2(3m−1 + 1)/3m−1 . From (5.4), (3m−1 + 1)c − 2d(3m−1 + 1) = (3m − 1)(3m−1 + 1). Dividing both sides by 3m−1 + 1, we get c − 2d = 3m − 1 = ξ3m − 1. Hence c = 2d + 3m − 1. Thus A = 3m + 3d 3m (3m + 1)/2. (2) If ξ = −, r = t, then by Lemma 5.1, r = −3m−1 , l = (3m + 1)(3m−1 − 1), and l/k = 2(3m−1 − 1)/3m−1 . From (5.4), (−3m−1 + 1)c + 2d(3m−1 − 1) = (3m + 1)(3m−1 − 1). Dividing 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 = +, r = t, then by Lemma 5.1, r = −3m−1 , l = (3m − 1)(3m−1 + 1), and l/k = 2(3m−1 + 1)/3m−1 . From (5.4), (−3m−1 + 1)c + 2d(3m−1 + 1) = (3m − 1)(3m−1 + 1) 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 0, 2d(3m−1 + 1) (3m−1 + 1)(3m − 1), or 2d 3m − 1. Now 2A = 2 + 2c + 2d 2 + 2c + 3m − 1 3m + 1. (4) If ξ = −, r = s, then by Lemma 5.1, r = 3m−1 , l = (3m + 1)(3m−1 − 1), and l/k = 2(3m−1 − 1)/3m−1 . From (5.4), 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). RANK 3 PERMUTATION CHARACTERS 13 As d A Since 0, we have c (3m−1 − 1)(3m + 1)/(3m−1 + 1). Thus 1 + (3m−1 − 1)(3m + 1)/(3m−1 + 1) = (32m−1 − 3m−1 )/(3m−1 + 1). 0, 32m−1 − 3m−1 3m + 1 (3m − 1 + 3m+1 (3m−2 − 1) − = m−1 + 1 3 2 2(3m−1 + 1) we obtain A (3m + 1)/2 > 3m−1 . 5.2. Permutation characters of maximal subgroups in C. By [21, Proposition 2.6.2], 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 M Ω M MI . 5.2.1. The reducible subgroups C1 . The reducible subgroups in C1 (Ξ) are all the groups MΞ of the forms NΞ (W ), where dimW = α, 1 α 2m + 1 and W is either non-degenerate or totally singular. The corresponding subgroups are of type ε2 ε Oα1 (3) ⊥ O2m+1−α (3) or Pα , where ε1 = sgn(W ), ε2 = sgn(W ⊥ ). The subgroup + MΞ is maximal in Ξ except when MΞ is of type O2 (3) ⊥ O2m−1 (3), as it is contained ± in the subgroups of type O1 (3) ⊥ O2m (3). ε2 ε Proposition 5.3. Assume M is of type Oα1 (3) ⊥ O2m+1−α (3). There is an M orbit on Eξ (V ) such that equation (3.1) does not hold unless M is of type O1 (3) ⊥ ε2 O2m (3). In this case M is in Table 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 = sgn(Wi ). Write Xi = X(Wi ), where X ranges over the symbols Ω, S and I. By [21, Lemma 4.1.1], we have MI = I1 × I2 and Ω1 × Ω2 MΩ . Without loss of generality, we can assume dimW1 = 2b + 1, where 0 b < m. If b = 0, then the proposition holds. Assume b 1. Then W1 contains non-singular points of both types. Let xi , i = 1, 2 be non-singular points in W1 of diﬀerent types. By [21, Lemma 2.10.5], 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 parameters for 1 Eη (W1 ). By Lemma 5.1, d = 2 3b−1 (3b −η1), c = (3b −η1)(3b−1 +η1), 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 (5.1) holds then η3b − 1 = ξ3m − 1. This implies that b = m, a contradiction. Thus (5.2) holds. Then (3m−1 + ξ1)(3m − ξ2 + ξη3b ) = 2(3b − η1)(3b−1 + η1). Observe that m − 1 b. Assume ﬁrst that m − 1 = b and ξ = −η. We have 3m−1 + ξ1 = 3b − η1 and 3m − ξ2 + ξη3b = 3b+1 + η2 − 3b = 2(3b + η1). Clearly 2(3b + η1) > 2(3b−1 + η1), and hence equation (5.2) does not hold in this case. Next assume that m−1 = b and ξ = η. Then (3m−1 +ξ1)(3m −ξ2+ξη3b ) = (3b +ξ1)(3b+1 − ξ2 + 3b ) = 2(3b + η1)(2 · 3b − η1) > 2(3b−1 + η1)(3b − η1), a contradiction. Finally assume that m − 1 > b so that m − 1 b + 1. Then 3m−1 + ξ 3b+1 + ξ1 > 3b − η1 and 3m − ξ2 + ξη3b 9 · 3b − ξ2 − 3b = 8 · 3b − ξ2 > 2(3b−1 + η1). Multiplying these two inequalities side by side will lead to a contradiction. Thus equation (5.2) does not hold. Therefore b = 0 and so W is a point or a hyperplane. Let Fq be a ﬁnite ﬁeld of size q and let W be a totally singular subspace of V of dimension α. By Witt’s Lemma, we can assume that W has a basis 14 HUNG P. TONG-VIET {e1 , . . . , eα }, where the vectors ei are taken from a standard basis of V. Let X = eα+1 , . . . , em , fα+1 , . . . , fm , a and Y = f1 , . . . , fα . 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 5.4. Let (V, Fq , Q) be a classical orthogonal geometry with sgn(Q) = +. Assume that dim(V ) = 2m and β = {e1 , . . . , em , f1 , . . . , fm } is a standard basis of V. Let W1 = e1 , . . . , em , W2 = f1 , . . . , fm , and T0 = NI(V ) (W1 , W2 ), where V = β . Then (i) T0 ∼ GLm (q) and T0 acts naturally on W1 . = (ii) As T0 -modules we have W2 ∼ W1 . = ∗ (iii) T0 ∩ Ω(V ) = {x ∈ T0 | detW1 (x) ∈ (F∗ )2 }. q Proof. This is a special case of [21, Lemma 4.1.9]). The next lemma is also a special case of [21, Lemma 4.1.12]. Lemma 5.5. Let W be a totally singular subspace of V. Keeping the notation above, we have: (i) MI = U : N ; (ii) N = T0 × I(X), where GLα (q) T0 I(W ⊕ Y ), and T0 acts naturally on W ; and as T0 -modules we have Y ∼ W ; = (iii) U is a p-group and U Ω(V ). It follows from (i) and (iii) of Lemma 5.5 that MΩ = U (N ∩ Ω(V )). Proposition 5.6. Assume M is of type Pα . Then M has at most two orbits in Eξ (V ) so that 1G 1G by Corollary 3.7 and so M is in Table 1. P M Proof. We will show that MΩ has at most two orbits on Eξ (V ), and hence we deduce that M also has at most two orbits on Eξ (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 [21, Proposition 2.6.1], 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 contains both plus and minus points. Otherwise, X has no minus points. As Ω(X) MΩ , and U MΩ , we have x(Ω(X)U ) ⊆ xMΩ . For any v ∈ xΩ(X), w ∈ W, we will show that there exists u ∈ U such that vu = v + w, 1 which implies that | x Ω(X)U | = |Eξ (X)||W |. Thus | x MΩ | 2 |xΩ(X)U | = 1 1 |W ||Eξ (X)| = 3α 2 3m−α (3m−α + ξ1)) = 3m (3m−α + ξ1). Therefore 2 1 m m−α 3 (3 + ξ1). (5.5) | x MΩ | 2 Let B = [f ]β . Then 0 B=0 Iα  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 RANK 3 PERMUTATION CHARACTERS 15 that 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 5.4, SLα (3) T N ∩ Ω MΩ , and so T U MΩ , where T0 , N are as in Lemma 5.5. 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 = (η1, 0, . . . , 0) and F1 = (1, 0, . . . , 0). Then [z]β = (E1 , 0, F1 ). As zg = z, we α α have D (E1 , 0, F1 )  B D∗ A  0 Is D∗ C  0 0  = (E1 , 0, F1 ), D∗ or (E1 D + F1 D∗ A, F1 D∗ C, F1 D∗ ) = (E1 , 0, F1 ), hence  = F1  F1 D ∗ F1 D ∗ C = 0  E 1 D + F1 D ∗ A = E 1 Since F1 D∗ = F1 , D∗ = 1 b 0 ∗ ,D = D1 1 −bt D1 , and D−1 = 0 D1 1 bt −1 , 0 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 C= 0 c0 0 ,A = C1 a11 a0 a , A1 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 , we have (ξ, 0) 1 −bt D1 0 D1 + (1, 0) a11 a0 a A1 = (η1, 0). It follows that (η1 + a11 , a − ηbt D1 ) = (η1, 0). Therefore a11 = 0 and a = ηbt D1 . Finally as A + At = −λCC t , we have 0 t a0 + ηD1 b at + ηbt D1 0 A1 + At 1 = −λ−1 0 0 t , 0 c0 ct + C1 C1 0 16 HUNG P. TONG-VIET t t hence a0 = −ξD1 b, A1 + At = −λ(c0 ct + C1 C1 ) and 1 0 A= 0 t −ηD1 b ηbt D1 . A1 0 Is 0 0 c0  0 Iα 0  B D∗ A 0 ,A = C1 0 Is C  0 0 , Iα ξbt D1 , A1 In summary, for any g ∈ (T U )z , we have    D 0 0 D Is 0  = 0 [g]β =  B D∗ A D∗ C D∗ 0 where D= 1 −bt D1 , D∗ = 0 D1 1 0 ∗ ,C = b D1 0 t −ξD1 b B = −λC t ∈ Ms,α (3), with C1 ∈ Mα−1,s−1 (3), A1 ∈ Mα−1 (3), b, c0 ∈ Mα−1,1 (3), t A1 + At = −λ−1 (c0 ct + C1 C1 ), A ∈ Mα (3), D, D∗ ∈ SLα (3), C ∈ Mα,s (3). 1 0 We see that the subgroup of SLα (3) generated by all matrices D is isomorphic to U0 : SLα−1 (3), where U0 is an elementary abelian subgroup of order 3α−1 . Given 1 such D, there are 3α−1+(α−1)(s−1) choices for C and 3 2 (α−1)(α−2) choices for A. Therefore |(T U )z | = 3α−1 |SLα−1 (3)|3α−1+(α−1)(s−1)+ 2 (α−1)(α−2) . Since |U | = 3αs+ 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). (5.6) | z MΩ | 2 1 m m−α 3 (3 + It follows from (5.5) and (5.6) that |Eξ (V )| | x MΩ | + | z MΩ | 2 1 2m−α α 1 m m ξ1)+ 2 3 (3 −1) = 2 3 (3 +ξ1) = |Eξ (V )|. 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 ). 5.2.2. The imprimitive subgroups C2 . Let V be a vector space over a ﬁnite ﬁeld Fq with n = dimV. 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 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 (Ξ) are the stabilizers in Ξ of α-decomposition D of V such that D is non-degenerate and if α = 1 then q = p, a prime. Lemma 5.7. Let MΩ be the stabilizer in Ω(V ) of a non-degenerate α-decomposition D and n = bα. Then 1 1 RANK 3 PERMUTATION CHARACTERS 17 (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 . = Proof. This is Propositions 4.2.14 and 4.2.15 in [21]. Lemma 5.8. Assume D is a 1-decomposition of V with q odd. For 1 k |Dn | = (q − 1)k−1 n . k k n, Proof. Without loss of generality, we can assume 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 vectors taken from the basis β, with coeﬃcients in F∗ . Clearly, there are n q k choices for k-sets, and for each k-set, there are (q − 1)k−1 points of length k. The result follows. Lemma 5.9. Assume G is nearly simple primitive rank 3 of type Ωn (3) and M is of type O1 (3) Sn with n = 2m + 1. Let β = {x1 , . . . , xn } be an orthogonal basis for V. (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 I(V )(D) = ri |1 i n ∼ = n ∼ 2n−1 . For 1 2 , and Ω(V )(D) = ri rj |1 i, j n = 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 gen= erated by rxi −xj rxr −xs , with i = j, r = s is isomorphic to An . By Lemma 5.7(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 the point x1 , and x1 + x2 Ω(V )(D) = { x1 + x2 , x1 − x2 }, as (x1 + x2 )r2 r3 = x1 − x2 . By [21, (4.2.17)], 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. 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 | = |{ x2 , . . . , xn }| = n − 1, 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 . Thus by Lemma 5.8 2 1+c2 +d2 = | x1 +x2 MΩ | = Dn = n(n−1). 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 5.8 again, 2 d2 = | x1 + x2 ⊥ ∩ x1 + x2 MΩ | = Dn−2 + 1 = n2 − 5n + 7, and so c2 = 4n − 8. Proposition 5.10. 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 two orbits on Eξ (V ) so that 1G 1G by Corollary 3.7, and hence M is in Table 1. P M Proof. Retain the notation in the previous lemma. By Propositions 2.5.10, and 2.5.13 in [21], sgn(x⊥ ) = (−)m and sgn(x1 + x2 )⊥ = (−)m+1 , as the discriminant 1 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: 18 HUNG P. TONG-VIET (i) x is a plus point. If m is even then we choose x = x1 . By Lemma 5.9 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 by Lemma 5.9 again d = d2 = (n − 2)(n − 3) + 1, c = c2 = 4n − 8. Then c−2d = 14n−2n2 −22. As n 5, 2n2 +22 > 14n so that c−2d < 0. Therefore, equation (5.1) cannot hold. If equation (5.2) 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 5, then 3m−1 + 1 > 8(2m − 1), hence this equation cannot hold. For 2 m 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, then x = x1 + x2 and d = (n − 2)(n − 3) + 1, c = 4n − 8. We have 2d − c = 2n2 + 22 − 14n. If equation (5.2) 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, for any m 3, (3m−1 − 1)(3m + 1 + 2n2 + 22 − 14n) > (3m−1 − 1)(3m + 1) > 8(2m − 1); when m = 2, (3m−1 −1)(3m +1+2n2 +22−14n) = 24 = 8(2m−1). If equation (5.1) 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. Then m = 2, and |Eξ (V )| = 1 3m (3m + ξ1). In this case, x1 MΩ and 2 x1 +x2 +x3 +x4 MΩ are two orbits of plus points with orbit sizes 5 and 23 5 = 40, 4 1 respectively. As |E+ (V )| = 2 32 (32 + 1) = 45 = 5 + 40, 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 = 20 and 24 5 = 16, respectively. Since 5 2 1 |E− (V )| = 2 32 (32 − 1) = 36 = 20 + 16, there are exactly two orbits of minus points. 1 Finally, suppose that n = 7, 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 7 = 42 and 24 7 = 336. Since 336 + 42 = 378, there 2 5 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 5.11. 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 1-3. 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 MΩ . By [21, Lemma 4.2.8], we have MI = I1 Sb . Thus Ω1 Sb MΩ MI . Since α > 1 is odd, α 3 and so V1 contains both 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 MΩ permutes the Vi s, and Ωi centralizes V1 , for all i > 1, xξ MΩ = b xξ N Sb = xξ Ω1 Sb = Eξ (V1 )Sb = ∪b Eξ (Vi ). Thus A = | xξ MΩ | = 2 3a (3a +ξ1) i=1 1 a a by Lemma 5.1(i). Hence A 2 b · 3 (3 + 1). In view of inequality (5.3), it suﬃces RANK 3 PERMUTATION CHARACTERS 19 1 1 a a to show that 2 3m 2 b · 3 (3 + 1). Since m = ba + b1 , this inequality is equivalent ba+b1 a a to 3 b3 (3 +1). As b 3 and a 1, 3ba 33a = 3a ·32a 3·32a > 32a +3a . b−1 If we can prove that 3b1 = 3 2 b, then clearly 3ba · 3b1 b3a (3a + 1), and we b1 are done. To show that 3 b, we will argue by induction on b 3. When b = 3 √ (b+1)−1 b−1 b−1 b−1 √ then 3 2 = 3 3 = b. Suppose that 3 2 b. Then 3 2 = 3 2 3 b 3, by induction assumption. We have 3b2 = b2 + 2b2 b2 + 6b > b2 + 2b + 1 = (b + 1)2 , √ b+1−1 as b 3. Thus b 3 b + 1. Hence 3 2 b + 1. The result follows. 5.2.3. The ﬁeld extension subgroups C3 . 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 5.12. 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  2 T (ζ1 )  T (ζ2 ζ1 )  fβT =  . .  . Let    X=  α−1 T (ζ1 ζ2 ) 2 T (ζ2 ) . . . ... ... .. .  T (ζ1 ζα ) T (ζ2 ζα )  . . .  . 2 T (ζα ) T (ζα ζ1 ) T (ζα ζ2 ) . . . ζ1 p ζ1 . . . α−1 ζ2 p ζ2 . . . p ζ2 α−1 ... ... .. . ... α−1 ζα p ζα . . . p ζα i α−1      p ζ1 and E = diag(λ, λp , . . . , λp ). As T (a) = i=0 ap for any a ∈ F , X t X = fβT , hence 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 subﬁelds 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 τ = ζ + 1. Then τ 2 = −1, F = F(τ ) and T (τ ) = 0. Choose βT = {1, τ }. Then fβT = T (1) T (τ ) T (τ ) T (τ 2 ) = 2 0 . 0 −2 20 HUNG P. TONG-VIET v Hence det(fβT ) = −1. This proves (i). Let β = {ζ1 v , ζ2 v , . . . , ζα v } and W = F . As (ζi v , ζj v ) = T (f (ζi v , ζj v )) = T (ζi ζj f (v , v )) = T (λζi ζj ), we have   2 T (λζ1 ) T (λζ1 ζ2 ) . . . T (λζ1 ζα ) 2  T (λζ2 ζ1 ) T (λζ2 ) . . . T (λζ2 ζα )   fβ =  . . . . .. . . .   . . . . T (λζα ζ1 ) T (λζα ζ2 ) . . . 2 T (λζα ) Obviously X t EX = fβ . Therefore det(fβ ) = det(X)2 N (λ) = D(F )N (λ). n Proposition 5.13. 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 hence M is in Table 1. Proof. Let q = 3α , and µ, ν be the generators for F∗ and Gal(F /F), respectively. n As n is odd, it follows that α is also odd. Write α = 2α1 + 1 and α = 2b + 1. Then m = bα + α1 , where n = 2m + 1. Multiplying by a suitable constant to the quadratic form Q , we can assume that D(Q ) = . By [21, Proposition 2.6.1], 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 α−1 βn = {ζ, ζ 3 , . . . , ζ 3 } be a normal basis of F over F, and βi = βn ⊗ wi , Since j j+1 (ζ 3 wi )φ = ζ 3 wi , we obtain   0 1 0 ... 0 0 0 1 . . . 0   . . . . (φ )βi =  . . . . . . .  . . . . . 0 0 0 . . . 1 1 0 0 ... 0 As det(φ )βi = (−1)α−1 = 1, det(φ ) = 1. So 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 [21, (4.3.11)], MI = I φ ∼ I Zα . Since Ω is perfect, Ω L, hence = Ω MΩ NL (F ) MI = I φ . By [21, Proposition 4.3.17], [MΩ : Ω ] = α. As φ ∈ Ω ∩ I φ , φ ∈ MΩ , and hence MΩ = Ω φ . It follows from [21, Lemma 2.10.5] 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 . α−1 (2) z MΩ = { w ∈ V | Q (w) ∈ {γ, γ ν , . . . , γ ν }}, where z ∈ V with γ = Q (z). 2b+1 2b+1 For (1), assume that w = i=1 (λi wi )φ = i=1 λi wi ∈ V . Then wφ = 2b+1 ν 2b+1 2 ν ν i=1 λi wi , hence Q (wφ ) = ( i=1 −λi ) = Q (w) . For (2), from (1) we have α−1 Q (zφ ) = γ ν , hence {Q (zφj )}α = {γ, γ ν , . . . , γ ν }. Thus z MΩ = { w ∈ j=1 V | Q (w) ∈ {γ, γ ν , . . . , γ ν }}. For a non-zero vector w ∈ V , consider spanF (w) as an α-subspace in V. (a) Case α > 3. Let z ∈ {w1 , w1 + w2 }. Then Q (z) = 1 and Q(z) = T Q (z) = α = 0 so z is non-singular in V. Also as Q (z) = 1 is ﬁxed under ν, by (2) 1 we have z MΩ = z Ω , and by Lemma 2.2, | z MΩ | = | z Ω | = 2 (q 2b + εq b ), α−1 RANK 3 PERMUTATION CHARACTERS 21 ⊥ 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 sgn(zV ) = ε, by Lemma 2.2, there are q 2b−1 −εq b−1 vectors w0 with Q(w0 ) = Q (z)−1 (1−ϕ2 ) = 0. Also dimF (KerT ) = α− 1 ⊥ 1, we conclude that dz = | z MΩ ∩zV | = 2 3α−1 (q 2b−1 −εq b−1 ) = 1 (q 2b −εq b ). Thus 6 1 2b b bα bα−1 cz = 3 (q + ε2q ) − 1 = (ε3 − 1)(ε3 + 1), and cz − 2dz = εq b − 1 = ε3bα − 1. ⊥ Assume (5.1) holds. Then cz −2dz = ε3bα −1 = ξ3m−1 −1, where ξ = sgn(zV ). The 1 latter equation yields m = bα+1. Recall that m = bα+α1 , hence α1 = 2 (α−1) = 1, this forces α = 3, a contradiction. Suppose (5.2) holds. Then (3m−1 +ξ1)(3m −ξ1+ ξ(c − 2d)) = 2c, hence (3m−1 + ξ1)(3m − εξ3bα − ξ2) = 2(3bα − ε1)(3bα−1 + ε1). We will show that 3m−1 +ξ > 2(3bα −ε1) and 3m −εξ3bα −ξ2 > 3bα−1 +ε1, so that after multiplying these two inequalities side by side, we get a contradiction. For the ﬁrst inequality, we have 3m−1 + ξ 3bα+α1 −1 − 1 3α1 −1 3bα − 1 3 · 3bα − 1, as α1 2. Now 2(3bα − ε1) 2(3bα + 1). 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, as 3bα − 3 > 0, we have 3m − εξ3bα − ξ2 3bα+α1 − 3bα − 2 = (3α1 − 1)3bα − 2 > 2 · 3bα − 2 = (3bα + 1) + (3bα − 3) 3bα−1 + ε1. (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 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 , so that D(xi ⊥ ) = V V V , D(yi ⊥ ) = , and hence by [21, Proposition 2.5.10, 2.5.13], sgn(xi ⊥ ) = (−)b V 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 V V the type 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 ⊥ = V V ωw1 , (ω 2 − ω)w1 ⊥ spanF (w2 ) ⊥ U, x3 ⊥ = (ω + 1)(w1 + w2 ), (ω 2 − 1)(w1 + V w2 ) ⊥ spanF (w1 − w2 ) ⊥ U, and similarly y1 ⊥ = (w1 + w2 ), ω 2 (w1 + w2 ) ⊥ V spanF (w1 − w2 ) ⊥ U, y2 ⊥ = ω(w1 + w2 ), (ω 2 − ω)(w1 + w2 ) ⊥ spanF (w1 − V w2 ) ⊥ U, y3 ⊥ = (ω + 1)w1 , (ω 2 − 1)w1 ⊥ spanF (w2 ) ⊥ U. By Lemma 5.12, V D(spanF (wi )) = N (f (wi , wi )) = N (1) = 1 = and D(spanF (w1 − w2 )) = N (f (w1 − w2 , w1 − w2 )) = N (−1) = −1 = . Thus D(xi ⊥ ) = and D(yi ⊥ ) = V V for all i = 1, . . . , 3, and so by [21, Proposition 2.5.11], sgn(xi ⊥ ) = (−)m−1 = V (−)b and sgn(yi ⊥ ) = (−)m = (−)3b+1 = (−)b−1 . Let z ∈ {xi , yi } and γ = Q (z). V 3 We have z MΩ = { w ∈ V | Q (w) ∈ {γ, γ 3 , γ 9 }} = j=1 zφj Ω . Observe that ⊥ in (V , F , Q ) all vectors zφj , j = 1, . . . , 3 have the same type, say ε = sgn(zV ). 1 It follows from Lemma 2.2 that 1 + cz + dz = 3 2 (q 2b + εq b ) = 1 (36b+1 + ε33b+1 ). 2 3 9 ⊥ For any w ∈ z MΩ ∩ zV , Q (w) ∈ {γ, γ , γ } 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 ). 22 HUNG P. TONG-VIET ⊥ 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 nonzero on these values, we get a contradiction. Thus Q (w0 ) = 0. By Lemma 2.2, 1 ⊥ dz = 3 2 (q 2b−1 −εq b−1 )·3α−1 = 1 (q 2b −εq b ) as |kerT | = 3α−1 = 32 , dimF (zV ) = 2b, 2 ⊥ 2b b b and ε = sgn(zV ). Then cz = q + ε2q − 1. Hence cz − 2dz = ε3q − 1 = ε3m − 1. Therefore equation (5.1) holds. ⊥ ⊥ Assume z ∈ {x3 , y3 }. Then sgn(zV ) = −sgn(zV ) = −ε, γ = ±ω 8 and Q (w0 ) = −1 2 γ (γQ (w) − ϕ ). 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 2.2, 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 )]. After simplifying, we get 2dz = 36b +ε33b+1 , and hence cz = 36b −1, so that cz − 2dz = −ε33b+1 − 1 = −ε3m − 1. Thus equation (5.1) holds. 3 Let ξ = (−)b = sgn(xi ⊥ ) and η = sgn(yi ⊥ ), i = 1, . . . , 3. As i=1 | xi MΩ | = V V 1 6b+1 1 6b+1 1 3b+1 3b+1 1 6b+1 3b+1 3b+1 3b+1 + ξ3 ) + 2 (3 + ξ3 ) + 2 (3 − ξ3 ) = 23 (3 + ξ1) = 2 (3 1 m m 2 3 (3 + ξ1) = |Eξ (V )|, MΩ has exactly three orbits on Eξ (V ). Thus equation (5.1) holds for all points in Eξ (V ). Similarly MΩ has three orbits on Eη (V ), and so equation (5.1) holds for all points in Eη (V ). 5.2.4. The tensor product subgroups C4 . Let Vi be vector spaces over F 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 extends linearly. Now suppose that fi is a non-degenerate symmetric bilinear form on Vi , so that (Vi , F, fi ) is an orthogonal geometry. We next deﬁne the bilinear t form f = f1 ⊗· · ·⊗ft on V1 ⊗· · ·⊗Vt by f (v1 ⊗· · ·⊗vt , w1 ⊗· · ·⊗wt ) = i=1 fi (vi , wi ) and extend linearly. We write (V, f ) = (V1 ⊗· · ·⊗Vt , f1 ⊗· · ·⊗ft ) for such a structure and call a tensor decomposition and denote by D. The members of C4 (Ξ) is the stabilizer of a 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 3, εi = sgnVi , and dimV1 < dimV2 . Proposition 5.14. 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 1-3. Proof. Let v = v1 ⊗ v2 ∈ V, where vi ∈ Vi are non-singular vectors. Then v is a non-singular point. By [21, (4.4.14)], MI = I1 ⊗ I2 . We have Ω1 × Ω2 MΩ I1 ⊗ I2 = MI . As all Ωi act transitively on EρVi (vi ) (Vi ), it follows that v MI = v MΩ = v1 ⊗ v2 (Ω1 × Ω2 ) = v1 Ω1 ⊗ v2 Ω2 . Therefore v MΩ = v M since MΩ M MI . Write n1 = 2a + 1, n2 = 2b + 1. Then | v MΩ | 1 a a 3 (3 + 1)3b (3b + 1) < 32(a+b) as 3a + 3b + 1 < 3a+b and | v1 Ω1 || v2 Ω2 | 2 b > a 1. Since dimV = 2m + 1 = (2a + 1)(2b + 1), m = 2ab + a + b. Then m − 1 − 2(a + b) = a(b − 2) + b(a − 1) + a − 1 0 so that 3m−1 32(a+b) > | v MΩ |. This violates (5.3) so that (3.1) cannot hold. RANK 3 PERMUTATION CHARACTERS 23 5.2.5. The tensor product subgroups C7 . Let V1 be an α-dimensional vector space over F, and assume that f1 is a non-degenerate symmetric bilinear form on V1 . 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 = f1 ⊗ · · · ⊗ fb . Let Xi = X(Vi , fi ) for 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. Proposition 5.15. Assume M is of type Oα (3) Sb with α 5. There is an M -orbit on Eξ (V ) such that (3.1) does not hold so that M is not in Tables 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, i=1 the stabilizers of vi in Πb Ωα (3) contain a subgroup which is isomorphic to i=1 Πb Ωα−2 (3). Thus | vi MΩ | = | vi (Πb Ωα (3))| [Ωα (3) : Ωα−2 (3)]b < 3(4a−3)b , i=1 i=1 where a = α−1 2. Now m − 1 = 1 ((2a + 1)b − 3). Consider the following function 2 2 in variable x ∈ [2, +∞), where b 2, f (x) := 1 ((2x + 1)b − 3) − (4x − 3)b. We have 2 f (x) = b(2x + 1)b−1 − 4b b(2x + 1) − 4b b > 0. Hence f (x) f (2) = 1 g(b), 2 where g(b) = 5b − 10b − 3 and b 2. By induction on b 2, g(b) > 0. Thus 3m−1 > 3b(4a−3) > | vi MΩ |. This contradicts (5.3). Thus (3.1) cannot hold. These are all the maximal subgroups in C(G) of G. We next consider maximal subgroups in S(G). 5.3. Permutation characters of maximal subgroups in S. In this section, we consider the maximal subgroup M ∈ S(G). By deﬁnition of S, 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, that is, ind(V ) = +. We note that some of the small cases in this section will be handled by using GAP [10]. We will describe how to do this at the end of the paper. 5.3.1. Embedding of alternating and symmetric groups. Recall the construction of the fully deleted permutation module for alternating groups in [21, p. 185]. Let {ε1 , . . . , εn } be a standard basis for Fn , and let w0 = ε1 + · · · + εn ∈ Fn . Put p p ⊥ U = w0 , W = Fp w0 , and V = U/(U ∩ W ). Deﬁne ei = εi − εi+1 , i = 1, . . . , n − 1. Then {ei }n−1 is a basis for V if p does not divide n, and {ei + U ∩ W }n−2 is a i=1 i=1 basis for V if p|n. Deﬁne εp (n) to be 1 if n is divisible by p, otherwise, εp (n) = 0. Then dimV = n − 1 − εp (n). Let Q be the quadratic form on V induced from the quadratic form associated to the natural bilinear form on Fn . Then (V, Fp , Q) is a p classical orthogonal geometry and An Ω(V ). To simplify the notation, we always write ei instead of ei + U ∩ W. Lemma 5.16. Assume M is almost simple of type An with n 10 and V is the fully 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; 1 (2) | w M | = 8 n(n − 1)(n − 2)(n − 3), cw = 2n3 − 25n2 + 111n − 172 and 1 dw = 2 + 4(n − 4)2 + 8 (n − 4)(n − 5)(n − 6)(n − 7). 24 HUNG P. TONG-VIET Proof. Observe that v Sn = { εi − εj | i = j ∈ {1, . . . , n}} and w Sn is the set of all points of the form εi + εj − εr − εs , where i, j, r, s ∈ {1, . . . , n} and pair-wise distinct. As n 10 and An is (n − 2)-transitive on the index set {1, . . . , n}, we have x An = x Sn for any x ∈ {v, w}. Hence x M = x Sn for x ∈ {v, w}. 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 partitions {1, 2}, {3, 4}. Thus g ∈ S2 × S2 × Sn−4 . Therefore 1 |M : M v | = 2 [Sn : Sn−2 ] = 1 n(n − 1), and |M : M w | = 1 [Sn : S2 × S2 × Sn−4 ] = 2 2 1 8 n(n − 1)(n − 2)(n − 3). (i) Parameters for v. We have u ∈ v M ∩ v ⊥ if and only if u = εi − εj ∈ F3 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 such points u . Thus dv = | v M ∩ v ⊥ | = 2 1 (n − 2)(n − 3), and cv = 2n − 4. 2 (ii) Parameters for w. We will show that c = 2n3 − 25n2 + 111n − 172 and d = | w M ∩ w⊥ | = 2 + 4(n − 4)2 + 1 (n − 4)(n − 5)(n − 6)(n − 7). For any u ∈ w M ∩ w⊥ , u = εi + εj − εr − εs , 8 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) = ∅. Then 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. (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 n−4 points respectively. Thus there are 4(n − 4)(n − 5) points 2 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) = 4(n − 4) such points. (5) |supp(u) ∩ supp(w)| = 4. There are just 2 points in this case: {ε1 − ε2 + ε3 − ε4 , ε1 − ε2 − ε3 + ε4 }. 1 Therefore dw = 8 (n−4)(n−5)(n−6)(n−7)+2(n−4)(n−5)+4(n−4)(n−5)+2 = 1 1 2 8 (n−4)(n−5)(n−6)(n−7)+2(n−4) +2, and cw = 8 n(n−1)(n−2)(n−3)−d−1. Proposition 5.17. 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 1-3. 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 1 only if n = 6k +2, n = 6k +3 or n = 6k +4. By Lemma 5.16(1), | v M | = 2 n(n−1). 1 Assume that n 13. Then m − 1 = 1 (n − 2 − ε3 (n)) − 1 2 (n − 5), as ε3 (n) 1, 2 m−1 (n−5)/2 and so 3 3 > n(n − 1)/2 = 1 + c + d, violating (5.3) and so equation (3.1) cannot hold. Hence we can assume that 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 = ξ81 − 1 = −40, or (ξ27 + 1)(ξ81 − 41) = 32. RANK 3 PERMUTATION CHARACTERS 25 Table 5. Small degree representations of ssome alternating groups. An dimDλ λ m(λ) A6 A7 A7 A8 A8 A8 A9 A9 9 13 15 7 13 21 21 7 (4, 2) (5, 2) (5, 12 ) (7, 1) (6, 2) (6, 12 ) (7, 12 ) (8, 1) (22 , 12 ) (3, 2, 12 ) (3, 22 ) (4, 3, 1) (32 , 12 ) (32 , 2) (4, 3, 2) (42 , 1) These equations clearly cannot hold with ξ = ±. By Lemma 5.16(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 by (5.3). Thus we can assume that 10 n 22. Then n ∈ {10, 14, 15, 16, 20, 21, 22}. (a) n = 10. Then n − ε3 (n) = 9, m = 4, d = 191, c = 438 and c − 2d = 56. (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 5.18. 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 1-3. Proof. As n 12, by Lemma 2.6, we have dim(V ) = 2m+1 (n2 −5n+2)/2 so that 2 m (n2 − 5n)/2. However when n 12, 3m−1 3(n −5n−4)/4 > n! = |Aut(An )|. Thus equation (3.1) cannot hold in view of (5.3). Let λ = (λa1 , λa2 , . . . , λah ) be a p-regular partition. Then λ is called a JS1 2 h partition if λi − λi+1 + ai + ai+1 ≡ 0 (mod p). Proposition 5.19. 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 1G = M P by Corollary 3.7 and hence (L, S) = (Ω7 (3), A9 ) is in Table 2. Proof. Using information on the p-modular representations of alternating groups and their covering groups in [20], we need to consider the cases given in Table 5. (i) Let λ = (8, 1). Then m(λ) = (42 , 1) = λ. By [7, Theorem 2.1], Dλ ↓A9 is irreducible. Thus A9 S9 Ω7 (3), and there are two classes of S9 in Ω7 (3). As 8 − 1 + 1 + 1 = 9 ≡ 0(mod 3), λ is a JS-partition, and hence by [23, Theorem 0.3], 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 = (7,1) D and D(7,1) ↓A8 is irreducible. 26 HUNG P. TONG-VIET 2 (iii) If λ = (6, 12 ) or (32 , 2), then A8 < S8 < S9 < Ω21 (3), since D(7,1 ) ↓S8 = 2 (6,12 ) D and 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 1 n! n! points of diﬀerent types in Dλ . We have |vSn | = 2 2(n−4)! and |wSn | = 1 2(n−5)! . 2 We then get contradictions by using (5.3). (vi) If λ = (4, 2) then dimDλ = 9, m = 4, and we have an embedding A6 S6 Ω9 (3). As A6 ∼ L2 (9) < A10 < Ω9 (3), 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 [10], S8 has two points w1 , w2 of diﬀerent types with the same orbit sizes 315. We have (c, d) = (230, 84) or (c, d) = (212, 102). We see that equations (5.1) and (5.2) cannot hold. 5.3.2. Embedding of groups of Lie types in cross-characteristic. Proposition 5.20. Assume M is almost simple of type S, where S is a ﬁnite simple group of Lie type 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 on Eξ (V ) so that 1G 1G by Corollary 3.7, and so (L, S) is in P M Table 2. Proof. Suppose equation (3.1) holds for some r ∈ {s, t} and some M -orbit x M 1 m with x ∈ Eξ (V ). Then | x M | 2 (3 + 1), by (5.3). On the other hand, using the lower bounds for degrees of cross-characteristic representations of ﬁnite simple groups of Lie type (see [24] and [31]), we have 2m + 1 e(S), so that (3m + 1)/2 e(S)−1 (3(e(S)−1)/2 )/2. Moreover | x M | |M | |Aut(S)|. It follows that (3 2 +1)/2 |Aut(S)|. By this condition, we get a ﬁnite list as in Table 6. Next we shorten the list by using information on cross-characteristic representations of small groups together with [20] and [10]. 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 is a prime power, we can assume that 7 q 67. Case q ≡ 1 (mod 4). By [14, Table 2] and the fact that dimV is odd, we have dimV ∈ {(q + 1)/2, q}. Using (5.3) 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 [5] 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 RANK 3 PERMUTATION CHARACTERS 27 Table 6. Small groups in cross-characteristic. S L2 (q) Ln (q), n 3 P Sp2n (q), n + 1)/2 |Aut(S)| L2 (q), 3 q 68 L3 (2), L4 (2), L5 (2), L3 (4) S4 (5), S4 (7), S6 (5) S4 (2), S6 (2), S8 (2), S4 (4) Un (2), 3 n 7, U3 (4), U3 (5) Ω+ (2) 8 Ω− (2) 8 (3 e(S)−1 2 2 Un (q), n 3 P Ω+ (q), n 4 2n P Ω− (q), n 4 2n Ω2n+1 (q), n 3, q odd E6 (q) E7 (q) E8 (q) F4 (q) 2 E6 (q) G2 (q) 3 D4 (q) 2 F4 (q) Sz(q) 2 G2 (q) F4 (2) 3 2 D4 (2) F4 (2) Sz(8) 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, the latter inequality yields | uξ M | |Out(S)|| uξ S| 4 · 8400 < 3m−1 . In view of (5.3), equation (3.1) cannot hold. Case q ≡ 3 (mod 4). As in previous case, by [14, Table 2], we have dimV = q and 3 q + 1. Using (5.3) again, we get q ∈ {7, 19}. If (S, dimV ) = (L2 (7), 7) then L2 (7) < Ω7 (3) but Ω7 (3) has no maximal subgroup with socle L2 (7) by [5]. 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, this implies | uξ M | |Out(S)|| uξ S| 2 · 342 < 3m−1 . In view of (5.3) equation (3.1) cannot hold. Case q ≡ 0 (mod 2). As in previous case, by [14, Table 2], we have dimV ∈ {q − 1, q + 1}. Using (5.3) again, we have q = 8, 16. By [20], the only possibility is q = 8 and dimV = 7. However by [5], 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), L4 (2) ∼ = = A8 , which have been done above, we can exclude these groups. 28 HUNG P. TONG-VIET If S = L3 (4), then Out(S) ∼ 2 × S3 ∼ D12 . By [20], dimV ∈ {15, 19, 45, 63}. By = = using (5.3), 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 non-singular 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 non-singular 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 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 155. But (3m + 1)/2 (377 + 1)/2 > |Aut(L5 (2))| 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. [14]), L cannot embed in Ω13 (3) and Ω63 (3). By [12, Theorem 2.1], if Φ is a representation of S which is not the smallest representation, then dimΦ (q n − 1)(q n − q)/(2(q + 1)), which are 40 and 1240, respectively. But then inequality (5.3) 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. [14]), 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 (5.3). 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 = 51 and (3m + 1)/2 (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 [5], 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 order 5 and (c1 , d1 ) = (96968, 48183), (c2 , d2 ) = (47912, 24663). RANK 3 PERMUTATION CHARACTERS 29 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 [20], the smallest odd degree non-trivial 3-modular representation of S has degree 55. Thus (3m + 1)/2 (327 + 1)/2 > |Aut(S)|. If S = U3 (4), then dimV ∈ {13, 39, 75}. However if dimV 39 then (5.3) cannot hold, and if dimV = 13, then by [14], U3 (4) ﬁxes no quadratic form. If S = U7 (2), then by [14], we have dimV > 250 and so (5.3) cannot hold. For the remaining cases, using [20], 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 [20], the smallest odd degree non-trivial 3-modular representation of S has degree 35, and the second smallest odd degree one has degree 147. Using (5.3) 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 equation (3.1) cannot hold. − (vii) Case S = O8 (2). By [20], dimV 203. Then inequality (5.3) cannot hold. (viii) Case S = F4 (2). By [14], dimV > 255. Then inequality (5.3) cannot hold. (ix) Case S = G2 (4). By [20], dimV 649. Clearly, 3m−1 > 415 |Aut(S)|. (x) Case S = Sz(8). By [20], dimV = 35. Then inequality (5.3) cannot hold. (xi) Case S = 3 D4 (2). By [20], either dimV = 25 or dimV 351. If dimV 351, then m 174 and clearly 3m−1 3174 > 3 · 229 |Aut(S)|. When dimV = 25, the group S is not maximal in Ω25 (3) as 3 D4 (2) F4 (3) Ω25 (3) (see [26]). (xii) Case S = 2 F4 (2) . By [20], 2m + 1 77. Then 3m−1 337 > |Aut(S)|. 5.3.3. Embedding of groups of Lie type 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. If q is a power of p, then we put ci q − 1}. A p-restricted dominant weight is a Xq = { i=1 ci λi | ci ∈ Z, 0 weight that lies in Xp . 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 30 HUNG P. TONG-VIET Z-form of U (see [15, 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 ([15, 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. We will label the Dynkin diagram as in [28]. 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 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, that is, L(λ) ∼ S c (N ). In particular, dimL(λ) = ( +c)! ([30, 1.14]). = !c! i (2) If > 1, then L(λi ) ∼ N and dimL(λi ) = +1 (see [3]). = 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 is even then L(λi ) leaves invariant no non-degenerate bilinear form, and if is odd then L(λi ), i = +1 does not preserve any such form. Let λ = λ( +1)/2 . If ≡ −1 (mod 2 4) then L(λ) ﬁxes a symmetric bilinear form and it ﬁxes an alternating bilinear form if ≡ 1 (mod 4) (see [2] Chapter VIII §13 Table 1, p. 217). The following constructions for adjoint modules of groups of type A and 2 A are taken from [27], 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 i < j + 1, Eii − Ei+1,i+1 , i = 1, . . . , − εp ( + 1). (5) Let S = SUn (q) and λ = λ1 + λ , where n = + 1. Let V2 = {aIn | a ∈ Fq2 }, t 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 = q basis consisting of Ei,j + Ej,i , µEi,j + µEj,i 1 i<j + 1, Eii − Ei+1,i+1 , i = 1, . . . , − εp ( + 1). Proposition 5.21. Assume M is almost simple of type S, where S is simply connected of 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 1-3. Proof. Assume (3.1) holds for some r ∈ {s, t} and any M orbits in Eξ (V ). We consider the case f = 1 and f > 1 separately. We use the notation Lε (q) to denote n Ln (q) when the sign ε is + and Un (q) when ε = −. Case f = 1. We can assume that 2. By [16, 31.1] and [32, §13], there exists a 3-restricted dominant weight λ ∈ X3 such that V ∼ L(λ). As |Aut(S)| = = 2 2 |Aut(Lε+1 (3))| 3( +1) , where ε = ±. It follows from (5.3) 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 RANK 3 PERMUTATION CHARACTERS 31 2( + 1)2 + 3 and has odd degree. If 18, then 3 /8 2( + 1)2 + 3, and so by [28, Theorem 5.1], 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, the only possibility for λ is λ1 + λ . If < 18, then by Theorem 4.4, Appendix A6 through A21 in [28], either λ = 2λ2 when = 3 or λ = λ1 + λ for 2 17. Suppose that 4. We have dimL(λ1 +λ ) = 2 +2 −εp ( +1). As dimL(λ1 +λ ) is odd, it follows that = 6b1 + 1, 6b1 + 2 or 6b1 + 3 for b1 1. Consequently 7. As constructed 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 = ±, 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 non-singular 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 S xξ , the stabilizer of xξ in S. We have ε 1 + c + d |Aut(S) : U | = [Aut(L +1 (3)) : SLε−1 (3)] 2 · 32 −1 (3 + 1)(3 +1 + 1) < 2 34 +2 . As 2m + 1 = 2 + 2 − ε3 ( + 1) + 2 − 1, m − 1 ( 2 + 2 − 4)/2. We 2 have ( + 2 − 4)/2 − (4 + 2) = ( ( − 6) − 8)/2. If 8 then ( − 6) − 8 > 0, hence m−1 > 4 +2. If = 7 then 2m+1 = 2 +2 = 63, or m = 31 and m−1 = 30 = 4 +2. Thus m−1 4 +2 for any 4. Hence 3m−1 34 +2 > 1+c+d. This contradicts to inequality (5.3). 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 3 [5], Ω7 (3) has no maximal subgroup with socle Lε (3). 3 Assume that (S, L) = (L4 (3), Ω15 (3)). For each extension of S, using [10], 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 exist two nonsingular 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 non-singular 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 2. 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 [21, Theorem 5.4.5], 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. As |Aut(L2 (3f ))| = f 3f (32f − 1) < 34f , it follows from 32 HUNG P. TONG-VIET (5.3) that (3f − 3)/2 4f, with f 3. However by induction on f 3, this is not true. Thus equation (3.1) cannot hold. Consider case 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 [21, Theorem 5.4.5] and [30, 1.11] 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+ f m−1 ε f 2 − ε3 ( + 1)) . We will show that 3 > |Aut(L +1 (3 ))|. Then (5.3) cannot ε f f ( +1)2 hold. As |Aut(L +1 (3 ))| < 3 and m − 1 (( 2 + 2 − ε3 ( + 1))f − 3)/2 2 f (( + 2 − 1) − 3)/2, it suﬃces to show that (( 2 + 2 − 1)f − 3)/2 > f ( + 1)2 . This is true by induction. The proof is complete. 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 ), 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 (d, 1, d−1 )) = ti . Then {γi }i=1 form an orthonormal basis for E. 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 Zbasis of the fundamental dominant weights is {λ1 , . . . , λ }, deﬁned as following, for 1 i = 1 . . . , − 1 λ1 = 2 (γ1 + s + γ ), and λ +1−i = γ + γ −1 + · · · + γ +1−i , Proposition 5.22. 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 so that M is not in Tables 1-3. Proof. Assume (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. 2 2i From (5.3), we have 3m−1 |Aut(S)| = |Aut(Ω2 +1 (3))| = 3 i=1 (3 − 1) 2 2 + 2 3 . Hence dimL(λ) = 2m + 1 4 + 2 + 3. Notice that if 5 then 3 − 2 2 2 2 3 2 (4 + 2 + 3) 5 − 4 − 2 − 3 = ( − 1) − 4 > 0, and so > 4 + 2 + 3. If > 11, then dimL(λ) 4 2 + 2 + 3 < 3 , and hence by [28, Theorem 5.1 ], λ is either λ −1 or 2λ . For 2 11, by [28, Theorem 4.4] and the upper bound for dimension of L(λ) above, again, λ is one of the weights above or = 3, λ = 2λ1 , and dimL(2λ1 ) = 35. So 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) RANK 3 PERMUTATION CHARACTERS 33 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 ru : N → N is deﬁned by vru = v − (v,u) u, Q(u) for any v ∈ N. Let S act on N ⊗ N by (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 = = = 2 i=1 (ei ⊗ fi + fi ⊗ ei ) + x ⊗ x ∈ S (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 ∼ Ω2 +1 (3). It follows that H ∼ Ω2 −1 (3), and H ﬁxes v. By (5.3) we have Ω(N ) = = 3m−1 1+c+d |Aut(Ω2 +1 (3)) : Ω2 −1 (3)| = 2 · 32 −1 (32 − 1) 34 . Hence m − 1 < 4 , so that 2 2 + = 2m + 1 < 8 + 3. As is odd and > 1, the above inequality holds only when = 3. In this case, we have Ω7 (3) Ω27 (3). Using [10], 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), for ξ ∈ {±1}, let v = e1 ⊗ e1 + ξf1 ⊗ f1 + w ∩ w⊥ . 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 dimL(2λ ) = 2 2 + 3 − ε3 (2 + 1), it follows that 2 2 + 3 − 1 2 2 +3 −ε3 (2 +1) < 8 +3. If 4 then 2 2 +3 −1 24 +3 −1 = 8 +(3 −1) > 8 + 11 > 8 + 3, and if = 3, then 2 2 + 3 − ε3 (2 + 1) = 27 = 8 + 3. Since 3 in this case, (5.3) cannot hold. Finally = 3 and λ = 2λ1 . In this case, we have Ω7 (3) Ω27 (3). Using [10], there are two non-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. Case f 2. By [21, Theorem 5.4.5] and [30, 1.11], dimV = (dimΨ)f , for some self-dual irreducible k S-module Ψ. As dimV is odd, so is dimΨ. Firstly, suppose that f 3. Since dimΨ is at least 2 + 1, it follows that 2m + 1 (2 + 1)f . By 2 2 (5.3), we have 3m−1 |Aut(Ω2 +1 (3f ))| < f · 3f (2 + ) 3f (2 + +1) . As 2m + 1 1 (2 + 1)f , we have 2 ((2 + 1)f − 3) < f (2 2 + + 1). Clearing fraction, we get f 2 (2 + 1) − 3 − f (4 + 2 + 2) < 0. By induction, this inequality cannot happen. 34 HUNG P. TONG-VIET Secondly, suppose 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 + + 1) = 2(4 2 + 2 + 2). As 2, f (2 + 1)f − 3 − 2(4 2 + 2 + 2) 22 (2 + 1)2 − 2(2 + 1)2 + 2 4 − 5 2(2 + 1) + 3 > 0. Hence f (2 + 1)f − 3 > 2(4 + 2 + 2), a contradiction. Finally, suppose 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 p ap + bp = 0 or 1, correspondingly. This holds 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 11 2 1,2 2 ap + ap +1,i = 0, 1 < i < 2 + 1. Therefore, u(1) g = ug ν = u(ap ) = u = u(1) . This 1i ij 2 means that if g ∈ S ﬁxes 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 Sv . = 2 Hence by (5.3), 3m−1 |Aut(Ω2 +1 (32 )) : Ωε (32 )| |Aut(Ω2 +1 (32 )) : Ω+ (32 )| 2 2 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 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 form on N has the form B= 0 −I I 0 . From the isomorphisms Sp2 (q) ∼ SL2 (q), and Sp4 (q) ∼ Ω5 (q) for q odd, we can = = assume that 3. Proposition 5.23. 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 1G by Corollary 3.7 and hence (L, S) = (Ω13 (3), P Sp6 (3)) is P M in Table 2. The last case is in Table 3. Proof. Assume (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. = 2 2i We ﬁrst get an upper bound for dimV. As |Aut(P Sp2 (3))| = f 3 i=1 (3 − 1) 2 2 32 + , by (5.3), 3m−1 32 + , and hence 2m + 1 4 2 + 2 + 3. If 5 then 4 2 + 2 + 3 < 3 , and so dimV < 3 . If > 11, then dimL(λ) < 3 , and hence by [28, Theorem 5.1], λ is either λ −1 or 2λ . For 2 11, by [28, Theorem 4.4 ] and the upper bound for dimension of L(λ) above, again, λ is one of the weights above or = 4, λ = λ1 , and 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 = RANK 3 PERMUTATION CHARACTERS 35 w = e1 ∧f1 +s+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 [10], 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. Thus we assume that 5. Let H be the centralizer in S of the subspace generated by {e1 , f1 }. Then H ∼ Sp2 −2 (3). By (5.3), we have 3m−1 |Aut(P Sp2 (3)) : = 2 2 P Sp2 −2 (3)| = 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 contradiction. Next consider case λ = λ −1 . As dimL(λ −1 ) = 2 2 − − 1 − εp ( ) is odd, = 6k + 2, 6k + 3 or 6k + 4. 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 . ⊥ ∼ Sp2( −2) (3), Let H, K be the centralizers in S of N1 , N1 , respectively. Then H = K ∼ Sp4 (3), and H, K commute. In the basis β1 = {e1 , e2 , f1 , f2 }, let =     1 1 0 0 1 0 1 0 0 1 0 0 0 1 0 0    g= 0 0 1 0 , h = 0 0 1 0 . 0 0 −1 1 0 −ξ 0 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 Kv . Thus E × H Sv , hence 1 + c + d |Aut(S) : (E × H)| = |Aut(P Sp2 (3)) : (E × P Sp2( −2) (3))| < 38 −7 . Hence 2m + 1 < 16 − 11. Since 2m + 1 = 2 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 2 2 − 16 − ( − 8) + 1 = (2 − 1)( − 8) + 1 > 0, a contradiction. Thus 4. If = 4, then dimL(λ −1 ) = 27, by using [10], equation (3.1) cannot hold. If = 3, then dimL(λ −1 ) = 13. Using [10], 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 are left with case = 4, λ = λ1 , dimL(λ1 ) = 41. Case f 2. It follows from case f = 1 that if λ is a 3-restricted dominant weight such that L(λ) is self-dual and is of odd degree then dimL(λ) 2 2 − −1−εp ( ) 2 2 − − 2. By [21, Theorem 5.4.5] and [30, 1.11], 2m + 1 = (dimΨ)f , for some self-dual irreducible k S-module Ψ of odd degree. Thus 2m + 1 = (dimΨ)f (2 2 − 2 − 2)f . By (5.3), 3m−1 |Aut(P Sp2 (3))| 3f (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 this inequality cannot happen. 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 the form B= 0 I I 0 . 36 HUNG P. TONG-VIET From the isomorphisms Ω+ (q) ∼ SL2 (q) ◦ SL2 (q), Ω− (q) ∼ L2 (q) and P Ω± (q) ∼ = = = 4 4 6 L± (q), we can assume that 4. 4 Proposition 5.24. 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 1-3. Proof. Assume (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. = 2 By inequality (5.3), 3m−1 |Aut(P Ωε (3))| 32 − +2 . Thus 2m+1 4 2 −2 +7. 2 By [28, Theorem 5.1 and 4.4], either 4 and λ = λ −1 , 2λ or = 4 and λ = 2λ1 , 2λ2 . If λ = λ −1 , then dimL(λ) = 2 2 − and L(λ) ∼ ∧2 N. Since dimL(λ) is odd, = must be odd. Thus 5. Let a = e1 − ξf1 , b = e2 + f2 ∈ N and z = a ∧ b ∈ ∧2 N, 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, dim(N1 ) = 2, and sgn(N1 ) = ξ. ⊥ ⊥ ⊥ Since N = N1 ⊥ N1 , dim(N1 ) = 2 − 2 and sgn(N ) = sgn(N1 ) · sgn(N1 ), it ⊥ follows from [21, Proposition 2.5.11] that sgnN1 = εξ. Since the discriminant D(Q) ≡ detB = (−1) = −1 (mod(F∗ )2 ), as is odd, by [21, Proposition 4.1.6], ⊥ H = Ω(N1 ) ∼ Ωεξ−2 (3) Ω(N ) centralizes N1 . Hence H M z . Therefore = 2 εξ 2 −2 −1 ε (3 − ε)(3 + εξ) 34 −1 . Since 1 + c + d |Aut(P Ω2 (3)) : Ω2 −2 (3)| = 4 · 3 m−1 (2 2 − −3)/2 4 −1 5, 3 =3 >3 > 1 + c + d, a contradiction to (5.3). If λ = 2λ , then dimL(λ) = 2 2 + − 1 − ε3 ( ) and V = L(λ) ∼ w⊥ /(w⊥ ∩ w ), = −1 2 where w = i=1 (ei ⊗ fi + fi ⊗ ei ) ∈ S N if S is of type D and w = i=1 (ei ⊗ fi + fi ⊗ ei ) + x ⊗ x + y ⊗ y otherwise. Let zξ = e1 ⊗ e1 + ξf1 ⊗ f1 + w ∩ w⊥ , ξ = ±1, and N1 = e1 , f1 N. Then zξ is non-singular in V. Let N1 = e1 , f1 N and H be ⊥ the centralizer of N1 in Ω(N ) ∼ Ωε (3). Since sgn(N1 ) = ε, H ∼ Ωε −2 (3) and H = 2 = 2 ﬁxes zξ . Thus 1+c+d |Aut(P Ωε (3)) : Ωε −2 (3)| 12·32 −2 (3 −ε)(3 −1 +ε) 2 2 2 34 . If 5, then 3m−1 3(2 + −5)/2 > 34 > 1 + c + d, contradicts to (5.3). If = 4, then 2m + 1 = 35, hence m − 1 = 16. As 3m−1 = 316 = 34 > 1 + c + d, we also get a contradiction to (5.3). Case f 2. It follows from case f = 1 that if λ is a 3-restricted dominant weight such that L(λ) is self-dual and is of odd degree then dimL(λ) 2 2− . As V is an absolutely irreducible k S-module, by [21, Theorem 5.4.5] and [30, 1.11], 2m+1 = (dimΨ)f , for some self-dual irreducible k S-module Ψ of odd degree. Thus 2 2m + 1 = (dimΨ)f (2 2 − )f . By (5.3), 3m−1 |Aut(P Ωε (3))| 3 · 3f (2 − +1) , 2 so that 2m + 1 < 2f (2 2 − + 1) + 5. Combining these two inequalities, we have (2 2 − )f < 2f (2 2 − + 1) + 5, where 4, f 2. By induction, this cannot happen. Assume that S is simply connected of exceptional type deﬁned 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 . Proposition 5.25. Assume M is almost simple of type S, where S is of exceptional type above and is deﬁned 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 1G by Corollary 3.7 and so they are in Table 2 or P M (L, S, λ) = (Ω77 (3), E6 (3), λ2 ), (Ω133 , E7 (3), λ1 ) and they are in Table 3. RANK 3 PERMUTATION CHARACTERS 37 Proof. Assume (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). (a) Case G2 . By (5.3) 3m−1 |Aut(G2 (3e ))| 315e . Thus 2m + 1 30e + 3. Assume ﬁrst that e = 1. Then 2m + 1 33. Let λ ∈ X3 be a 3-restricted dominant weight with V ∼ L(λ). From Appendix A.49 in [28], 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 [5], these are maximal embeddings and G2 (3) has only one orbit in Eξ (V ). If λ = 2λ1 or 2λ2 then dimV = 27. We have G2 (3) Ω27 (3). But 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 5.22). Assume that e 2. By [21, Theorem 5.4.5] and [30, 1.11], 2m+1 = (dimΨ)e , for some self-dual irreducible k S-module Ψ of odd degree. It follows that 2m + 1 7e . Combining with 2m + 1 30e + 3, we have e = 2. Then dimV = 72 = 49. However G2 (9) is not maximal in Ω49 (3) since G2 (9) Ω7 (9) Ω72 (3) where the ﬁrst embedding arises as in previous case, while the second is the twisted tensor product embedding. (b) Case F4 . By (5.3), 3m−1 |Aut(F4 (3e ))| 353e . So 2m + 1 106e + 3. Assume that e = 1. From Appendix A.50 in [28], λ = λ4 and dimL(λ) = 25. In this case, we have an embedding F4 (3) Ω25 (3). By [4], 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 2. we have 2m + 1 25e , so that 25e 106e + 3. But this cannot happen for any e 2. (c) Case ε E6 . By (5.3), 3m−1 |Aut(ε E6 (3e ))| 379e . Thus 2m + 1 158e + 3. Assume that e = 1. From Appendix A.51 in [28], 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 E6 over F3 . If e 2, then 77e 158e + 3. However this cannot happen for any e 2.. (d) Case E7 . By (5.3), 3m−1 |Aut(E7 (3e ))| 3134e . Thus 2m + 1 268e + 3. Assume that e = 1. From Appendix A.52 in [28], 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 2. We have 133e 268e + 3. This cannot happen for e 2. (e) Case E8 . By (5.3), 3m−1 |Aut(E8 (3e ))| 3249e . Thus 2m + 1 498e + 3. Assume that e = 1. From Appendix A.53 in [28], dimL(λ) 3875 > 498.1+3 = 501. Assume that e 2. Clearly 3875e > 498e + 3 for any e 2. (f ) Case 3 D4 . By (5.3), 3m−1 |Aut(3 D4 (3e ))| 330e . Thus 2m + 1 60e + 3. Assume that e = 1. From Appendix A.53 in [28], 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 2. Clearly 35e > 60e + 3 for any e 2. (g) Case 2 G2 . By (5.3), 3m−1 |Aut(2 G2 (32e+1 ))| 38(2e+1) . Thus 2m + 1 32e + 19. By [21, Theorem 5.4.5], 2m + 1 72e+1 , and so 72e+1 32e + 19. This cannot happen. 5.3.4. Embedding of Sporadic groups. Let S be a sporadic simple group and let S be the universal covering group of S. Deﬁne g3 (S) := [2log3 (|Aut(S)|) + 4] and 3 (S) to be the minimal degrees of irreducible faithful representations of S and its covering groups over F3 . Lemma 5.26. Let S be a simple sporadic group. Then in Table 7. 3 (S) and g3 (S) are given 38 HUNG P. TONG-VIET Table 7. Minimal degrees of representations for sporadic groups in characteristic 3. S M11 M12 2.M12 J1 M22 2.M22 J2 2.J2 M23 HS 2.HS J3 M24 M cL He Ru 2.Ru M 3 (S) 5 10 6 56 21 10 13 6 22 22 56 18 22 21 51 378 28 196882 g3 (S) 20 26 26 25 28 28 29 29 33 37 37 37 39 42 45 50 50 229 S Suz 2.Suz ON Co3 Co2 F i22 2.F i22 HN Ly Th F i23 Co1 2.Co1 J4 F i24 B 2.B 3 (S) g3 (S) 64 54 12 54 154 54 22 53 23 61 77 63 176 63 133 65 651 74 248 75 253 82 276 82 24 82 1333 87 781 106 4371 144 96256 144 Proposition 5.27. 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 1-3. Proof. Assume equation (3.1) holds for some r ∈ {s, t} and for any M orbits in Eξ (V ). Then by (5.3), |Aut(S)| 3m−1 . It follows that 3 (S) 2m + 1 2log3 (|Aut(S)|) + 3 g3 (S). Recall that V is an absolutely irreducible F3 S-module with dimV = 2m + 1 and the Frobenius-Schur indicator of V is +. By [14] and Lemma 5.26, we need to consider the following cases: (S, dimV ) = (M22 , 21), (M cL, 21), and (Co2 , 23). If (S, dimV ) = (M22 , 21), then M22 < A22 < Ω21 (3), 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. 5.3.5. Computation with GAP. In this section we brieﬂy demonstrate how to compute the parameters c and d deﬁned in Section 3 by using GAP [10]. We still assume the hypotheses and notation from Section 4. Recall from section 5.1 that if x ∈ Eξ (V ) then ∆(x) = Eξ (V ) ∩ x⊥ and Γ (x) = Eξ (V ) ∩ (V − x⊥ − { x }). RANK 3 PERMUTATION CHARACTERS 39 Let M be a subgroup of G, where G is almost simple with socle L = Ω2m+1 (3), m 3. Note that Eξ (V ), ξ = ±, is an L-orbit of non-singular points of type ξ. The parameters c and d are deﬁned as follows: d = | x M ∩ ∆(x)| and c = | x M ∩ Γ (x)| = | x M | − d − 1. As an example, let M ∼ S8 , a symmetric group of degree 8 and L = Ω13 (3). Then = M embeds into L via the Specht module V corresponding to the partition (6, 2) (see Proposition 5.19). Let U be the permutation module for S8 in characteristic 3. We know that V is a composition factor of the tensor product U ⊗ U. Using MeatAxe to decompose U ⊗ U to obtain V. From this module, we can obtain the embedding of M into Ω(V ) and also the quadratic form of V. Using this information, it is very easy to compute c and d. >U:=PermutationGModule(SymmetricGroup(8),GF (3));; GF (3) is the ﬁeld of size 3. >T:=TensorProductGModule(U,U); The tensor product U ⊗ U. >decomp:=MTX.CompositionFactors(T); >V:=decomp[Position(List(decomp,x− >MTX.Dimension(x)),13)]; ﬁnd the module of dimension 13. >f:=MTX.InvariantBilinearForm(V); The bilinear form on V. >gens:=MTX.Generators(V); The generators for M when embedded in Ω(V ). >M:=Group(gens);; >ob:=OrbitsDomain(M,GF(3) (MTX.Dimension(V)));; all orbits of M on V. >leng:=List(ob,x− >Size(x));; The lengths of all orbits of M on V ; >Positions(leng, 315); ﬁnd the positions of orbits of length 315. for example, the positions are [54, 65, 72, 73]. These positions are not ﬁxed. > x := ob[54][1]; ; x ∗ f ∗ x; x is a representative for the orbit ob[54]. x ∗ f ∗ x; is the norm of x. The diﬀerent norms mean the points are in diﬀerent L-orbits. > op:=Orbit(G,GF(3) (MTX.Dimension(V)),x);; op is the M -orbit x M. > d:=0;;for i in [1..Size(op)] do if op[i] ∗ f ∗ x = 0 ∗ Z(3) then d := d + 1; ﬁ;od; Z(3) is a generator for the ﬁeld GF (3). > c:=Size(op)-1-d; For matrix groups, given the matrix generators ‘gens’, which can be obtained from the package ‘atlasrep’ or from the Atlas website [6]. The module is then constructed by using the GAP command ‘V:=GModuleByMats(gens,GF(3))’. > LoadPackage(‘atlasrep’); > DisplayAtlasInfo(‘G2 (4)’); > gens:=AtlasGenerators(‘G2 (4)’,9).generators;; This is the 64-dimensional representation of G2 (4) in characteristic 3. > V:=GModuleByMats(gens);; > M:=Group(gens); 40 HUNG P. TONG-VIET Acknowledgment. This work is part of my Ph.D thesis. I would like to thank my thesis advisor Professor Kay Magaard for his guidance and his help. References 1. M. Aschbacher, R. Guralnick, and K. Magaard, Guralnick-Magaard Problem, in preparation (2009), 1–8. 2. N. Bourbaki, Lie Groups and Lie algebras, Springer, Berlin, 2005. 3. E. Cline, B. Parshall and L. Scott, Cohomology of ﬁnite Groups of Lie type, Publications Mathematiques de L’I.H.E.S. 45 (1975), 169–191. 4. A. Cohen and B. Cooperstein, The 2-spaces of the standard E6 (q)-module, Geometriae Dedicata 25 (1988), 467–480. 5. J. H. Conway et al., Atlas of ﬁnite groups, maximal subgroups and ordinary characters for simple groups, with computational assistance from J. G. Thackray, OUP, Eynsham, 1985. 6. R.A. Wilson et al., A world wide web atlas of ﬁnite group representations-V3, http://brauer.maths.qmul.ac.uk/Atlas/v3. 7. B. Ford, Irreducible representations of the alternating group in odd characteristic, Proc. AMS 125 (1997) no. 2, 375–380. 8. B. Ford and A.S. Kleshchev, A proof of the Mullineux conjecture, Math. Zeitschr 226 (1997), 267–308. 9. D. Frohardt, R.M. Guralnick and K. Magaard, Incidence matrices, permutation characters, and the minimal genus of a permutation group, Journal of Combinatorial Theory, Series A 98 (2002), 87–105. 10. The GAP Group, GAP-Groups, Algorithms, and Programming, http://www.gap-system.org, Version 4.4.7, 2006. 11. L.C. Grove, Classical groups and geometric algebra, Graduate Studies in Mathematics, Vol. 39, American Mathematical Society, 2002. 12. R. M. Guralnick, K. Magaard, J. Saxl, and P.H. Tiep, Cross characteristic representations of symplectic and unitary groups, Journal of Algebra 257 (2002), 291–347. 13. D.G. Higman, Finite permutation groups of rank 3, Math. Zeitschr, 86 (1964), 145–156. 14. G. Hiss and G. Malle, Low-dimensional representations of quasi-simple groups, LMS J. Comput. Math. 4 (2001), 22–63. 15. J.E. Humphreys, Introduction to Lie algebras and representation theory, Graduate Texts in Mathematics, no. 9, Springer-Verlag, 1972. 16. J.E. Humphreys, Linear algebraic groups, Graduate Texts in Mathematics, Springer-Verlag, New York, Heidelberg, Berlin, 1987. 17. I.M. Isaacs, Character theory of ﬁnite groups, AMS Chelsea Publishing, AMS, Province, Rhode Island, 2006. 18. G.D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, Vol. 68, Springer-Verlag, Berlin, Heidelberg, New york, 1978. 19. G.D. James, On the minimal dimensions of irreducible representations of symmetric groups, Math. Proc. Camb. Phil. Soc. 94 (1983), 417–424. 20. C. Jansen, K. Lux, R. Parker and R. Wilson, An atlas of Brauer characters, LMS Monographs, New Series 11, Oxford Science Publications, 1995. 21. P. Kleidman and M.W. Liebeck, The subgroup structure of the ﬁnite classical groups, LMS Lecture Note Series 129, Cambridge University Press, 1990. 22. A.S. Kleshchev and J. Sheth, Representations of the alternating group which are irreducible over subgroups, Proc. London Math. Soc. 84 (2002), no. 3, 194–212. 23. A.S. Kleshchev, Branching rules for modular representations of symmetric groups I, Journal of Algebra, 178 (1995), 493–511. 24. V. Landazuri and G. Seitz, On the minimal degrees of projective representations of the ﬁnite Chevalley groups, Journal of Algebra, 32 (1974), 418–443. 25. M. W. Liebeck, C.E. Praeger and Jan Saxl, The maximal factorizations of the ﬁnite simple groups and their automorphisms, Memoirs of the AMS, vol. 86, no. 432, AMS, Province. Rhode Island. USA, July, 1990. 26. M. W. Liebeck and Gary M. Seitz, On ﬁnite subgroups of exceptional algebraic groups, J. Reine Angew. Math. 515 (1999), 25–72. RANK 3 PERMUTATION CHARACTERS 41 27. M.W. Liebeck, The aﬃne permutation groups of rank three, Proc. London Math. Soc. 54 (1987), no. 3, 477–516. 28. F. L¨ beck, Small degree representations of ﬁnite Chevalley groups in deﬁning characteristic, u LMS J. Comput. Math. 4 (2001), 135–169. 29. K. Magaard and G. Malle, Irreducibility of alternating and symmetric squares, Manuscripta Mathematica, 95 (1998), no. 2, 169–180. 30. G. Seitz, The maximal subgroups of classical algebraic groups, Memoirs of the AMS, vol. 47, no. 365, American Mathematical Society, Province. Rhode Island. USA, 1987. 31. G. Seitz and A.E. Zalesskii, On the minimal degrees of projective representations of the ﬁnite Chevalley groups II, Journal of Algebra, 158 (1993), 233–243. 32. R. Steinberg, Endomorphisms of linear algebraic groups, Memoirs of the AMS, vol. 80, American Mathematical Society, Province, Rhode Island, 1968. 33. H.P. Tong-Viet, Rank 3 permutation characters and maximal subgroups, Ph.D-Thesis, University of Birmingham, 2009. E-mail address: Tong-Viet@ukzn.ac.za School of Mathematical Sciences, University of KwaZulu-Natal, Pietermaritzburg 3209, South Africa

Rank 3 permutation characters and maximal subgroups more

Log In

Reset Password