The maximal subgroups of the Chevalley groups F4(F) where F is a finite or algebraically closed field of characteristic not equal to 2,3 Citation Magaard, Kay (1990) The maximal subgroups of the Chevalley groups F4(F) where F is a finite or algebraically closed field of characteristic not equal to 2,3. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/D2GB-VK65. https://resolver.caltech.edu/CaltechETD:etd-06132007-094324 Abstract NOTE: Text or symbols not renderable in plain ASCII are indicated by [...]. Abstract is included in .pdf document. We find the conjugacy classes of maximal subgroups of the almost simple groups of type F4(F), where F is a finite or algebraically closed field of characteristic not equal to 2,3. To do this we study F4(F) via its representation as the automorphism group of the 27-dimensional exceptional central simple Jordan Algebra J defined over F. A Jordan Algebra over a field of characteristic not equal to 2 is a nonassociative algebra over a field F satisfying xy = yx and [...] = [...] for all its elements x and y. We can represent Aut(F4(F)) on J as the group of semilinear invertible maps preserving the multiplication. Let G = F4(F) and [...]. We have defined a certain subset of proper nontrivial subalgebras as good. The principal results are as follows: SUBALGEBRA THEOREM: Let F be a finite or algebraically closed field of characteristic not equal to 2,3. Let H be a subgroup of [...] and suppose that H stabilizes a subalgebra. Then H stabilizes a good subalgebra. The conjugacy classes and normalizers of good subalgebras are also given. STRUCTURE THEOREM: Let H be a subgroup of [...] such that [...] is closed but not almost simple. Then H stabilizes a proper nontrivial subalgebra or H is contained in a conjugate of [...]. The action of [...] on J is described and it is shown that [...] is unique up to conjugacy in G. THEOREM : If L is a closed simple nonabelian subgroup of G, then [...] is maximal in [...] only if L is one of the following: [...]. For each member [...] we identify those representations [...] which could give rise to a maximal subgroup of G and show the existence of [...] in G. Up to few exceptions we also determine the number of G conjugacy classes for each equivalence class [...]. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: Mathematics Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Aschbacher, Michael Thesis Committee: Aschbacher, Michael (chair) Wales, David B. Defense Date: 16 April 1990 Record Number: CaltechETD:etd-06132007-094324 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-06132007-094324 DOI: 10.7907/D2GB-VK65 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 2575 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 06 Jul 2007 Last Modified: 21 Dec 2019 02:04 Thesis Files Preview PDF (Magaard_k_1990.pdf) - Final Version See Usage Policy. 3MB Repository Staff Only: item control page

The Maximal Subgroups of the Chevalley Groups F,(F) Where F is a Finite or Algebraically Closed Field of Characteristic 4 2,3 Thesis by Kay Magaard In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1990 ( Submitted April 16 , 1990 ) ii Acknowledgements I offer my heartfelt thanks to my advisor, Michael Aschbacher, for suggesting the thesis topic, for teaching me many group theoretic techniques, for sharing his ideas with me, and for providing me with many opportunities to interact with the leading figures of Group theory. Through his generosity this thesis has benefited a great deal; I also thank him for that. Additional thanks go to David Wales for explaining various aspects of ordinary and modular representation theory to me. I would also like to thank Peter Kleidman for offering valuable advice, for explaining some of his work to me, and for teaching me how to use the Atlas. Finally, I wish to thank all those people who made my stay at Caltech enjoyable. As they know who they are, I will omit a list. iii Abstract We find the conjugacy classes of maximal subgroups of the almost simple groups of type F4(F), where F isa finite or algebraically closed field of characteristic # 2,3 . To do this we study F4(F) via its representation as the automorphism group of the 27- dimensional exceptional central simple Jordan Algebra J defined over F. A Jordan Algebra over a field of characteristic # 2 is a nonassociative algebra over a field F satisfying xy = yx and (x2y)x = x2(yx) for all its elements x and y . We can represent Aut(F4(F)) on J as the group of semilinear invertible maps preserving the multiplication. Let G=F,(F) and G<P ¢ Aut(G). We have defined a certain subset of proper nontrivial subalgebras as good . The principal results are as follows: SUBALGEBRA THEOREM: Let F be a finite or algebraically closed field of characteristic # 2,3. Let H beasubgroup of [ and suppose that H stabilizes a subalgebra. Then H__ stabilizes a good subalgebra. The conjugacy classes and normalizers of good subalgebras are also given. STRUCTURE THEOREM: Let H be a subgroup of [ such that HM G is closed but not almost simple. Then H_ stabilizes a proper nontrivial subalgebra or H_ is contained in a conjugate of Ny(3%: SL4(3)) . The action of 3°: SL4(3) on J is described and it is shown that 33: SL,(3) is unique up to conjugacy in G. THEOREM : If L._ is a closed simple nonabelian subgroup of G, then Ny(L) is maximal in [ only if L is one of the following : S= {F 4(F)sPSLo(F),Go(F),°D4(2),PSL3(3),PSU3(3),PSL9(q): q€{8,9,13,17,25,27}} For each member L € ¥ we identify those representations x, which could give rise to a maximal subgroup of G and show the existence of x (L) in G. Up to few exceptions we also determine the number of G conjugacy classes for each equivalence class TL: iv Table of Contents Acknowledgements li Abstract ill Table of Contents iv Introduction 1 Chapter I: A Structure Theorem for I 8 Section 1: Facts about Composition Algebras ....... 0... eee eee eee 8 Section 2: Definition of J ; Facts about J and Aut(J) ...........4.. 12 Section 3 : 3 - Forms, the Dickson 3-Form and some of its Properties 2.0... ce et ee eee eee eee eens 18 Section 4 : Generalities on Subalgebras of Jo... 1c ee tenes 29 Section 5 : Nondegenerate Subalgebras ... 1... ee eee re eet ee eee eens . 32 Section 6 : Brilliant Subgroups... 1.6... eee eee ee eee eee ee eee et wee 43 Subalgebra Theorem for G.... 0... eee eee eee ees 49 Section 7 : Local Subgroups of Gow... ee ee ee tee ee tees 50 Section 8 : Non Local Subgroups with Non Simple Socle.............-.. . 63 Section 9 : Outer Automorphisms, Field Extensions and a Structure Theorem for To... ccc eee cee ee ee ee eee eee eee eee 66 Section 10 : More about Subalgebras .. 1.0... 0... eee ee eee eee 70 Chapter II : Almost Simple Subgroups of Lie Type Defined over Fields whose Characteristic equals char(F) 83 Section 11 : Initial Reductions .......... 02 cee ee eee ee eee eee eens 83 Section 12 : Groups of Lie Type Ag and 2A5 ce eee ee ee eee eens 85 Section 13 : Groups of Lie Type Ay seem eee ee eee eee ene nee Leeeees 89 95 Section 14: The Existence and Uniqueness of certain A,’s and Go’s vee eee v Chapter III : Cross Characteristically Embedded Simple Subgroups, Alternating and Sporadic Simple Subgroups 108 Section 15 : Initial Reductions .. 0... 06. ee eee eee eee eens 108 Section 16 : Establishing Nonmaximality of Certain Simple Subgroups of f... 115 Section 17 : Proof of the Simple Subgroup Theorem... ......-- 20+ ee eues 122 References 124 Introduction The study of a finite group almost always necessitates some knowledge of its subgroups. In many instances some knowledge of its maximal subgroups is desirable, if not crucial. This is, in part, because the maximal subgroups of a group determine its primitive permutation representations and vice versa. Also, in certain applications of group theory to other fields of mathematics, one needs information about maximal subgroups. One such application is the inverse problem of Galois theory, i.e., when can a finite group be realized as a Galois group of a Galois extension of the rational numbers. See for example Matzat [M]. It is for these and other reasons that the maximal subgroup problem has existed for as long as the subject of group theory itself. However, up to the late 1970’s the maximal subgroup problem was only solved for special classes of groups. In 1985 Aschbacher and Scott [As10] showed amongst other things that the maximal subgroup problem can be reduced to two well defined problems: Let G bea finite group with L < G < Aut(L) for some simple nonabelian simple group L, and let V_ be a faithful irreducible G-module over some field of prime order (dividing |G). (1) Determine H1(G,V). (2) Determine the maximal subgroups of G not containing L. Any group H satisfying the assumptions of G is called an a/most simple group. Over the course of the past ten years, significant progress has been made on the second problem. For example, O’Nan-Scott and Aschbacher proved powerful structure theorems for the Alternating and Symmetric groups, the classical groups, respectively, 2 which in some sense make up the bulk of the finite simple groups. The structure theorems go as follows: Let G _ be an almost simple group. Suppose that G is represented as a group of automorphisms of some mathematical object X(G). Then any proper subgroup H of G either stabilizes some member of the set C(G,X(G)) of "natural structures” on X(G) or H is an almost simple group acting irreducibly on X(G). The natural structures C(G,X(G)) are usually substructures, coproduct or product structures on X(G). For example X(G) is the set {1,....n} when G is the Alternating group of degree n, and when G_ is a classical group over a field F, then X(G) is a pair (V,f) where V_ is a vectorspace over F and f_ is trivial or a nondegenerate symplectic, hermitian or quadratic form on V. Once a structure theorem is proved for a particular group G, enumeration of its maximal subgroups involves listing the almost simple groups that act irreducibly on X(G) and then deciding which of these occur as maximal subgroups of G. At this- stage the classification of the finite simple groups and their representation theories are often used to decide which groups possesses suitable irreducible representations on X(G). For most almost simple groups we have either structure theorems or know explicitly the conjugacy classes of maximal subgroups. Presently, for the families of exceptional groups of Lie type F,(F), E7(F), E,(F) as well as the sporadic groups Baby Monster and the Monster, we have neither. The goal of this thesis is to find the conjugacy classes of maximal subgroups of the almost simple groups of type F,(F), where F is a finite or algebraically closed field of characteristic 4 2,3 . I believe it’s best to study F,(F) via its representation as the automorphism group of the 27- dimensional exceptional central simple Jordan Algebra defined over F. Recall 3 that a Jordan Algebra over a field of characteristic # 2 is a nonassociative algebra over a field F satisfying xy = yx and (x2y)x = x2(yx) for all its elements x and y. A homomorphism of a Jordan Algebras is a linear map preserving the multiplication. An algebra is called simple when it has no homomorphic images other than 0 and itself. The algebra is called central simple when it’s simple modulo the center. Recall here that an element z is in the center of a Jordan algebra iff zx =Ax for all x in the algebra and AE F is some fixed element. The results of Albert [A1], Jacobson and Jacobson [Jal], Jacobson [Ja6], and Schafer [Schal] classify the central simple Jordan algebras over F. It turns out that all but one family of examples is constructed from some associative algebra A over F by redefining the product of A as ab = 1/2 ( a*b + bxa) for all a,b € A where * denotes the associative product. The examples that can’t be constructed from an associative algebra in this way are called exceptional When F is finite or algebraically closed there is, up to isomorphism, a unique example J of exceptional type. Its dimension over F is 27. It’s automorphism group is F,(F). We can represent Aut(F4(F)) on J as the group of semilinear invertible maps preserving the multiplication. From now on we will let G=F,(F) and G<T < Aut(G). By a subalgebra of J we mean a subspace containing id (the identity of J) which is closed under multiplication. By a proper nontrivial subalgebra we mean a subalgebra A such that <id> < A < J. We now give a rough definition of the class of good subalgebras. By a composition algebra we mean a pair (V,N) where V an F-algebra with a multiplicative identity and Nis a nondegenerate quadratic form on V such that N(xy) = N(x)N(y) for all x,y € Vv. 4 Albert and Jacobson [A2] have characterized J as 11 ¢1 €9 { fy 2 63 : where 1% € F and c.€O \ Cg 3 13 where O denotes the unique eight dimensional composition algebra over F, and — denotes the unique automorphism of order 2 of O such that c€ = N(c) 1 for all cé QO. We define multiplication on J by xy := 1/2(x*y + y*x) , were * denotes ordinary matrix multiplication. Note that this multiplication is commutative. We define Q to be the quadratic form Q(x) = 1/2 tr(x”) on J. An idempotent is an element x € J satisfying x2 = x. An idempotent x is primitive if Q(x) = 1/2. Let Ay = <id,x> where x is a primitive idempotent. Let Ay = <x,y,z> where x,y,z are pairwise orthogonal (wrt. Q ) primitive idempotents. Let Ag be a three dimensional extension field of F contained in J. 7% 44 dy do d3 78 where D is a proper nondegenerate (wrt. N ) subalgebra of O. We remark that up to conjugacy in Aut(Q) there are four (three) choices for D when F is finite (algebraically closed) respectively. Finally let Ap be the set of subalgebras B such that xy = 0 forall x,y € BN iat plus some technical conditions described in section 6. We remark that the stabilizers of members of Ap are the maximal parabolic of I. 5 Now we define a subalgebra to be good if a is conjugate to a member of We prove the following: SUBALGEBRA THEOREM: Let F be a finite or algebraically closed field of characteristic # 2,3. Let H be asubgroup of [ and suppose that H_ stabilizes a subalgebra. Then H_ stabilizes a good subalgebra. The structure of the stabilizers of good subalgebras can be found in propositions 2.7 and 2.10 in lemma 5.2 and 5.13 and proposition 6.4. STRUCTURE THEOREM: Let H be a subgroup of I such that H NM G is closed but not almost simple. Then H_ stabilizes a proper nontrivial subalgebra or H_ is contained in a conjugate of N,(3°: SL4(3)) . In lemmas 7.6 , 7.7 and 7.9 we describe the action of 3°: SL3(3) on J and prove that this 33; SL3(3) is unique up to conjugacy in G. SIMPLE SUBGROUP THEOREM: If L is a closed simple nonabelian subgroup of G then N,(L) is maximal in [I only if L is on the following list and J affords the representation p of L: L P # of Gconj. cl. nec. cond. for p F4(Fo) M(0)®M(Aq) 1 [F:Fo] is a prime PSLO(F) M(0)@M(8A,)®M(16A,) 1 p>l7orp=0 PSLo(F) M(0)@M(8,)/M(A,)°@M(3A,)/M(8A,) 1 p=13 , [FI] #13 L p # of G conj. cl. nec.cond. for ex. of p PSLo(13) M(0)@ M(8Aq)\M(2A,)®M(4A,)\M(8A,) >1 p=13 Go(F) M(0)®M(2A,) 1 p=7 PSLo(7) M(0)®M(6A,)®M(4.}) ®M(4A,)\M(0)@M(2A,)\M(4A,) 1 p=7 3D 4(2) 1+26 1 p27 3D, (2) 1+26 >1 p=7 PSL,(3) 1+26 >1 all p PSU4(3) 1426 >1 =7 PSL»(27) 1+26 >1 all p PSL.(25) 1+26 >1 p#5,-1/e F PSL.(17) 14+9+16 >1 p#17 PSL,(13) 1412414 >1 =7 PSL,(9) 148+9+49 >1 p#5 PSL,(8) 14+8+9+9 >1 p#7 The symbol ? means that it is not known wether an appropriate embedding exists. However, if it did one would get an example of a maximal subgroup. The symbol > 1 means that at least one embedding of the specified type into G exists. However, in case L € {PSL,(3), PSU3(3), PSLj(13)} the only known embedding does not lead to a maximal subgroup. More precise information about the groups on the list can be found in Sections 14, 16 and 17. Now some words about the strategy and organization. Section 1 is a collection of facts about composition algebras that are needed in the subsequent discussion of the exceptional 27-dimensional Jordan algebra of section 2 . Section 3 discusses Aschbacher’s representation of Eg: Also a proof of the fact that Aschbacher’s Eg- 7 form is isometric to the trilinear form arising from J is given in section 3. In sections 4, 5 and 6 the subalgebra stabilizers of G are analyzed leading to a proof of the Subalgebra Theorem for G_ in section 6. In section 7 we analyze the normalizers of semisimple abelian subgroups of G. In section 8 we analyze nonlocal subgroups whose generalized Fitting subgroup is not a finite simple group and prove the Structure Theorem for G. In section 9 we generalize the Subalgebra Theorem and the Structure Theorems from G to [. Starting from section 10 we deal exclusively with almost simple subgroups of G. In section 10 we show that when a simple subgroup L of G stabilizes a proper nontrivial subalgebra then N,(L) is not maximal in [. The sections 11 - 14 deal with the simple groups of Lie type defined over the same characteristic as G. The remaining sections deal with the cross- characteristically embedded simple groups of Lie type, the Alternating, and the Sporadic simple groups. 8 CHAPTER I: A STRUCTURE THEOREM FORT SECTION 1: FACTS ABOUT COMPOSITION ALGEBRAS Definition: A composition algebra C over F is an (not necessarily associative) F-algebra defined by the following properties: 1. C has a multiplicative identity 1 2. C admits a nondegenerate quadratic form N such that N(xy) = N(x)N(y) for ail x,y € C (N will be called the associated quadratic form of C.) By a homomorphism of composition algebras we mean a linear map which preserves multiplication. The following is well known and can be found for instance in [Spl] or [Ja2]. COMPOSITION ALGEBRA THEOREM: 1. Every composition algebra over F is 1, 2, 4 or 8 dimensional over F. 2. Two composition algebras are isomorphic iff their associated quadratic forms are equivalent. 3. If a composition algebra C contains an isotropic vector (i.e., an x € C such that N(x) = 0 ) then the associated quadratic form has maximal Witt-index. In particular if F is algebraically closed, then two composition algebras over F are isomorphic iff their dimensions are equal. 4. If F is finite or algebraically closed, there is one isomorphism class of m dimensional composition algebras over F, where m = 1,4,8 . There are two isomorphism classes of two dimensional composition algebras over F as there are two nonequivalent 9 choices for N. When F is algebraically closed there is a unique isomorphism class of two dimensional composition algebras over F . 5. There exists an involutory automorphism of C called conjugation and denoted by —,such that x¥ =X x = N(x) 1 Proof: Part 1 can be found in [Ja2] pg. 425. Part 2 is Ex 2, pg. 428 [Ja2]. Part 3 is ex. 3, pg. 428 [Ja2]. Part 5 is on pg. 422 [Ja2]. Part 4 is a consequence of part 3 and the classification of quadratic forms over finite and algebraically closed fields. By part 5 above we know that 1 is a nondegenerate vector. Thus C = <1> + <i>t. We denote by (x) the projection of x€C onto <1> with respect to that decomposition. Definition: Let F be finite or algebraically closed. We will denote by O the eight dimensional composition algebra over F. O will also be called an octave. The following is also well known . Aut(Q) Theorem: 1, Aut(O) is the Chevalley group Go(F). 2. If A,B C O are isomorphic N-nondegenerate composition subalgebras, then there exists a g ¢ Aut(O) such that Ag = B. Proof: Part 1 is proved in [Spl] Pg 15 - 20. Part 2 is exercise 2, pg. 428 in [Ja2]. For future reference we give a multiplication table (table 1.1) for O, found in [Ja2]}ch8 . 10 Table 1.1 1 iy ip ig iy is ig iz 1 i I ig ig ig is ig i7 iy iy cyl ig Cyi9 is, C414 “iz -Czig ip ip - ig Col ~Coly ig iz Col4 Coir, ig ig “Cyn Coly -€) Col iz Cyig “Cols “C4Coi4 i4 i4 “i, -ig “iz Cal “Cgiy ~Cgig ~Cgig is is ~Cyi4 -i7 -Cyig Cgiy -€yCgl Cgig Cy Cgi9 ig ig iz ~Coig Cais Caio ~Cgig “CoCgl “CoCgiy iz iz Cyig -Cgig = Cy Cig Cgig -€4Cgin CoCgiy Cy CgCgl The associated quadratic form N of O is given by: N(w) = x? - c,x? - coxd + €4Cx2 - cgx4 + c44x? + Cotgxe - €19¢3x4 , where we O is w = Xgl + 35 x; iy. We will choose c= 1 and Co = -l= Cg to assure that N will have maximal Witt- index. Note also that <igs ’ ig; +1 > j€{0,1,2,3} is a hyperbolic pair with respect to N. Fact 1.2: With the choice of c, as above the set { 1, i, : 1< k <7} becomes an orthogonal (i.e., all basis elements are pairwise orthogonal) basis of O with respect to the quadratic form N. Moreover N_ is nondegenerate. Fact 1.3: There are three (four) Aut(Q) conjugacy classes of nondegenerate composition subalgebras when F is algebraically closed (finite). The following four subalgebras are representatives: <1>, <L,ij>, <1, ip,19,13>, ( <l,w>, where -N(w) ¢ F? and we 1 4 ). We will denote these by 1, F2, Ft, K, respectively. 11 Fact 1.4: The following are true: 1. X= (x,1) 1- (x - (x,1)1) 2. (xy,z) = (x,2¥) = (y,Xz) Proof: This is immediate from the equation N(xy) = N(x)N(y). A proof can be found in [Sp1] pg. 2. Note that O is not an associative algebra, however, the following is true. Proposition 1.2 : Any proper composition subalgebra of O generated by two vectors is associative, and every pair of vectors generates a proper composition subalgebra. In particular the subalgebra generated by two basis vectors of table 1.1 is associative. Proof: This is well known and a proof can be found in [{Sp1] pg. 7 . Corollary 1.3: Recall the projection map # from above. Then for all] triples c,EO (cy (egeg)) = 4((cyC¢)cg)) - Proof: The composition algebra multiplication is bilinear and 7 is a linear map. So it is enough to show that the claim holds for any triple of basis vectors {1 ly seesiz}- Now observe, using table 1.1, that (ir(igi,))# 0 iff igi,€ <ip> . So the only triples that contribute something nonzero lie in a subalgebra generated by two vectors, i.e., an associative subalgebra. The claim follows. Proposition 1.4: Let V be a nontrivial eight dimensional Aut(O) module and N an Aut(Q) invariant quadratic form on V. Then (V,N) admits a unique Aut(O) invariant composition algebra structure with identity element 1 and with respect to the quadratic form N. Proof: Recall that Aut(O) ~ Go(F). We observe also that V = M(0) ® M(Aj) as an 12 Aut(O)-module. Suppose * is another composition algebra multiplication. Then (x,y,2) == (xy,z) is a nontrivial alternating Aut(O) invariant trilinear form on M(A,) ~it , where 1 is taken with respect to N. The same is true for (- ,- ,-), defined analogous to (- ,- ,-) but using *. So by [As7] 10.9 we have (x,y,z), = a (x,y,z) for all x,y,z € 1+ and 04#a€F. Therefore the following holds: (x#y,z) = a(xy,z) for all x,y,z € it. From Fact 1.4.2 we get (xy,1) = (x,y) = (x«*y,1). So for x,y € 1+ we can write the following: (+) xey - (x#y,1)1 = £ (xevii = a(xysis)is = a (xy - (xy,1)1). If we apply N to (+) after we get the following: (++) N(xey) = (1-a)?(xy,1)?+ a? N(x)N(y). Now we choose x,y € 1+ such xLly and N(x) # 0 # N(y) and substitute into (++) we get: N(x)N(y) = N(x#y) = a N(x)N(y). So a=+1. Now we choose x,y so that (xy,l) = (x,y) #0 and that N(x) #0 # N(y). Also set a =-1. Substituting this into (++) gives the following: N(x)N(y) = N(xty) = 4(xy,1)? + N(x)N(y), contradicting (xy,1) # 0. The only way out is that a = 1. Then the claim follows from (+). SECTION 2: DEFINITION OF J; FACTS ABOUT J AND Aut(J) Let F be a finite or algebraically closed field of characteristic # 2,3 and G = F4(F). Let O be the octave algebra over F. Let J be the exceptional 27 dimensional central simple Jordan Algebra over F (c.f. introduction for definitions). 13 Proposition 2.1: J can be characterized as 1 %1 2 C1 19 ¢3 : where 406 **F and c € O \ 2 ©3 73 We define multiplication on J by xy := 1/2(x*y + y*x), were * denotes ordinary matrix multiplication. Proof: See [A2]}. Note that J is a commutative nonassociative F-algebra. We denote by Aut(J) the set of invertible linear maps g:J — J satisfying (xy)g = (xg)(yg) forall x,y ¢€ J. Proposition 2.2: The following are true: 1. G= Aut(J). 2. J= M(0) ® M(A4) asa G module, where M(A) means the irreducible G module of high weight A wrt. some fixed Cartan subgroup H of G. Proof : When F is algebraically closed, this is shown in [Spl] pg. 100 Satz 20. For F finite this is shown in [Ja4]. Definitions: Q(x) := (1/2)tr(x?) » (%y) := Q(x+y) - Q(x) - Q(y). Let x € J and regard J as an algebra of matrices as in proposition 2.1 and recall 7 from before the definition of octave in section 1. Also recall N the quadratic form associated with the octaves. Then define: Det(x) := 717973 - 73N(cy) - ¥gN(cg) - 13N(cq) + (Cy (Co¢3)) + m(€4(Eqe3)). Let (x,y,z) be the symmetric trilinear form such that Det(x) = (1/6) (x,x,x). 14 Jacobson in [Ja3] shows that Det(x) is a cubic form over F. As we are working over fields of characteristic # 2,3 note that (x,y,z) is uniquely given by: (x,y,z) = Det(x+y+z) - Det(x+y) - Det(x+z) -Det(y+z) + Det(x) + Det(y) + Det(z). Proposition 2.3: 1. Q is a nondegenerate quadratic form on J N . Q(id) = 3/2 Q(x?) = Q(x)? for all x eid eo aN . (xy,z) = (x,yz) for all x,y,z € J *) ou . (xy)x?2 = x(yx . xP (x,id) x? + (Q(x) - (1/2) (x,id)?) x + Det(x) id for all xeJ a Proof: [Sp1] pages 64, 65 and 68. Remark: Proposition 2.3.5 and the commutativity of the product of J show that J is indeed a Jordan Algebra. Furthermore it follows from proposition 2.3.5 that J is power associative, i.e., xi is a well defined element of J for all x € J and ieé z*; see for example [Ja2] ch. 7.4. From now on we will use the symbol L to mean orthogonal with respect to Q. Definition: By a subalgebra of J we mean a subspace of J containing id which is closed under the J multiplication. Note that the equation in Proposition 2.3.6 is referred to as Hamilton's equation. One easy consequence of Hamilton’s equation is: 15 Lemma 2.4: Let A be the subspace generated by {id, xian integer greater than O}. Then A is a subalgebra of dimension at most 3. Proof: That A is a subalgebra is a consequence of proposition 2.3.5. The statement about the dimension follows from Hamilton’s equation. Theorem 2.5: 1. ge GL(J) isin G iff (xg,yg) = (x,y) and (xg,yg,zg) = (x,y,z) for all x,y,z € J. 2. ge GL(J) isin G iff idg = id and (xg,yg,zg) =(x,y,z) for all x,y,z € J. 3. { g « GL(J) : (xg,yg,zg) = (x,y,z) for all x,y,z € J} is the universal Chevalley group of type E¢(F). Proof: Parts 1 and 2 are the contents of Lemma 1 and Theorem on pg 186 of [Ja3]. Part 3 can be found in [Ja5]. Definition: By x#y we denote the unique element of J satisfying (x,y,z) = 2( x#y,z) for all z € J. Lemma 2.6: x#y = xy - (1/2)(x,id) y - (1/2)(y,id) x - (1/2)(x,y) id + (1/2)(x,id)(y,id) id. Proof: See [Sp1] pg 64. Definition: An element x ¢ J is a primitive idempotent if x? =x and Q(x) = 1/2. Proposition 2.7: 1. G is transitive on primitive idempotents of J. 2. Let x= . Then x is a primitive idempotent. oo oF oo ©& eo °c 16 3. J = <id,x> 1 E(x) 1 E,(x) where E.(x) = {yeJ: xy = i(1/2) y}. 0c,¢ 0 0 0 4. Ex(x)={ [F,9 9 |: « €O} and Ep(x) ={ 0-7 C3 |: ye F,cg € O}. Cy 9 0 0T37 5. The stabilizer of x in G is Sping (F) with Eg(x) a nondegenerate 9-dimensional space with respect to Q and G,/ Cg, (E(x) = Qg9(EQg(x),Q). Gx acts faithfully on E(x). 6. If A is a Q-nondegenerate subalgebra containing x, then A = <id,x> 1 Ep a(x) LE; q(x) where E; q(x) = E,(x) 1 A. Proof: See [Sp1] for parts 1, 3,5 and 6. Parts 2 and 4 are easy computations. Lemma 2.8: Let yEJ. Then y#y = 0 iff either 1. y is a scalar multiple of a primitive idempotent. 2. (y,id) =Oand y* =0. Proof: This is the content of Satz 10 pg 77 in {Spi}. Lemma 2.9: If ye J and y2 =0,then ye id+ and y#y = 0. Proof: We use Hamilton’s equation (prop. 2.3.6) to deduce that Det(y) id = -(Q(y) - (1/2)(id,y)”) y. Now Q(y) = (1/2)(y,y) = (1/2)(y?,id) = 0. Therefore, since {y,id} is a linearly independent set, Det(y) = Q(y) = (y,id) = 0. So now, using lemma 2.6 y#y = y”-(1/2)(y,id) y - (1/2)(y,id) y - (1/2)(y,¥) id + (1/2)(y,id)? id = 0. 17 Proposition 2.10: 1 0 0 000 0 0 0 Let x, = 0 00 Xq =| 0 1 0 and Xg =| 9 0 0 and 0 0 0 0 0 0 001 1. id = x, + Xq + Xg, (x4 5%5) = by where by is the Kronecker delta. 2J= <X1.Xq.Xg> ae Wi Lt Wo + W3.- 3. CQ(<x1%99%3>) = Sping (F) and CQ(<xpXq%3>) induces 2(W; ,Q) on W.. 4. N GQ(<x1.X9X3>) = Syms semidirect product CQ(<x1.x9.%3>)- Symg, permutes the set {x;:i = 1,2,3}. 5. <X4.XgqsXg> is a nondegenerate subalgebra of J. Proof: Parts 1 and 2 are easy calculations using proposition 2.7. Parts 3 and 4 can be found in [Ja4] section 6 or alternatively in [Spl]. Part 5 is an easy calculation using proposition 2.1.. Let e;(c) denote the matrix of J whose ij entry is c, whose ji entry is T, and all of whose other entries are zero. Then we can identify O with W, via cr €o3(c) Vc € O. We will now describe briefly how the (noncommutative) octave multiplication of Wi (see section 1 for definition of octave) can be recovered from the (commutative) Jordan multiplication of J (see also [Sp1] pg. 75). Let v = 2e,;9(1) and w= e€33(1) and €93(¢),€93(4) € W,. Define y¥4(eg3(c)) = 2e,9(€) and y_(e€93(c)) = €13(¢) and observe that vy_(e93(c)) = wy 4(€93(¢)) = c. Similarly define y 4(€93(4)) = 2e19(4) and y_(€93(4)) = €13(d). Now the octave multiplication A of Wy, can be gotten as follows: €53(¢) Aeg3 (4) = y4(e93(c)) y_(e93(4)) = 2e19(€) €13(4) = €93(cd). 18 Convention: When we speak of a composition subalgebra of W,, we mean the image of a composition subalgebra of O under the above identification of O with Wi. Observation 2.11: Under the identification of W, with O above we have Qly, = 1 N, where N_ is the quadratic form associated to O from section 1. Proof: Let ey3(c) € W,- Then tes Q(e93(¢)) = (1/2) tr(egg(c)”) = (1/2) tr(} 0 cé o}) 0 0 ce (1/2) (2N(c)) (as cé = N(c)L by part 5 of the Comp. Alg. Thm. of Section 1) Remark: We can identify Ws with O in the same way as we identify Ww, with QO. Also we can identify Woy with O via e€43(€) m ¢ VcEeO. SECTION 3: 3-FORMS, THE DICKSON 3-FORM AND SOME OF ITS PROPERTIES All the notation and results of this section up to the definition of © before proposition 3.2 are due to Aschbacher and can be found in [As1] - [As5]. Throughout this section let V be a vectorspace over F. Definition: A triple 3(T,P,f) is a 3-form iff (F1) fis a trilinear form on V. (F2) P: V x V + F is linear in the first variable and satisfies: P(x,ay) = a” P(x,y) P(x,y+z) = P(x,y) + P(x,z) + f(x,y,z) for all x,y,z 6 Vand ae F. 19 (F3) T: V & F satisfies: T(ax) = a°T(x) T(x+y) = T(x) + T(y) + P(x,y) + Ply,x) for all x,y ¢ Vandae F. Remark: As char(F) # 2,3, each of the three forms {f,P,T} in the 3-form determine the remaining members of the 3-form. For example T(x) = (1/6) f(x,x,x) and P(x,y) = (1/2)f(x,y,y). Definition: Px is the quadratic form whose associated bilinear form is fy := f(x,-,-), where Px(y) = P(x,y). Definition: By xA we denote the radical of fy , and UA:= A xA. xe Definition: UO := { veV: Py(u) = 0 for all ueU }. Note that UA and U® are linear subspaces of V. Moreover if an isometry of the 3- form fixes the subspace U it must also fix UA and UO. Definitions: If UC UA wecall U singular. If UC UO wecall U_ brilliant. Note that singular implies brilliant as we assume char(F) # 2,3. Let S = {xy,...,xp} be a set of singular points. Definition: S is called a special r-tuple if x; + x; is brilliant nonsingular and x; ¢ (x; + x,)8 for all pairwise distinct i,j,k. A subspace is special if it is generated 20 by a special r-tuple. Now we come to the definition of the Dickson 3-form. Let V be a 27-dimensional vector space with basis X = { Xj 5X5 Xj :1<ij <6, i<j }, subject to the convention Xi = OX Let f be the symmetric trilinear form which has . t . . « monomials XPXGXGs 1< ifj<6 and X1d 2d *3a 4d *5d 6a? where d € Coset. Coset is some set of coset representatives for Altp in Alt, where Alt is the alternating group on {1,...,6} and P is the partition 12|34|56. The table below lists the monomials with sign +. Table 3.1 12 34 56 14 23 56 16 25 34 12 35 64 14 25 63 16 23 45 12 36 45 14 26 35 16 24 53 13 24 65 15 26 43 13 26 54 15 24 36 13 25 46 15 23 64 The Dickson 3-form is the 3-form (T,P,f) where f is the trilinear form above and P and T- are defined as follows: P(x,y) := 1/2 f(x,y,y) T(x) := 1/6 f(x,x,x) In [Asl] we find two proofs that the isometry group of the 3-form is the universal Chevalley group of type Eg. Moreover the set X is a set of weight vectors for some 21 Cartan-subgroup H of Eg. Definition: We define H as the subgroup of Eg generated by the isometries h(t), l(a), ae(F*)® such that the product of a’s coordinates is 1, t€F, where h(t) maps x¢ to t? 37 -1 -2 : > +s Xe Xj to t Xp X36 to t 6 and fixing % for i,j < 5, and l(a) ' ’ -l Ps maps x; to ax;, x; to ax, and Xj to (ajas) Xj where a; denotes the i-th Xg> x6 to t coordinate of a. Note that H is a Cartan subgroup of E¢(F) and that X is a basis of weight vectors for X. Definitions: V; := < x; :j <i> forl<i<6 but i¥5 Visi=< xy :ij > Us m< Var X5> Vs =< V4, X56> = << Us, xg: 151 <5 > Vigi=<Vg.x: 1Si<g6 > U3 '= < X16>%25%347 Ug = < 16% 25%34) X12 %*56-%2T X51 %6 > Vg = < *16>%25%34%12%56%2*5 1 %6> Visi= <Xjj? 1<i<j<6 > Let u; v i denote the Eg-conjugacy class of U; resp. V;- Observe that Vv; 1<i<6 and Us are singular, that Vio and Vio are brilliant, and that Vi59 = V49: Definition: A brilliant subgroup of Eg is a subgroup which is contained in the stabilizer of a member of ¥; where i € { 1, 2, 3, 5, 6, 9, 10, 12 }. 22 Definition: A subspace U_ will be called totally dark if T has no nontrivial zeros in U. Definition: A subspace U is a member of Ug if there exists a quadratic field extension K of F such that uK is a member of vk and U ¢ T9- Definition: A 3-decomposition of V is a decomposition of V into the direct sum of three subspaces A, ® Ay ® Ag such that each summand is a member of V9 and A; © Ay = Aj, for all {ij,k} = {1,2,3}. Suppose that F* contains an element of order 3. Aschbacher shows that there exists a unique class E of elementary abelian groups of order 81 in A = Eg(F): Diagonal which is not contained in a maximal torus of the algebraic group over E,(F). Under the action of a member of E the space V decomposes into a direct sum of 27 one dimensional Eigenspaces each of which is dark. We will call such a decomposition of V a 27-decomposition of V. Let d = X16 + X95 + Xg4° Aschbacher shows that V is the Pa orthogonal direct sum of d and dO and that the centralizer of d in Eg is Fy. Let ®: J + V_ be defined as follows: B(e)4(1) ) = X1g P(CgQ(1)) = X95 (egg (1)) = * 34 (e19(1)) = Xy9° X5g 2 B(€y3(L)) = xQt+ X5 ®(e93(1)) = X4- XG B(eqo(ig)) = Xyqt%sg (yg 7 4)) = *_-%5— B(€gg (iy) = 1+ XG ®(ey9(ig)) = Xygt%og P(€yg( 1 9)) = %X_-%B_— P(Egalig)) = *1 + Xs (ey 9(ig)) = Xq5 -Xo9g —P(€qg( 1g) = XQt XH (Egg lig) = *1 -%6 B(e,5(i4)) = xg -xq —B(€yg (1 4)) = Xyqt%gq «(Cag ig)) = X45 X23 23 ®(ey9(i5)) = xgtxy «(ep (7 5)) = Xy4- X 3g P(Cg(i5)) = x45+X03 B(e,o(ig)) = xgtxq «(73 1g)) = X19" X4g — PCE93(ig)) = XQ4+%35 ®(eyo(i7)) = %3-Xq Bley (1 7)) = XygtX4g — P(€ggliz7)) = Xo4q -%35 Proposition 3.2: ® is an isometry of the trilinear forms (, , ) and f(,, ). Proof: First we observe that there exists an isometry of the trilinear forms as both of them are Eg- forms and the E¢ forms of V are unique up to conjugacy in GL(V) . Recall first the definition of Ww; from proposition 2.10. We show first that Bly: Wo @(W) is an isometry, where W = < e,(1) , Wj, €13(1) >. Then we observe that (W,, N) (see observation 2.11) is isometric to ( ®(W;) , “Px;). Finally we will show that ® is uniquely determined by Plw> i.e., given Pw there is exactly one map ¥: win o(w)t such that S| + W is an isometry of Eg-forms. The proof of this last claim is constructive, and in fact we constructed © using this construction. Hence we omit further details. Since we are working over fields of characteristic # 2,3 it is enough to show that for every x € W_ we have Det(x) = T(®(x)) to prove that | is an isometry of trilinear forms. Let xeé J. Now we observe first that x can be written as x = e,4(@,) + €99(29) + e33(ag) + €39(C1) + €13(€9) + €93(cg) where a; € F and c;€ O. So then Det(x) = a,a9a@4 - La; N(cq 3) + #( cycgeg + TC o€3)- On the other hand T(@(x)) = ayagag +2, PCy ’ B(ers(cy ;)) + f($(e19(c1)),#(e13(€)),P(e93(¢3))) where ij € Par(d) and {i,r,s} = {1,2,3}. A straightforward calculation using the orthogonal basis of table 1.1 and using table 3.1 shows that N(w) = - Px, ((w)) for all w € W, subject to our identification of W, with O. And so (W,, N) is isometric to ( o(W;), -Px;,). 24 Ifx € W then in particular c, = 0, and hence ®(e;5(c,)) = 0. Thus *( cycoeg + €EoC3) = O = f( P(e; (cz)), B(€13(&)), Bleg3(cg))). So clearly Det(x) = T(@®(w)) and [yy is an isometry. Now it remains to show that © is uniquely determined (in the sense above) by its restriction to W. To do this we will first prove the following: €19(i)) is the unique element of V with the property that . ty 2. (ez 9(i;), e€13(1) ’ €93(i;)) =2 N(i}). Where A = { we Wy: (e93(i;), €13(1), w) = 0} and ty means the perp with respect to N. If we show that ey 9(i)) is unique with respect to (1) and (2) and if y is an isometry of E¢- forms, then ¥(ey9(4;)) must be the unique element with respect to 1. and 2. as w preserves properties 1. and 2. The following holds: (e93 (ij), €13(1), ey9(w)) = m(igw + Wis) (*) Now observe using table 1.1 that (+) = 0 iff weé ij7 and that (*) = 2 N(i;) when w= Me So (1) and (2) hold for ey 9(i)). Uniqueness follows from the linearity of the map x Now let W be an isometry of Eg forms such that W|y = ®|y. Assume for now that ¥ exists. Then ¥(e19(i;)) is the unique element of V with the property that 1. <¥(e,5(i:))> = A 34h 6(W.) . 12\j ~ 37° 2. f(W(ez9(i;)), ¥(ey3(1)), ¥legg(i))) = 2 N(ij). Where A = { we ®(W3) : f( ¥(eg3(i;)), ¥(e;3(1)), w) = 0} and ts, means the perp with respect to Px. 4 Similarly ¥(e;3(i;)) is the unique element of V with the property that 25 . +95 1. <¥(e13(i;))> =A N (Wo). 2. f(U(e1 3(4)), ¥(e,(1)), ¥(eg3(i;))) =2 N(i). Where A = { we ®(Wo) : f( ¥(eg3(ij))s ¥(e,9(1)), w) = 0 } and tos means the perp with respect to Pxos: So we can compute ¥(e13(i))) and ¥(e,9(i;)) for all j € { 0,... ,7}. Now a straightforward but lengthy computation will show that W = ®, and the claim follows once we show the existence of V. Now we turn to the existence of ¥. We note that W _ is generated by the special triple {e,,(1): 1<i<3}, a vector e}3(1) € We such that Peo (1) (213) =-landa basis B,= {e93(1),e93(i)e93((1/2)(igg tins 4 1) : 1¢ j < 3} of W,. Moreover it is shown in (Ja4] theorem 9 that Cry (teii(D}ega(2),e93(1))™ = Go(F). Also it can be shown using the identification of W, with O that {e93(iy),e93((1/2)(ig5 tig: 4 1) 21 <j<3} is a basis of standard vectors (in the sense [As7]) for the 7-dimensional Go(F) module <{e93(i)) : 1 <j<7}> with respect to the Aut(O) (~ Go(F)) invariant trilinear form of proposition 1.4. We indicate here a correspondence. First let us recall the setup of [As7]. Let Y = {Y9,Y)¥; : 1 < i< 3} be a basis for a 7-dimensional vectorspace over F. We define an alternating trilinear form h by YoY iY tVoVa¥otYoYgvgty1Yo¥gtYiYoy3, and a bilinear form B_ by -2y2 + ¥1¥ 1 +Yo¥oty¥3y3- The stabilizer of the pair of forms (h,B) in GL(<Y>) is Go(F). The set Y is called a standard set. Now we give a correspondence of Y to a basis of 1+< 0. Let Yo ip, ¥y > (1/2)(igtig), ¥4 > (1/2)(ig-ig), Yq > (1/2)(i4+ig), vb + (1/2)ig-ig)s vg > (1/2) igriz)s ¥g + (1/2)ig+iz). Now ®(W) is generated by the special triple {X16 >%95>%34}, by a basis #(B,) of XonA4 x44 , and a vector XotX5 E xg 4M Xg44 such that Pxo5(x9+X5) = -l1. Moreover we observe that ® maps {e;;(1): 1<i<3} into {x16,xX95,x34}, ®(ey3(1)) = xo+x5, and ®(e99(1)) = X1-Xg- Furthermore by {[As3] 5.5 we have 26 Go(F) ~ Cg (*16%25%34%2t%5-*1-%6)- Finally a long but straightforward calculation shows that ®(3,(S)\ €93(1)) is a basis of standard vectors for the CH g(*16%25°%34%2+%5*1-%6) - module <#(B,(S)\ egg(1))> . If we can show that Eg is transitive on triples {S,B,(S),u(S)} , where S is a special triple {81559,83}, B,(S) is a basis of 8A Nn 834, u(S) € 8,4 n 834, and B,(S) contains a vector b, such that Pg (b3) = -l, Cp, (Sb,u(8)) ~ Go(F), and B,(S) \ b, is a basis of standard vectors for the pg (Ssb1 u(8))- module <,(S)\b,>, then we will have the existence of a YW such that Vw = Oy. Now we show that Eg is transitive on the set of triples {S,B,(S),u(S)}. Aschbacher showed in [Asl] 6.9 that Eg is transitive on special planes. The content of (As1] 3.15.3 and 4 is that NE.(S)™ ~ Sping (F), (W;, Ps, ) is a hyperbolic orthogonal space, Ng,(S)™ acts as O(W;) on (W,, Ps,), and Nx,(S) induces a triality outer automorphism on Ng, (5). Now it is well known that Nx, (5) is transitive on the points of Wo of Ps," value -1. Also well known is the fact that M := Npg(S-u(S))™ ~ Sping(F) and that M_ acts irreducibly in a spin representation on Ww: It is known that M is transitive on the set of vectors w € W, such that Ps, (w) = -1 and that Ny(w) ~ Go(F). To complete the proof we observe that the standard bases of the 7-dimensional Go(F) module correspond to the apartments in the building of Go(F), and that Go(F) is transitive on the set of apartments in its building. We define the multiplication on V in such a way that © is an isomorphism of algebras. Define Q (x) := Q(@)(x)) for all x€ V and P(ad+w) = a” for all we dO and aéF. Lemma 3.3: 1. Q(x) = - Py(x) + (9/2) P(x). 27 2. U ¢ dO is a singular subspace with respect to Q iff U is a singular subspace with respect to Pa: 3. (id) = d and (id +) =do. Proof: We know that G_ stabilizes a two dimensional space of quadratic forms of J as J is the direct sum of two absolutely irreducible G modules, both of which are nontrivial under Q. (see prop. 2.2.2) The space must therefore be spanned by Py and P ;i.e., (*) Q(x) =a P q(x) + BP(x) for all xEV . We solve for a and # by substituting d and X16 into the equation (*). Note that X4g = (1/3) d + w where wed@. Then using proposition 2.3 we get: (3/2) = Q(d) = a Py(d) + BP(d) = 3a + 8 and (1/2) = Q(x16) = aP (x19) + BP( (1/3)d + w) = (6/ 9). Solving this yields a =-1 and 6 = (9/2). This proves part 1. Now part 2 is an easy consequence of part 1. For part 3 we note first that dim(id+) = dim(d®) = 26. Let (,), f(d,,) and (, ) denote the bilinear forms associated to Q, Pa and P respectively. Now let x € d@. From part 1 we get: (id,@1(x)) = - f(d,d,x) + (9/2) (dx) = 0+0,a8 (dx) = P(d +x) - P(d) - P(x). This gives part 3. In view of Proposition 3.2 and Lemma 3.3 we will identify V and J via ®, Det with T, Q with G, id with d,and id+ with do. Conventions: We will call x €J or a subspace U of J singular (brilliant or dark) if x resp. U is singular (brilliant, dark) with respect to the 3-form T. We call x resp. U Q-(non)singular resp. P a (non)singular if x resp. U is (non)singular with respect to 28 the quadratic form Q resp. Py. Lemma 3.4: 1. Let x EJ. Then x is singular iff x#x = 0. 2. UC J is singular iff x#y =0 forall x,y € U. Proof: Let z € J andx,y € U. Then (x,y,z) = 2 (x#y,z) by definition of #. If x#y =O then (x,y,z) = 0 for all z€J and hence by definition U is singular. If U_ is singular then (x,y,z) = 0 for all zeJ and x,yE€U. Therefore (x#y,z) = 0 for all zEJ and hence x#y = 0 as Q is a nondegenerate quadratic form and char(F) # 2. Corollary 3.5: Primitive idempotents are singular. Proof: By proposition 2.7 G_ is transitive on primitive idempotents. So every primitive idempotent is conjugate to x listed in Proposition 2.7. It is routine to check that x#x = 0. Now Lemma 3.4 applies. Lemma 3.6: Let S_ bea brilliant subspace of J such that SO is a hyperplane, and sv € J\ SO. 1. Then Pg Ig is similar to Pylg: 2. A subspace W of §S is singular iff W is singular with respect to Pg. Proof: This is the content of [As1] sec. 2.13. Definition and Lemma 3.7: If S is a nonsingular subspace satisfying the conditions of lemma 3.6, then S is contained in a unique member of Tj, denoted by ¥49(S). S is called a hyperbolic subspace. Proof: This is the content of [As1] sec. 7.2. 29 SECTION 4: GENERALITIES ON SUBALGEBRAS OF J Lemma 4.1: G has two orbits on singular points of J. These are the scalar multiples of the primitive idempotents and the vectors satisfying x= 0 and x # OQ. (Call the latter Class two nilpotents.) Proof: Note first that by Lemma 3.4, x is singular iff x#x = 0. If x ¢dO then by lemma 2.8 x is singular iff x is a multiple of a primitive idempotent, and by 2.7.1 G is transitive on primitive idempotents. If x € dO then by lemma 2.8 x is singular iff x2= 0. Moreover Aschbacher showed in [Asl] 8.6 that G is transitive on singular points of dQ. Lemma 4.2: The G orbits on brilliant nonsingular vectors of dO are parameterized by the Q-values. Representatives for the orbits are Xy+ X09 5Xz, + Xe where a # 0. Proof: If every brilliant x € dO is of the form s+t where s, t are class two nilpotents, then the claim follows from [As1] 9.2. Now we claim that every x is of the right form. Define A := Wyg (x) MN dO, where Wj 9(v) is the unique member of Yi0 containing v. Now let s € J \ x© be singular. Then Pg, induces a nondegenerate quadratic form on Vig(x)- Now by lemma 3.6 a€A is singular iff a is singular with respect to Pg. The claim follows as each nonsingular point of A is contained in a Ps - hyperbolic line contained in A. Lemma 4.3: x2 =0,(x, + X19)" = X5 , (xy +a xg) # (xy +0 xg) = oxy Proof: A straightforward computation. Lemma 4.4: The # square of a brilliant vector is singular. Proof: By [Sp1] pg. 77 (x#x)#(x#x) = Det(x) id. When x is brilliant then Det(x) 30 = 0, by definition and lemma 3.3.4. So then either x#x = 0, or x#x is singular by lemma 3.4. The claim follows. Definition: Let U bea subspace of J. Then $2(U) m= < {xy:x,y € U}> and s#2y) i= < {x#y: x,y €U} >. Lemma 4.5: Let U_ be a subspace of J, then <U,id> is a subalgebra iff $2(U) Cc <U,id> = iff s#2(y) Cc <U,id>. Proof: The first equivalence is obvious. The second one follows from lemma 2.6. Lemma 4.6: Let U be a subspace of J , then s#2(y) = vet. Proof: Let v € UO. By definition of #, 0 = (s,t,v ) = 2(s#t,v) for all s,te U. So vec s*(u)t. Now let v es? (yt, Then 0 = (s#t,v) = (1/2)(s,t,v) for alls,t € U,sov € UO. Thus S09 = s#2(y)4, so the lemma holds. Lemma 4.7: If U is a Q-singular subspace of dO, then xy = x#y for all x,yeE U. In particular $2(U) = S*(v) . Proof: Let x,y € U. Then by lemma 2.6 xy = x#y + (1/2)(x,id) y + (1/2) (y , id)x + (1/2)(x,y)id - (1/2)(x,id)(y,id) id. Now (x,y) = 0 by assumption, and (x,id) = (y,id) = 0 in view of lemma 3.3.3. Thus xy = x#y for all x,yE€U. Now the claim follows easily. Lemma 4.8: Let U be a subspace of J. If for every s€U we have s*= 0 (i.e., s*(U) = 0) then U is singular. Proof: From lemma 2.9 and lemma 3.4 we know that every sEU is singular and that U 31 Cc iat. Now let s,vES. To complete the proof we need to show that v € sA. So let zéJ_ be any vector. From the singularity of s,v,s+v it follows that 0 = (stv, s+v ,z ) = (s,8,z) + 2(s,v,z) +(v,v,z) = 0 + 2s,v,z) + 0. So it follows that (s,v,z) = 0 for allz EJ. Thus v € sA. Definition: Let A be a subalgebra of J. We will denote the Q-radical of A by R(A). Note that R(A) c idt as id € A. Lemma 4.9: Let A be a subalgebra of J, then 1. R(A) is brilliant. 2. Either R(A) or S2(R(A)) is generated by singular points. 3. Both R(A) and $2(R(A)) are invariant subspaces of NQ(A) . 4. If R(A)# 0 then N(A) is contained in a brilliant subgroup of E¢. 5. S2(R(A)) C R(A). Proof: Let r €R(A). Then by proposition 2.3.4, 0 = (ab,r) = (a,br) for all a,be A. Thus bre R(A) for all be A. Thus S2(R(A))C R(A), proving 5. Since R(A) is Q-singular $*( R(A)) = S2(R(A)) by lemma 4.7. So R(A) c R(A)~ c $2(R(A))+c R(A)O by lemma 4.6. This shows part 1. For part 2 note that every vector x€R(A) is brilliant and Q-singular. So by lemmas 4.4 and 4.7, S2(R(A)) is either generated by singular points or trivial. In the latter case R(A) is singular by lemma 4.8 , so 2 holds. Part 3 is self evident and part 4 follows from [As2] Theorem 2 which states that the normalizer in Eg of a brilliant subspace which is generated by singular points is brilliant. Because of lemma 4.9.4 we will now call subalgebras with nontrivial Q-radical brilliant. 32 Also the following is self evident: Remark: If H <G and H stabilizes a subalgebra A. Then either: 1. A is nondegenerate with respect to Q, or 2. H is a brilliant subgroup of G. A subalgebra A will be called nondegenerate if R(A) = 0. SECTION 5: NONDEGENERATE SUBALGEBRAS Springer’s Lemma: Let A be a nondegenerate subalgebra of J then: 1. A contains a primitive idempotent iff there exists 0 # x € A such that Det(x) = 0. In particular if dim(A) > 3 then A contains a primitive idempotent. 2. If dim (A) > 3 and A contains a primitive idempotent x then Eo a(x) #0. (See prop. 2.7.6 for the definition of Eo a(*)-) 3. If dim(A) > 3 and Ey a) # 0 then A is isomorphic to 7% 4, do Jp := { qy 19 43 : d-€D, where D is a nondegenerate subalgebra of O \ do 43 73 4. dim(Jp) = 3 + 3 dim(D) Proof: See [Sp1] pg. 76. Lemma 5.1: Let A be a two dimensional nondegenerate subalgebra of J, then A = <id,x> where x is a primitive idempotent. G is transitive on two dimensional nondegenerate subalgebras of J. Proof: Assume that A does not contain a primitive idempotent. Then in view of 1 of 33 Springer’s Lemma, A must be totally dark. By [As3] sec 7 there exists a cubic field extension K of F such that AK contains exactly three brilliant points, none of which are singular. Moreover it is clear that AK is a nondegenerate subalgebra of JK, Therefore, by 1 of Springer’s Lemma, AK must contain singular points (recall from corollary 3.5 that primitive idempotents are singular): a contradiction. Thus A contains a primitive idempotent and proposition 2.7.1 completes the proof. Lemma 5.2: If F is finite, G is transitive on nondegenerate three dimensional subalgebras of J containing no primitive idempotents. If A is such a subalgebra then Cg(A) = 3D4(F) and Ng(A)/ Cg(A) = Z/3Z. Moreover Cg(A) acts irreducibly on at, When F is algebraically closed, J does not contain three dimensional nondegenerate subalgebras containing no primitive idempotents. Proof: Let A satisfy the hypothesis ; then by 1 of Springer’s lemma, A is a totally dark plane containing id. In [As3] sec 7 it is shown that Eg is transitive on totally dark planes, and that the normalizer in Eg of a totally dark plane is transitive on the points contained in that plane. This shows that G is transitive on totally dark planes containing id. [As3] sec 7 shows that the centralizer of a totally dark plane is 3p 4(F) and that the normalizer in Eg modulo the centralizer is teagan by Z.. The stabilizer of a point in a totally dark plane must therefore be Zz. This is the first claim. Now the minimal dimensional faithful FC Q(A) module is 24 dimensional, proving the second claim. Now Det( ) is a cubic polynomial and hence has a nontrivial zero on any subspace of dimension > 2 if F is algebraically closed. So the third claim follows from part 1 of Springer’s lemma. Remark: A is a totally dark subalgebra iff Ais a cubic field extension of F. To see 34 this observe that the only nontrivial totally dark subalgebras are the ones in lemma 5.2. So any such algebra must be cyclic, i.e., generated by a single element. Now every cyclic subalgebra is associative as a consequence of proposition 2.3.5. Let A be as in lemma 5.2 and Aid #4 x € A. Then by Hamilton’s equation: x3 = (x,id) x2 + ( Q(x) - (x,id)”) x + Det(x) id. Now as Det(x) #0 we have: x( x2 - (x,id) x - (Q(x) - (x,id)2)id) / Det(x) = id. So x is invertible and hence A is a field. For the converse let B be a subalgebra which is a cubic extension field of F. We need to show that Det(x) # 0 forevery 04x EB. Suppose x € B and Det(x) = 0. Then using Hamilton’s equation we get x(x2-(x,id)x - (Q(x) - (x,id))id) = Det(x) id = 0. Sothen x2 - (x,id)x - (Q(x) - (x,id)?) id = 0 as B is a field. Hence x generates a proper subfield of B. But B contains only <id> as a proper subfield, as (B: F] = 3. So x € <id> and hence x =0. This shows the converse. Lemma 5.3: Let x be a primitive idempotent and y,z€E g(x). Then yz € <id,x> and y? = Q(y) (id - x). Proof: See [Sp1] pg. 69. Lemma 5.4: If F is finite (algebraically closed) G has two (one) orbits of nondegenerate three dimensional subalgebras containing primitive idempotents, respectively. Let x be a primitive idempotent and y; € Eo(x) be a vector such that QAy;) =i, i € {1,non}, where non is a nonsquare of F. Representatives for the orbits can be chosen as follows: 1, A, = <id,x,yy>, Co(Ay) = Sping (F), Ng(Ay)/Ceg(A,) = Sym(3) . 2. Anon =<id , X , Ynon> » Cg(Anon) = Sping(F), Ng(Anon)/Cg(Anon) = Zo: Moreover Ay is a special plane containing id, and x is the unique primitive idempotent 35 contained in Anon. Hence the normalizer of Anon is not maximal in G. Anon does not exist when F is algebraically closed. Proof: We note first that in view of lemmas 5.1 and 5.3 each A; is indeed a subalgebra. Moreover an easy calculation shows that each A; is nondegenerate. Because of Springer’s Lemma we know that a three dimensional nondegenerate subalgebra A containing a primitive idempotent x decomposes as <id> © <x> ® Eg al*): Moreover from proposition 2.7.5 we know that N q(x) acts as ag (F) on Eo(x). Thus, depending on F, og (F) has two (resp one ) orbits on the points of Eq(F). The y; listed in the statement of the lemma can be chosen as orbit representatives. This shows that there are at most two (one) conjugacy classes of nondegenerate three dimensional subalgebras containing primitive idempotents and that {A;} contains a set of orbit representatives. We now investigate the centralizers of the A.’s. Now clearly CQ(A;) = Cox)» where C(x) = Ca(x), and therefore the centralizer structure follows from well known results about orthogonal groups. This also shows that the A,’s are not conjugate in G as their centralizers in G are nonisomorphic. Next we want to survey the primitive idempotents of A.. Let us first note that by lemma 5,3 y? = Q(y;) (id - x). So now let v = yid + 6x + ay; € A; be an 2 idempotent. We assume a,f,y # 0 to avoid trivial solutions. Then solving v“ = v by comparing coefficients yields the equations 2ya =a , g? + 276 - a7 Q(y,) =f, ¥? + a7 Qy,) = 7. Solving them leads to y = 1/2, Q(y;) = (1/4) a”, p a7 Q(y,). We now observe that the only time we have a nontrivial solution is when QA(y;) is a nonzero square in F. Thus we conclude that when i=non, x is the unique primitive idempotent in A;. Thus the normalizers of Anon lies in the normalizer of x. The nontrivial idempotents in A, are {(i/2)id + (1/2)x +(1/2)y,}. An idempotent is primitive if its Q-value is 1/2. Now Q(id +x+y;) = Q(id+x) + Q(y4) = 4 resp 2. 36 So the primitive idempotents in A, are {x,(1/2)(id - x +y,)}. A routine calculation shows that the idempotents are pairwise orthogonal. Now proposition 2.10.4 yields that the normalizer of Aj, is as claimed. The fact that A, is a special plane follows from [As1] sec 2. and corollary 3.5. Lemma 5.5: Let A be a nondegenerate subalgebra of dimension > 4. Suppose that Ey a = 0,where x €A isa primitive idempotent. Then: 1. Eo a(*) is nondegenerate with respect to Q. 2. <id-x> < S*(ANid®) < <id,x>. 3. <id,x> is an NG(A) - invariant subalgebra. 4. Ng(A) = Nog (x)Fo,a)- Proof: Part 1 is clear as J = <x,id>1 E(x) 1 E,(x) by proposition 2.7.3. Part 2 is a consequence of lemma 5.3, and the fact that A is nondegenerate. Parts 3 and 4 follow from part 2.as ANid@ and id are N(,(L) invariant. \ Recall the setup of proposition 2.10 and that W,; = Ei (x44) nN Ei(Xi49)) i € {1,2,3}. Also recall the definition of €;(¢) and the identification of W, with O after proposition 2.10. Lemma 5.6: Let A be a subalgebra of J which is isomorphic to Jp for some D. Then A is generated as an algebra by a set {X1,X9)X3} of three pairwise orthogonal idempotents which generate a three dimensional subalgebra, together with a vector weW 3 such that Q(w) = 1, a nonsingular vector veWo such that Q(v) = 1, and a composition subalgebra D C W, whose identity element is 2vw. )s oo © oo oO 1 Proof: Let 7: Jp A be an algebra isomorphism. Let x; = T(} 0 0 37 0 0 0 000 on Xg= T(| 0 101) , xg=7(}0 001), v=7(/ 10 0 }), 0 0 0 001 00 oo1 0 0 w=r(j| 0 0 0 |) and ={r(j oo * |):* €D}. Now the claim follows as 10 0 o * 0 any isomorphism of Jordan algebras is an isometry and preserves multiplication (see [Sp1] pg. 76 and Satz 17). Lemma 5.7: Let v,w and D be as in the last lemma. Then v and w determine a unique multiplication that makes (Wy>Px,) into an octave with identity element 2vw.. D is a composition subalgebra of O. Proof: It is shown in [Sp1] pg. 75 (and briefly outlined after proposition 2.10) that a CQ(pX9.%3,VW ) - invariant multiplication on W3 can be defined to make Wz into an octave with identity element 2vw. Aschbacher has shown in [(As3] sec. 5 that Go(F) = Cq(x1:X9.X3,V,w). Uniqueness follows from proposition 1.4. Lemma 5.8: If A is a nondegenerate subalgebra of J of dimension greater than three, x€ A is a primitive idempotent and Ex a(x) #% 0 then A is conjugate to Sp for some nondegenerate subalgebra D of O. Proof: Let A be a subalgebra of J satisfying the hypothesis. Then A_ is isomorphic to Jp by Springer’s lemma. In view of the previous two lemmas the claim will follow from the following four claims: 1. G is transitive on triples of pairwise orthogonal idempotents, which generate a three dimensional nondegenerate subalgebra. 2. CQ ({x15X95%3}) is transitive on vectors v of Wy whose Q-value is 1. 38 3. CQ({x1.x9.x3} »v) is transitive on vectors w of Wg whose Q-value is 1. 4. CQ ({x1,*9.%3},V,w) is transitive on isomorphic composition subalgebras of O = W). Claim 1 is the content of proposition 2.10.5 together with lemma 5.4. Claim 2 follows from proposition 2.10, lemma 5.4 and Witts theorem. To prove claim 3 we observe that CQ ({x15%95x3},¥) = Spin7(F), CQ ({x1.%9.x3},¥) acts irreducibly on W3, and that CQ({x1,x9,x3},v) acts on We as the image of the stabilizer of a nonsingular point under a triality outer automorphism of CQ ({x1X9%3})= Sping (F) (these facts are consequences of [Asi] 3.15.4). The claim follows Spin7(F) is transitive on vetors of Ws, of Q-value 1. To prove claim 4 we observe that CQ({x1>X9.X%3},v,W) ~ Go(F) c <Aut(O) by [Asi] section 5 and lemma 5.7 respectively. So CQ({x1,X9.X3},V5w) = Aut(Q). Now the Aut(Q) theorem of section 1 proves the claim. Now we can list all possible nondegenerate subalgebra types. Proposition 5.9: Let A be a nondegenerate subalgebra of J. Then A is G-conjugate to one of the following: 1. < id, x> where x is a primitive idempotent. 2. A totally dark plane containing id. 3. <id ,x,y;> where y; € Eg(x) satisfies Q(y,) =i, i€ { l,nonsquare € F}. 4. <id,x,E> where E Cc E(x) is a Q-nondegenerate subspace of dimension 2 or more. 5. ‘pb for some nondegenerate composition subalgebra D of O. Proof: If A is two dimensional then 1 holds by lemma 5.1. If A contains no primitive idempotents then lemma 5.2 and Springer’s lemma apply and 2 holds. So assume dim(A) > 3 and that A contains a primitive idempotent x. Then by Springer’s lemma, part 2, Eg a(*) is not 0. Moreover A = <id,x> © Eo a(x) ® E, a(x): If E, al) 39 = 0 then lemmas 5.4 and 5.5 apply and hence 3 or 4 must hold. If E, q(x) # 0 then by Springer’s lemma part 3 and lemma 5.8 5 must hold. The following will be useful. Lemma 5.10: Let U be a subspace of J containing id. Assume that J = U © UO, then U is a nondegenerate subalgebra of J. Proof: By assumption id € U. Let veUO, then (id,u,v) = 0 for all uEU. First we compute using lemma 2.6 and proposition 2.3 that 1. idf#tu = u - (1/2)(id,id)u - (1/2)(u,id)id - (1/2)(id,u)id + (1/2)(id,id)(u,id)id =u-(3/2)u - (u,id)id + (3/2)(u,id)id = (1/2) (-u + (u,id)id ) for all ueJ. 2. id#id = id Thus O=(id,u,v) = 2(id #u,v) = (-u +(u,id)id ,v) = -(u,v) +(u,id)(v,id) (+). Also 0 = (id,id,v) = 2(id#id,v) = 2(id,v) yielding (v,id) = 0. Substituting this into (*) gives: (u,v) = 0 for all ueU. This shows that UO C ut. By assumption dim(U®) = dim(U+) and therefore we actually have U@ = U+. Thus s#2(y) = U by lemma 4.6, and hence U is a subalgebra by lemma 4.5. As unut=0 Uis nondegenerate. Corollary 5.11: Let U be a subspace of type Ug, totally dark plane, Ug; Ug Vg oF Vi5 containing id. Then U is a nondegenerate subalgebra of J. Proof: By lemma 5.10 this is a matter of checking that for every choice of U, J=U @ UO. For the respective cases this is shown in [As1} sec 3.15.2, [As3] sec. 7, [As3] sec. 4.5.5, [As4] sec. 3.3.2, [As2] the remark at the end of sec. 3, [As1] sec. 3.8.1. 40 Definition: We call a subalgebra A of type U, or Y, if either A, or R(A) is of type U: or Y;. Lemma 5.12: We assume in the following that each U; resp. V; contains id. The following subalgebra isomorphisms hold: 1, U3 ~ <id,x, yj> where x and y are defined as in 5.4. 2. Ug = Ip where D= < 1>. 3. Ug ~ Jp where D is a quadratic field extension of F. 4. Vg = Ip where D is a two dimensional composition subalgebra of O with Witt index 1. 5. Vi5 = Jp where D is a four dimensional composition subalgebra of O. Proof: For part 1 observe that Ug, contains the three primitive idempotent idempotents X19 9X95 %34- So part 1 follows from lemmas 5.11 and 5.4. Parts 2 and 5 follow from proposition 5.9, corollary 5.11 and a comparison of dimensions (Springer’s lemma part 4) after we observe that E1,u,(*16) and F1v15*16) are not trivial. For part 3 recall from the definition of ug that uk is of type Vo for some quadratic field extension K of F. So part 3 will follow once we establish part 4. Recall the isometry ® of trilinear forms from section 3. Observe that @}(Vg) is <e;(1) , e;(1) ’ e;(i) >. Now observe from section 1 table 1.1 that D:= <1, i,> is a composition subalgebra of O and that the Witt index of Nip is one. The claim follows. Lemma 5.13: The normalizers in G of the Sp are as follows: 1.1f 0 =F, then N(Jp) = PSO3(F)xGo(F) ; Cg(Jp) = Go(F). 2. If D = F,, then N(Jp) = SLa(F)oSL4(F)Hp-<r> , Cg(Jp) = SL3(F) <r>. 41 3. If D = K (a quadratic field extension of F), then N(Jp) = SU3(F)oSU3(F).c . Cg(Jp) = SU.(F). 4. 1fD = F4, then N(Jp) = Sp¢(F)oSLo(F) Hp, CG(Jp)=SLo(F). Proof: The centralizers are given in [Ja4] theorem 9 and are as claimed. Let D= F*, then by lemma 5.8 and lemma 5.12.5 Jp = Vis: Aschbacher has shown in [Asi] that Vis is the alternating square of V, that Npg(Y is) = SL6(F)HSLo(F), where H is the Cartan subgroup of E¢(F) defined in section 3, Ce ,(Y 15) = SL. and that J= Vi, ® Vg © Vg as an SL¢(F) -module, where Ve= KX] oeeXQ>- Moreover Aschbacher has shown that d induces a nondegenerate “symplectic form on V¢ and on V6 and hence Nqg(Vj5) is as claimed as Ng(H) = Hp - Now let D = F. In view of lemma 5.8 and lemma 5.12.2 Jp = UG: Aschbacher shows in [As3] section 5 that NgQ(Ue) = SL3(F) x Go(F), J = Us,.2, W, as an SL3(F)-module, where each W, is a three dimensional irreducible SL3(F) module. Moreover because J is a self dual module there is an i such that Qlw, is nondegenerate. Now recall from lemma 3.3.2 that Qlw, is nondegenerate iff Palw, is nondegenerate. This shows that Ng(Ug)/CQ(Ug) is contained in SO(W; Qlw,) ~ PSO3(F). The opposite inclusion follows from lemma 3.3.2 and theorem 2.5.1. Now let D = F2. In view of lemma 5.8 and lemma 5.12.4 we need to compute Nq(Vog)- Aschbacher shows in {[As2]_ section 3 _ that NEQ(V9) = SL4(F)oSL3(F)oSLa(F)H<r> where 7 acts as a graph automorphism on CaQ(Vg) = SLo(F). Moreover Vg is a tensor product of two natural SL3(F)- modules and id is the sum of three fundamental tensors. Also NgQ(V9)” acts as SL3(F) x SLa(F) on Vg. Now let (m,n) € SL3(F) x SLa(F) and d = 3 view, € VOW, where V,W are natural SL4(F) modules and {v;} {w;} are bases for V resp. W. Let (aj 5) and 42 (b; 3) denote the matrices of m resp. n with respect to the bases {v.} resp. {w,}. Then d=d (m,n) = © vVm@wn = PEE a j bik VjOw, iff (0; 5) = ((a, 3) 1)" where t denotes the transpose map. So we conclude that Ng.(Y9)” NG acts as SL3(F) on Vg. Also from the definition of r in {As2] section 3 it is evident that r centralizes id. So NG(V9) is as claimed. Now let D = K. Then by lemma 5.8 and lemma 5.12.3 we need to compute Nq(Ug). Aschbacher showed in [As3]_ section 3 _ that NEQ(Ug) = SL3(K)oSU3(K/F).¢ where o induces a graph field automorphism on SL2(K). Moreover Uy = V@V" as an SL3(K) K - module and that id = © v,@ ve where {v;1l<is 3} is a K-basis for V and NE. (U9) acts as SLa(K) on Ug. So now let m € SL3(K) and let (a 5) denote the matrix of m with respect to the basis {v; }. af, viOvg iff (a) = (a)? Then id =idm=2Ov,m @v? m7 =E a. fi i ij,k ij where t is the transpose map. So m centralizes id iff m € SU,(K/F) and the claim follows as o centralizes id. Remark: With a little more work using [Jal] we could establish the Witt property for nondegenerate subalgebras of J namely: Let i: A + B_ be an isomorphism of nondegenerate subalgebras of J. Then i extends to an automorphism of J. We conclude this section with the following: Lemma 5.14: Let A be a nondegenerate subalgebra of J. Assume that N G(A) is maximal in G with respect to stabilizing a nondegenerate subalgebra. Then A is of type Ug, totally dark plane, Vg, Vis» Ug; Ug or <id,x>, where x is a primitive idempotent. Moreover in each case A = C y(CG(A)). Proof: By proposition 5.9, A is in the list of 5.9. The normalizers of algebras of type 43 <id,x,y;>, i=non, and type <id,x> © Eo a(x) were shown to stabilize the algebra <id,x> (see lemmas 5.4 and 5.5) and hence are not maximal in G. In view of lemma 5.12 this leaves only the subalgebras listed in this lemma. To prove the second assertion we check for each case that the complement AO of Ais a direct sum of nontrivial irreducible CGA) modules. For the case Ug this follows from 2.10.3 or alternatively from [As1] 3.15. For the case totally dark plane this follows from [As3] 7.3. For the cases Vo, Ug, Ug, V5 it follows from [As2] 3.5, [As3] 3.3.3, [As3] 5.7, {As1] 3.2 respectively. SECTION 6: BRILLIANT SUBGROUPS Recall from section 4 that a brilliant subalgebra is a subalgebra whose Q-radical is nontrivial. Recall from section 3 that a subgroup of G is brilliant if it stabilizes a member of {¥; > i €{1,2,3,5,6,10,9,12}}. In lemma 4.9.4 we proved that the normalizer of a brilliant subalgebra is a brilliant subgroup of G. In this section we will show that every brilliant subgroup of G_ is contained in the stabilizer of a nondegenerate subalgebra or a maximal parabolic subgroup of G. The maximal parabolics stabilize certain singular subspaces of id®. The next two lemmas show that we may regard the maximal parabolics as subalgebra stabilizers. Lemma 6.1: Let U be a singular subspace of id@. Then <id,U> is a subalgebra of J; its radical is U. Proof: In view of Lemma 4.1 every element of U is class 2 nilpotent and hence also Q-singular. Next we observe that (u,v) = Q(u+v) - Q(u) - Q(v) = 0-0-0=0 for all uy € U. So U is the radical of <id,U> as Q(id)#0. Now we observe that by 44 lemma 3.4.2 0 = u#v for all u,v € U. So by lemma 4.5 <id,U> is a subalgebra. Lemma 6.2: Let L be a subgroup of G and suppose that L stabilizes Ue Y. i€{1,....6}. Then L _ stabilizes the singular subspace UNid®@, and hence the subalgebra <id,(UNid@)>. Moreover Unid® #0 unless Ue Y, and U contains a primitive idempotent. Proof: The first part is evident in view of lemma 6.1 since the Y;, i€{1,...,6} are singular. If UNid®@ = 0 then dim(U) =1 and U is spanned by a singular point not in id®, which by lemma 4.1 must lie in the F-span of a primitive idempotent. Definition: Let x be a class two nilpotent. B(x) = Py - radical of xA. Let U be a singular subspace of id -. We define w(U): Ey pu). u Lemma 6.3: u(x) = Q- radical of xd. Proof: G is transitive on class two nilpotents so it suffices to check the claim for (x}). That’s a simple calculation. Definitions: Let U be a singular subspace of dQ. U is amber if Uc w(U). A. singular line lis scarlet if 1 is not contained in y(l). A singular subspace is scarlet if all of its lines are scarlet. A singular subspace U is tangerine if for some scarlet line | we have U =1@ (n(1) NU). Proposition 6.4: Let A be a subalgebra of J such that A = <id> @ R(A), R(A) is a singular subspace, and r = dim(R(A)) then: 1. If R(A) is amber then r = 1, 2 or 3. G_ has three orbits on amber subspaces and each orbit is characterized by r with representatives Vr. The stabilizer of an amber 45 subspace is a maximal parabolic subgroup of G. 2. If R(A) is tangerine then it is contained in a unique maximal tangerine subspace. Each maximal tangerine subspace is conjugate to V6: The stabilizer of a maximal tangerine subspace is a maximal parabolic subgroup P of G. Moreover P stabilizes a symplectic from on Vg which is induced by id. 3. If R(A) = 1 is a scarlet line then | is contained in the NG(A) invariant maximal tangerine subspace | @ (1). 4. If r > 3 and R(A) is scarlet then p(R(A)) #0 is amber. 5. If R(A) is neither amber, scarlet or tangerine then S C R(A) C S+ u(S) where S is scarlet of dimension 3 or 4. In this case 0 # w(R(A)) C w(S) and p(R(A)) is amber. 6. r < 6. Proof: Part 1 is the content of [As1] 9.11 and 9.12. Part 6 follows from the fact that Vs and Vg are the representatives for the E¢(F) orbits of maximal singular subspaces of J, see [As1] 6.5. Part 2 is the content of [As1l] 9.6, 9.12 and 8.4. For part 3 we note that by [Asl] 9.2.2 every scarlet line is G-conjugate to <X1.Xg>- Moreover in the proof of [As1] 9.6 it is observed that B(<xX}5Xg>) = <Xq)Xq.X 4X5 >- So B(<X1.Xg>) ® <X1X%g> = V6 which is maximal tangerine by part 2, and part 3 follows as p(U) is NQ(U) invariant for any singular subspace U of J. Part 4 is the content of [As1] 9.8 combined with the content of [As1] 9.10.2 and 3. The first half of part 5 is the content of [As1] 9.10.1. Now the representatives for S can be chosen as <X9,X4.X5g> and <X34X4.%5goX15>- Then ,»(S)is V 9 respectively <Xq>. Now an easy calculation shows Vo < “( < X3,X4;X5g> + Vo) < u(S)= Vo and <xg> < U(<xX9)X4.X59:X15> + <X_Q> ) < w(S) = <xg>. So part 5 follows. 46 Corollary 6.5: If A satisfies the hypothesis of proposition 6.4 then NGA) is contained in a maximal parabolic subgroup of G. Moreover the following are true: 1. If R(A) = Vy = <x,>, then NQ(A) ~ [F]'>: Spinz(F).F*, Cg(A) = FL: Spin,(F). 2. If R(A) = Vo = <x1X>, then NQ(A) ~ [F]??: SLo(F)xSL3(F).F*, Cg(A) = [F]?°:SL4(F). 3. If R(A) = Vg = <x1,x5,x3>, then NQ(A) ~ [F]?0: SL4(F)xSLo(F).F*, Cq(A)= (F]22: SLo(F). Moreover < id, VgA> and < id, v,A+> are N((A) invariant subalgebras of J and S*2( VA) = V3 and s¥(v..Vq+) =V3A. 4. If R(A) = Vg = <x,,...,%g>, then NG(A) & [F]!°: Spg(F).F*, Cg(A) = FD. Moreover Vet =~ Vg @ Vy5 is an Ng(A) invariant subalgebra and s* (VV 15) = V6: Where [F]" is some group with n composition factors all of which are isomorphic to the additive group of F. Proof: First we note that R(A) and u(R(A)) are NG(A) invariant. Unless R(A) is a scarlet line either R(A) or p(R(A)) is amber or contained in a unique maximal tangerine subspace, by proposition 6.4. In case R(A) is a scarlet line 6.4.3 applies and Ng(A) stabilizes a member of Y¢- Now the first claim follows from 1 and 2 of proposition 6.4. Part 1. is proved for example in ([Co2] or in [Asi] 8.9.2 and 7.8.4. The statements about NG(V¢) and C,(V¢) in part 4. can be found for example in [As1] 8.4 and 8.5. The rest of part 4, is a straightforward calculation. For parts 2. and 3. observe that Nn(ve)(Vi) i€{2,3} is isomorphic to R:SL,(F).F*, where R denotes the unipotent radical of Nycvg) Vi: Also observe that Cncv,) VD) = R. Now Aschbacher shows in [As1] 8.8 that CQ(Vo)/R(CQ(Vo))” = SLa(F), where R(CQ(Vo)) is the unipotent radical of Cq(Va), and so part 2. follows as R < CqQ(Vo). Also from {Asl] 8.7.2 it 47 follows that Cg(V3)/R(CQ(V3))° ~ SLo(F) so the first part of 3. follows. The second part of 3. is straightforward calculation. Orbits on Points Lemma: Let v €J. Then Nag(v) is contained in a Parabolic subgroup or in N(A) where A is a nondegenerate subalgebra of dimension less than or equal to three. Proof: Let veJ. Then by lemma 2.4 the subalgebra A generated by v is at most three dimensional. If this subalgebra is nondegenerate then the claim holds. If the radical of this algebra is singular, then corollary 6.5 applies. The remaining case is the brilliant nonsingular radical. In this case R(A) contains a point conjugate to <xX1+X19>- So then by lemma 4.3 and the fact that S2(R(A)) C R(A) (see lemma 4.9.5) R(A) contains a conjugate of <X1+X9.X%5g>- So as dim(R(A)) < 2, R(A) must be conjugate to <X1+X19.X5g>- Now <xse> = $2(R(A)) is a class two nilpotent, i.e., singular, and <id,x5¢> is an NG(A) invariant subalgebra with radical <X5g>. Hence <X5g> is amber by proposition 6.4.1 and hence N(v) is contained is a maximal parabolic. Lemma 6.6: G has two orbits on subspaces of type V10- If VEN 19 then NG(U) is contained in the stabilizer of a class 2 nilpotent or a primitive idempotent. Proof: Let be Un id+ be a brilliant nonsingular vector. Then b is conjugate to xy + X,9 or x4+ axe, by lemma 4.2. Now since U = ¥i9(d), U_ is conjugate to either Vig(x,+ X49) =< XX: 1Si #29 6> or Vio (xy + X6) = <x, x6 X45 XG» Xp Xk? {1,6,i,j,k,1} = {1,2,3,4,5,6}>. So G has at most two orbits on V 10° Now a simple computation shows that s¥2(y) = <x5;> resp s¥2(y) = < X,g>- Moreover x; is a class 2 nilpotent and X1g¢ isa primitive idempotent. Using lemma 4.1 this shows that G has two orbits on V10 and 48 that NQ(U) is contained in the stabilizer of a class 2 nilpotent or a primitive idempotent. Lemma 6.7: Let U be a subspace of type V9: Then NQ(U) stabilizes either a V15 type subalgebra or a point of idO . Proof: Let U €Vj5, then by [As]] 3.8, J = U ® T and TEY,, and T is Ng(U) invariant. Since U is brilliant id ¢U. If id €T then T is a subalgebra by corollary 5.12. Ifid ¢T then the two dimensional subspace S generated by the projections of id onto U resp T is an NQ(U) invariant subspace. Moreover S N <id>+ is an NQ(U) invariant point. This is the claim. Lemma 6.8: Let UEV, then NQ(U) either stabilizes a subalgebra of type Tg or a point of idQ. Proof: In [As2] section 3 Aschbacher shows that NQ(U) stabilizes a unique 3- decomposition of J. Let’s say J = U @ A @ B is that decomposition. Let S be the subspace generated by the projections of id onto the subspaces U, A and B. Then § is an NQ(U) invariant subspace containing id. If id projects nontrivially onto U, then the claim holds as the projection of id onto U is an Ng(U) invariant point. If id projects trivially onto U, then dim(S) < 2. If dim(S) = 1 then id € A or B and A or B is a subalgebra by corollary 5.12. If dim(S) = 2 then id+ NS isan Ng(U) invariant point. At last we have: Proposition 6.9: Let L be a brilliant subgroup of G. Then L is either contained in a maximal parabolic subgroup of G or L is contained in the stabilizer of some 49 nondegenerate subalgebra. In particular, if L stabilizes a brilliant subalgebra, the claim holds. Proof: If L is brilliant then L stabilizes a member of Y; ,1 € { 1,2,3,5,6,9,10,12}. Ifi < 6 corollary 6.5 applies as these Y; are singular subspaces. In the other cases lemmas 6.6, 6.7 and 6.8 and the orbit on points lemma apply. Now if L stabilizes a brilliant subalgebra then by lemma 4.9.4 L_ is a brilliant subgroup and the claim follows as above. Finally we can combine proposition 6.9, 5.14, and 6.4 to obtain the SUBALGEBRA THEOREM FOR G: If L be a subgroup of G stabilizing a proper nontrivial subalgebra, then L stabilizes one of the following: a. A two dimensional subalgebra containing a primitive idempotent. b. A_ special plane containing id. c. A totally dark plane containing id. d. A member of Ug containing id. e. A member of Vg containing id. f. A member of Ug containing id. g. A member of 15 containing id. h. An amber or maximal tangerine subspace of idO. Definition: A subalgebra of J is good if it is G-conjugate to one of the members listed in the subalgebra theorem. Note that the subalgebra theorem states that any subgroup stabilizing a proper nontrivial also stabilizes a good subalgebra. 50 SECTION 7: LOCAL SUBGROUPS OF G Lemma 7.1: Let N be a normal subgroup of M < G. Then C ,(N) is an M invariant subalgebra of J. Proof: Let x,y € C ,(N), and géN . Then (xy)g = xg yg = xy , so xyEC ,(N) . From now on throughout this section we will denote by N an abelian group of semisimple elements of G. We will also assume that all the Eigenvalues of N lie in F. Recall the definition of the subgroup H of section 3. Recall also that H is a Cartan subgroup of E¢(F) . We denote the set of weight spaces of a Cartan subgroup K of E,(F) by X(K). Recall from section 3 that X(H) is a set of 27 singular points. Lemma 7.2: If N < He, gE E,(F). Then N_ centralizes each member of Xg onto which id projects nontrivially. In particular each such member is a primitive idempotent. Proof: We observe first that (N,J] < id+ = id®. But N_ must fix every point x € x(H®) \ id® . But then [N,x] € id9 MN <x>. So [N,x] = 0 and the claim follows as x is a multiple of some primitive idempotent by lemma 4.1. Lemma 7.3: Every Cartan subgroup of G is of the form Cy(id) where K is a Cartan subgroup of Eg(F) and id is in the linear span of three members of X(K). In particular if N < K then C iN) is a subalgebra containing a subalgebra of type Ug. Proof: Aschbacher proved in (As1] 8.15 that Cy(id) is a Cartan subgroup of G, {x16 X95 X34} Cc X(H). Also id € <X1g» XQ5>X34>> which is a three dimensional 51 nondegenerate subalgebra of type Us (see lemma 5.4 and 5.12). Lemma 7.4: If N_ centralizes a primitive idempotent then N2 is contained in a Cartan subgroup of G. Proof: As in the previous lemma N < Cq(x) so N acts on Eq(x), and it suffices to show that N2 is contained in a Cartan subgroup of Q(E9(x),Q), as every Cartan subgroup of G_ stabilizing x is a Cartan subgroup of Ng(x) and vice versa. So it suffices to show that there is a subset of weight vectors of N2 which is a Q-hyperbolic basis of Eo(x) ‘(by a Q- hyperbolic basis we mean a basis consisting of a maximal number of pairwise orthogonal hyperbolic pairs). First we observe that Q(vn) = Q(vA(n))= A(n)2Q(v) = \(n2)Q(v) for every weight vector v of weight \ and every n€N . If X is a nontrivial weight of N?, then Q(v) =0 and v is Q-singular. Now let v_ be any weight vector corresponding to a nontrivial weight A of N2 then, as E(x) is Q-nondegenerate, there exists another weight vector w € Ep(x)\v> corresponding to a weight y of N2 with the property O (v,w). Then pp = xl and {v,w} is a Q- hyperbolic pair of weight vectors of N2. Moreover we can choose all other weight vectors to lie in <v,w>t N E(x). We write E9(x) = H 1 Wo where H is the sum of the Q-hyperbolic pairs H; and Wg is the zero weight space of N2. Now Wo is a nondegenerate subspace of E(x) and hence we can extend a Q-hyperbolic basis of H to a Q-hyperbolic basis of Eo(x)- As N2 acts trivially on Wo such a Q- hyperbolic basis is an N 2 _ invariant basis of weight vectors. So we are done. Lemma 7.5: If ne¢ 1, then N2 is contained in a Cartan subgroup of G. Proof: In view of lemma 7.4 it is enough to show that N_ centralizes a primitive idempotent. By assumption nox 1 so there exists a weight vector v of N_ such that N® acts 52 nontrivially on <v>. So there exists an n€N and a veéJ such that vn = a@ Vv and a341. Sothen T(v) = T(vn) = a®T(v). So v must be brilliant. So by [As3] 6.1 there must exist an N-invariant pair {s,U} where U € Yj and s is a singular point not in UO. Now by lemma 6.6 G has two orbits on V10: Representatives for the orbits are given in the proof of lemma 6.6. In case U_ is G-conjugate to ¥19(%1 +6) we find that s#?(q) is an N-invariant primitive idempotent which must be centralized by N (see prop.2.7.5). In case U is G-conjugate to ¥19(*1+%19) we conjugate so that Wi o(xy+X19) is N® invariant for some geG. Now (*) Wyo(xy+X,9) = <Xg5> © <xXg> @(Ey(X95) Wi 9(X1+%19))- To see this observe that Xo, is a primitive idempotent and that <x3> = Ep(xo5) A Wi 9(x1+xX 9) - Also observe that S2((x,+x)9)N id@) = <x,>. Now let x be any weight vector of N contained in Wy 9(xj+x 9) \ idO. As [N,J] < idO we conclude that x € C,(N). So x? € C(N) as well. Now we can normalize x so that x = Xo, + 8x5 + u with respect to the decomposition (*), with @ € F. Now x? = (x95) + B?(x5)* + u2 + 28 Xo5X5 + 2Xo5u + 26X5u = X95 +0 4+ u2 + 0+ u+ 0 (xz is a class 2 nilpotent, x5 € Ep(xo5), u € xpd ) = Xo5 + Axe + u (Xp generates 87( 1 9(%4 +19) nN id® )) 2)2 2 is an 2) Similarly we can show (x Xo, tu + u*. So either x or x idempotent of J. An easy computation using lemma 3.3.1 shows that Q(x) = Q(x 2 = 1/2. So by definition either x or x“ is a primitive idempotent. Note that the proof of 7.5 actually shows that any abelian group L such that L3 x 1 centralizes a primitive idempotent. In particular any elementary abelian 2-subgroup of G is contained in the centralizer of a primitive idempotent. 53 Lemma 7.6: Assume that F contains a primitive third root of unity. If N is not contained in a Cartan subgroup of G and n3 = 1, then |N| € {9,27} and one of the following holds: 1. C y(N) is a nontrivial totally dark subalgebra , and F is finite of order congruent to 4,7 mod(9). 2. NG(N) = 39:L5(3) and Ng(N) stabilizes a 27-decomposition containing id. Proof: If N_ is not contained in a Cartan subgroup of G, then in view of lemmas 7.2 and 7.4 it is not contained in an E¢(F) conjugate of H. Define M = <N,Z(Eg¢(F)>, where Z(Eg¢(F)) is the center of E,(F), which by our choice of F has order 3. Then Mis certainly not contained in an Eg(F) conjugate of H, hence M satisfies the hypothesis of [As3] 8.3 and hence |M|€{27,81}; the first part of the lemma follows. Moreover, by [As3] 8.3 the Eigenspaces of every element of N make up a 3- decomposition of J. Also the weight spaces of N are dark, else by [As3] 6.1 and 6.3.2 N is contained in an E,(F) conjugate of H. So no weight space of N can have dimension more than 3, by [As3] 7.3. Also observe that the nontrivial weight spaces of N are Q singular. Lastly, if |M|=27 then F is finite of order congruent to 4 or 7 mod(9) [As3] 8.3.1. If |M| = 27 then N_ has, at most 9 weight spaces, so at least one of them, say W, has dimension > 3. Now if W = C y(N) then 1 holds. So assume otherwise and let w € W, then by our earlier observation w is dark. So the map ly: J — J defined by lw(v) = w#v is injective. ( To see this observe that v € Ker(lw) iff v € wA and that wA = 0 when w is dark.) Moreover ly permutes the weight spaces of N nontrivially. Also lwolwolw: W > W is an isomorphism and lwolw(W) Cc C gfN)- So dim(C ,(N)) 23 and Cy(N) is a totally dark subspace, so 1 holds. If |M| = 81 and every proper subgroup of M containing Z(E¢(F)) is contained in an Eg(F)- conjugate of H, then the proof of [As3] 8.3 shows that M is an exotic 81 54 subgroup of Eg(F) whose Eigenspaces form a 27- decomposition. So now assume that some proper subgroup N, of M containing Z(Eg(F)) is not contained in an E¢(F) conjugate of H. In particular we may assume in this case that F is finite of order 4 or 7 mod(9) and that N,M G_ centralizes a totally dark subalgebra U of J. Now we may assume that N contains NjNG, so let née N\ (N,AG). If n centralizes U then N centralizes U and case 1 holds. So assume that n acts nontrivially on U and let {id,y,y7} be the Eigenspaces of n on U with respect to the Eigenvalues Lyw jw where w is a primitive cube root of unity. First we observe that (*) (id,y) = (id,y) = Q(y) = Q(y?) = 0 because n preserves the quadratic form Q. Also since < id,y,y?> is totally dark we may assume that T(y) is not a cube in F (else <id,y> contains brilliant points) so we may normalize y such that T(y) = w and hence T(y?) = (1/6) f(y?, y*,y”) by definition of T = (1/3)(y?#y?,y”) by definition of # = (1/3)(y4,y?) from lemma 2.6 and the facts (*) = (1/3)(y8,id) by proposition 2.3.4 = (1/3) ((y?)2,id) by power ass. of J (see remark after 2.3.5) = (1/3)( (T(y)id)?,id) by proposition 2.3.6 and the facts (*) = (1/3)(id,id)(T(y)) = T(y)” by proposition 2.3.2 and definition of (_,_) = w’. Now an easy calculation shows that T(id + y ty?) = T(id)+T(y)+T(y2) = 0; a contradiction to the total darkness of <id,y,y*>. Therefore N must centralize a totally dark subalgebra. Now it remains to show that if M is an exotic 81-group, then NaQ(N) = N:La(3). 55 Aschbacher showed that Ng.(M) = M:3°:SL9(3), CEN) = M. and that Ng (™) / M = Cgrcmy(2(Eg@)) = 39:SL3(3). Observe also that N is a complement of Z(Eg) in M. So if g € Eg and ne = N, then as Z(Eg)® = Z(Eg), we have Ng) < NE(M): And thus NEGO) / Nis the stabilizer of the pair of subspaces (Z(Eg),N) in SL(M) ~ SL,4(3). So NECN) = M: SL,(3), and N_ is the natural SL3(3)- module. Now we need to observe that every weight space of N_ is one dimensional. To see this let 0 #4w &€ Wy; A€ Hom(N,F). Then w must be a dark vector by our initial observations. So the linear map lw defined as before by lw(x) = w#x is injective and moreover I2,(W,) < Wo = Cy(N). So if for any weight the weight space has dimension greater than one, then C ,(N) has dimension greater than one and hence M is not an exotic 81- group, a contradiction. We conclude that the weight spaces of N are indexed by (correspond bijectively to) Hom(N,F). Now SL3(3) acts transitively on the nonzero elements of Hom(N,F) and fixes zero. But the zero weight space of N is the subspace of J spanned by id, so SL,(3) < G. As Z(Eg) is not contained in G, we conclude that NQ(N)= N: SL3(3). Lemma 7.7: G is transitive on 27-decompositions containing <id>. Proof: By [As3] 8.3 Eg has one or three orbits on 27-decompositions, parameterized by the equivalence class mod(F?) of the T -values of the points making up the decomposition. Thus Eg is transitive on the 27-decompositions containing id. Since the normalizer in Eg of a 27-decomposition is transitive on the weight spaces of the decomposition (see [As3] 8.3), the claim follows. Definition: dp(q) is defined to be the smallest d such that q4-1 = 0 mod(p). 56 Note that dp(q) is the dimension of the smallest GF gq - vectorspace V such that p divides |GL(V)| . Definition: Let K be an extension field of F. By yk we denote the exceptional central simple Jordan algebra over K and by GK its automorphism group. When K ~ F, the algebraic closure of F, we write G for GF and J for JF. From now on we will regard G_ as a subgroup of fixed points of some Frobenius automorphism of ck resp G. Definition: Assume that F* contains no element of order 3. A twisted 27-decomposition is a decomposition % of id+ into 13 pairwise orthogonal two dimensional subspaces such that there exists a quadratic field extension K of F so that gK 4 27- decomposition of Jk containing id. Moreover we require that the normalizer in G of this configuration is isomorphic to 33. SL3(3). Definition: Aj(U) is defined to be Ng(U)/CqQ(U). Definition: Let r be a prime. The r-rank of the group L_ is the maximum of the ranks of the elementary abelian r-subgroups of L and is denoted by m,(L). Lemma 7.8: Let r be a prime not equal to 2,3,p and R an r-subgroup of G. Then R is contained in a Cartan subgroup T of G and is hence abelian. Moreover Ag(R) is isomorphic to a section of the Weyl Group of G. Also m;(G) < 4. Proof: The first part of this statement is in [Sp3] which shows that R < T < G, 57 where T is a Cartan subgroup of G. Hence R_ is abelian and the r-rank of G is as claimed. We consider G = Fixa(e) where o is some field automorphism with fixed field F. Now in [Sp3] it is shown that Ca (R) is transitive on the Cartan subgroups of G contained in Ca (R). So by a Frattini argument Na (R) = Ca(R) S where § = Ny q(R) But T is self centralizing and Na(T)/T is the Weyl group. So as AG ~ S/ SNC&(R) and SNC (R) > T we have AG(R) is a section of the Weyl group. As G<G we have AQ(R) < AG(R) and the claim follows. Lemma 7.9: Assume that F* contains no element of order 3. Twisted 27- decompositions exist in J, and G is transitive on twisted 27-decompositions contained in J. Proof: By assumption, F* does not contain elements of order 3, F is finite and there exists a quadratic field extension K of F such that K* contains an element of order 3. Now let N < GK be an elementary abelian 3 group of order 27 whose weight vectors form a 27- decomposition of JK. since ick, and |G] have the same 3-share and G < GK we can chose N < G. Let o denote the field automorphism of Gk whose set of fixed points is G. Let M denote the normalizer of N in GK extended by ao. Now’ as go centralizes N we _ conclude using Lemma 7.6. that M ~ (N@<o>):SL3(3). Now M centralizes Oo(M) = <o>. Thus N: SL3(3) < G is the normalizer of N in G. Next we must show that N : SL,(3) stabilizes a twisted 27-decomposition. We already know that N : SL,(3) stabilizes a 27-decomposition of Jk containing id. Now if yeJ K ig a weight vector of N, then so is y’, moreover Q(y?) = Q(y) = 0 when y# id and <y, y’> is a Qk. hyperbolic pair. To see this observe that (y,y2) = (y?,id) by 2.3.4 and that y> = Q(y)y + T(y)id = T(y)id by Hamilton’s equation and the fact that Q(y) = 0. Combining this gives: (y,y”) = T(y)(id,id) = 3T(y) # 0 by 58 2.3.2 and the fact that y is dark. So JK is the sum of 13 hyperbolic pairs and <id>. Moreover this decomposition is also SL4(3) invariant. Now we claim that the N:SL.(3)-invariant decomposition of gk induces a decomposition on J. This is the twisted 27-decomposition. To see this we need to observe that o acts on < ysy"> and that C g_ (7) is a line of J. In <y.y"> section 9 there is a description of how o@ acts on Jk . Now for all né€N we have yon =yno = yAy(n)o =yody(n)? where Ay is the weight N corresponding to the vector y, showing o- invariance of <y,y?>. Now let’s consider JX as an N ®@ <o> 25 is an K(N ® <o>) submodule. Now apply module. Then we just saw that <y,y 25.7.2 of [As9], which states that in this situation there exists an FN- submodule U of J such that UK = <y,y?>. So U=C 9_(¢) and dimp(U) = dim, (US) = <y.y"> 2. So U isalinein J. Now we need to show that G is transitive on twisted 27-decompositions. Let Dy and Dy be _ twisted 27-decompositions whose stabilizers are N; : SL9(3) i=1,2 respectively. Their stabilizers in GK are conjugate, say Ny} = No. Since o centralizes G we conclude that o centralizes N, and No. So choh"! € CQK Ny) = -1 N,- So [oh]? =l= [o ,hjhoh forcing [c,h] = 1, ie., h € G. This shows conjugacy. Lemma 7.10: Let R <G bea 3-group. Then 1. Z(R) is contained in a Cartan subgroup T of G unless R = Z(R) is elementary abelian of order 27 and R stabilizes a 27 or twisted 27- decomposition of J. 2. If Z(R) is contained in a Cartan subgroup of G, then Ag(Z(R)) is a section of the Weyl group of G. 3. There exists a normal abelian subgroup A < R such that R/A is a section of the Weyl group of G. Hence R/A is a subgroup of an elementary abelian group of order 9. 59 4. dim(C ,(Z(R))) > 3 unless Z(R) = R, |R| = 27 and R_ stabilizes a 27 or twisted 27-decomposition. 5. Ng(R) stabilizes one of the following: a. A proper nontrivial subalgebra. b. A 27 - decomposition. c. A twisted 27-decomposition. Proof: The proof of 2. is the same as in lemma 7.8. For part 3. we use [Sp3] to embed R into Nq(T) where T is a c-invariant Cartan subgroup of G. Then let A = TOR. We observe that A is the kernel of he where the canonical projection 7: NQ(T) ++ W(G). The claim follows since the Sylow 3-group of W(G) is elementary abelian of order 9. Now we will prove part 1. If Z(R) is not elementary abelian Z(R) is contained in a Cartan subgroup of G by lemma 7.5. If Z(R) is elementary abelian and not contained in a Cartan subgroup of G, then by lemma 7.6 C,(Z(R)) is a nontrivial totally dark subalgebra of J or Z(R) stabilizes a 27-decomposition of J. If C y(Z(R)) is nontrivial then C y(Z(R)) OF is a proper nontrivial Z(R) invariant subalgebra of J containing brilliant points. So by Springer’s lemma part 1 C (Z(R)) OF contains a primitive idempotent and hence by lemma 7.4 Z(R) is contained in a Cartan subgroup of G. So now assume that Z(R) has order 27, is not contained in a Cartan subgroup of G, does not stabilize a 27-decomposition of J but stabilizes a 27- decomposition of J, and in particular F* does not contain an element of order 3. Let o be the Frobenius automorphism of F such that G = Fix(c). Then there exists a quadratic field extension K of F contained in F such that Fix(o”) = GX, By lemma 7.6 Z(R) must stabilize a 27-decomposition of JK. Now the first part of the proof of lemma 7.9 shows that Z(R) stabilizes a twisted 27-decomposition of J. 60 It remains to assert that when Z(R) is not contained in a Cartan subgroup of G then R=Z(R). Now R is contained in NQ(Z(R)) = Z(R):SL3(3), by the above and lemmas 7.6 and 7.9. Soif R# Z(R) then every g € R\Z(R) induces a nontrivial automorphism of Z(R) : a contradiction. Now part 4 is a consequence of part 1 and lemma 7.2 once we observe that dimy(C5(2(R)) = dimp(C )(Z(R)). Part 5 follows from parts 1 and 4 and lemmas 7.6 and 7.9. Lemma 7.11: Let R be an elementary abelian 2 group. Then: 1. R centralizes a primitive idempotent. 2. dim(C ,(R)) > 3. 3. If |R| = 2 then C,(R) € 15 or Cy(R) = <id,x,E9(x)> for some primitive idempotent x, and R = ZCo(C j(R))). G_ has exactly two conjugacy classes of involutions. 4. AQ(R) isa 2,3,7 group. 5. mo(G) = 5. Proof: Part 1 follows from the proof of lemma 7.5 (see remark after that lemma). So part 5 follows from proposition 2.7.5 and the fact that mo(Sping(F)) = 5. For part 3 we observe first that E,(F) has two conjugacy classes of involutions Ij and Ij, see [As3] 6.4.1. Moreover g€ Ijo iff C y(g) € ¥15 and gé€ Iy¢ iff Cy(g) = <p> + U where UE), and p is a singular point in J\UO. Now an involution g € E¢(F) lies in G iff id € Cy(g). Soif ge IjgN G then g centralizes a subalgebra of type V15- Now Ca(C4(g)) = SLo(F), by lemmas 5.12.5 and 5.13.4, and [C$Q(C 4(g), J] = [g,J]. Moreover because g is an involution, g acts as - Id on ([g,Jj. So g € Z(CG(C 4(g)))- Now Z(SLo(F)) has order 2 and it follows that G_ is transitive on 61 Ij9NG as G is transitive on subalgebras of type 15 . Now let g € Tg G. Then by conjugating g appropriately we may assume using lemma 6.6 that C (g) = <xyg> + Wyg(xyt x@ or <xg+w> + Wy 9(x]+x19) where wEU®. Now since id = x;g+Xo5+X34 € Cy(g) we see that the second possibility is out. Now the remaining parts of 3 follow from proposition 2.7. For parts 2 and 4 we will assume first that R is not contained in a Cartan subgroup of E¢(F). Then part 2 is the content of [As3] 6.3.4. For part 4 we use [As3] 6.7 to see that R = Ro ®J where RO = 1ygN R, that |[R|<4 |J|<8 and J centralizes a subalgebra of type Ug: So part 4 follows in this case and moreover Aqg(R) is a 2,3 group unless |J|=8. Now assume that R is contained in a Cartan subgroup of Eg(F). WLOG we may take R<H and even in Qo, the subgroup of elements in H_ of order 1 or 2, which we may identify with the six dimensional orthogonal space over GFy of Witt-index 2. Then by [As3] 6.4 Ag.(r)(R) is a section of 0 (2) and is hence a 2,3,5 group, so the same must be true for Ag(R). So now let g € Ag (r)(R) be of order 5, then |R|> 16, and we may also assume WLOG that [g,R] = R and Co, (8) = R+, where here the L is taken in Qe with respect to the unique quadratic form stabilized by the Wey] group of E,(F). Assuming [g,R] = R forces |R| = 16 and R+ is a two dimensional nonsingular nondegenerate subspace of No. Thus there are two possible isometry types for R and hence Age 2y(R) = OF (2). Only the order of O7(2) is divisible by 5 and hence R is uniquely determined up to conjugacy in E,(F). Now the Sylow 5 group of OF (2) has order 5. So up to conjugation in Eg(F) we may assume that g induces the permutation (X39 5% 30% 4 5%5 9% G )(XQ%B IN XH IXGOX jr Xin jp)? where p = (23456) € Symg, on the weight vectors of H, see section 3, and that R = So we can see that C y(R) = <X15X4>5 and is hence singular. Thus no E,(F)- 62 conjugate of R_ will centralize id (recall id is nonsingular) and part 4 follows. For part 2 observe that id is the sum of at least three weight vectors of every Cartan subgroup of Eg(F) containing R. If R < G then each summand of id must be centralized by R and part 2 follows. Recall that N_ is a semisimple subgroup of G and all the Eigenvalues of N lie in F. So either N contains a characteristic elementary abelian r-subgroup or N_ is infinite and Nisa characteristic subgroup of N which is contained in a Cartan subgroup of G. Thus lemmas 7.3 , 7.8 , 7.10 and 7.11 amount to the following: Lemma 7.12: One of the following holds: 1. C y(No) is a proper nontrivial subalgebra, and 0 4 No is characteristic in N. 2. <N, Z(Eg(F))> is an exotic 81-group and N(,(N) stabilizes a 27-decomposition. Lemma 7.13: If dr(q) # 1, r# 3 and R an r-subgroup of G , then dim(C ,(R)) > 2. Proof: Extend the field F to the smallest field K containing F such that r | |K]-1. Then Lemma 7.12 holds for R < GK. Now dimy(C 9K (R)) = dimp(C4(R)). So if dim (C 1K (R)) > 2 our claim holds. If dim (C «(R)) =1 then, by Lemma 7.12, NGk(R) must stabilize a 27-decomposition. So, by lemma 7.9, NQ(R) = N Gk(R) must stabilize a twisted 27-decomposition. We conclude this section with the following: Theorem 7.14: Let M be a closed subgroup of G that contains a normal solvable subgroup. Then one of the following holds: i. M_ stabilizes a proper nontrivial subalgebra. 63 ii. M_ stabilizes a 27-decomposition containing id. tii. M stabilizes a twisted 27-decomposition. Proof: By hypothesis M _ contains a normal abelian subgroup N. If N_ contains unipotent elements, then M has a unipotent radical. In this case the Borel-Tits theorem applies and M_ is contained in a maximal parabolic subgroup of G. By proposition 6.4 the maximal parabolic of G stabilize proper nontrivial subalgebras. So WLOG we can assume that N_ is a semisimple normal abelian subgroup. So lemmas 7.12 and 7.13 apply and the claim follows. Note that if case i holds, then M _ also stabilizes a good subalgebra by the subalgebra theorem. SECTION 8: NON LOCAL SUBGROUPS WITH NON SIMPLE SOCLE The following result of [As4] will be our major tool and starting point for this section. Theorem 8.1: If M is a nonlocal subgroup of G (hence also of Eg(F) ) with non simple socle, and M is a closed subgroup of G when F is algebraically closed, then one of the following holds: i. NG(M) is brilliant. ii . NQ(M) stabilizes a member of Ug or Ug. iii. Ng(M) stabilizes a 3-decomposition. We already saw in the section on subalgebras (proposition 6.8) that brilliant subgroups are subalgebra stabilizers. Lemma 8.2: Let M be as in the theorem and assume that N((M) stabilizes a 64 member U of Ug or Ug. Then NG(M) stabilizes a subalgebra. Proof: By hypothesis J = U © UO. If id projects nontrivially onto U with respect to this decomposition, then either U is an Ng(M) invariant subalgebra or Ng(M) stabilizes a point. So suppose that id €U@. We claim now that Cy (U,id) # 1. Suppose that this is true. Then the subalgebra generated by <U,id> # J and is N(M) invariant; hence the claim. Suppose that Cy,(U,id) = 1, then a faithful homomorphic image of M is contained in Ap, (U)- But Ag (U) is an SLs and hence M can not have more than one factor in F*(M). This contradicts our hypothesis on M. So the claim must hold. Lemma 8.3: Let S be a two dimensional subspace of id+. Then N,(S) stabilizes a proper nontrivial subalgebra or NG(S) is a local subgroup of G. Proof: We consider first the case CQ(S) # 1. In this case C y(CQ(S)) is a proper nontrivial N(°,(S)- invariant subalgebra. If CQ(S) = 1 then A<(S) = N,(S) < SL(S) ~ SLo(F). So as char(F)#2, NQ(S) does not contain simple subgroups. So NQ(S) contains a normal solvable subgroup and the claim follows. Corollary 8.4: Let M be as in theorem 8.1 and assume that Ng(M) stabilizes a 3- decomposition. Then Nq(M) stabilizes a proper nontrivial subalgebra or Ng(M) is a local subgroup. Proof: The space P generated by the projections of id onto the factors of the 3- decomposition is at most three dimensional and contains id. So PN id+ is at most two dimensional; hence the claim when PM idt¢ 0. If Pnidt+ =0 then id is contained in a member U _ of the 3-decomposition. Now U_ is a (proper nontrivial) 65 subalgebra by corollary 5.11. The collection of these last lemmas amounts to the following: Proposition 8.5: Let M be as in theorem 8.1, then Na(M) stabilizes a proper nontrivial subalgebra of J, or Nqg(M) is a local subgroup. Our efforts finally pay off: Structure Theorem: Let M be a closed subgroup of G. If F*(M) is not a simple group, then one of the following holds: i. M stabilizes a proper nontrivial subalgebra. ii. M stabilizes a 27- decomposition (containing id). iii. M stabilizes a twisted 27-decomposition. Proof: Recall first that char(F) = p (possibly 0). Recall also that F*(M) = E(M)F(M) and that F(M) is the Fitting subgroup of M. If F(M) # 1 then Theorem 7.14 applies and the claim follows. So now assume that F(M) = 1. Then E(M) is a direct product of more than one simple group. In this case M_ satisfies the hypothesis of proposition 8.5 and Ng(F*(M)) must therefore stabilize a proper nontrivial subalgebra or Ng(F*(M)) is local and hence satisfies the hypothesis of theorem 7.14. 66 SECTION 9: OUTER AUTOMORPHISMS , FIELD EXTENSIONS AND A STRUCTURE THEOREM FOR [. In this section we will prove the structure theorem for [. Proposition 9.1: Out(G) ~ Gal(F:Fo) where Fo ~ Z/pZ or Q and G_ has no diagonal or graph automorphisms. Proof: This is proved in [Cal] when F is finite, where it is also claimed that the result holds when F is a perfect field. Definition: Recall the basis X of weight vectors for H from section 3. Let o € Gal(F/Fp). Let yo be the semilinear map defined by: (> Arx)y¥¢ = DO APx. xEX xEX Notation: If F, is a subfield of F, then we denote by Jp, the set XF, = {= X05 a, € F,} and by Gpl= { My(g) : g€G and mj € F, for all ij }. Lemma 9.2: Let Fg denote the fixed field of c. 1. Yo preserves the multiplication of J. 2. Cy,(J) = Jr," Moreover Je, is the exceptional Jordan algebra over Fo. ed Aut(G) = <G,7¢: c€Gal(F:F9) >. > Cg(10) = Gp = Aut(Jp). Proof: Let x,y € X. Then we observe by using proposition 3.2 that xy is a linear combination of elements of X with coefficients in the prime field Fo: So clearly xy7g = XYgY7o, showing 1. Part 2 is now immediate from part 1. For part 3 we observe that — := (yey) Ero € GLlo7(F) and that g preserves the multiplication of J. So 67 @ € Gand hence <G,yg:cE€Gal(F:Fy)> C Aut(G). The reverse inclusion follows from proposition 9.1 , proving 3. Part 4 is clear given parts 1,2,3. Lemma 9.3: If N is normal in a subgroup L of I, then C,(N) is an L invariant subalgebra. Proof: As in the proof of lemma 7.1 we see that C,(N) is a subalgebra. Now let | € L,né€N and x€C,(N). Then xln = xml = xl, showing that xl € C j(N). Proposition 9.4: Aut(G) does not fuse G-orbits of nondegenerate, amber or maximal tangerine subalgebras. Proof: The idea of the proof is to observe that the invariants defining the various subalgebras are in the prime field and hence invariant under any field automorphism. Let y €Aut(G) \ G and A be a nondegenerate subalgebra. Then A is G-conjugate to one of the members of the list of proposition 5.9 and y = gyg where g€G and yg is as in 9.2. Now it is well known and easy to check that Q(x7)= Q(x)” and f(xy,.y7,z7) = f(x,y,2)%, for all x,y,zEJ. Soifz is a primitive idempotent of J then : (zy)? = 27 y (by 9.2.1) = zy (because z is an idempotent) Moreover Q(z7) = Q(z)” =1/2¢ = 1/2. So zy satisfies the definition of primitive idempotent. Now we observe that E;(z) y = E,(zy) as O and 1/2 € Fo. If W < E,(z) is nondegenerate and has Witt index b, then so does Wy, as Q(x)? = 0 iff Q(x)=0. We recall from section 5 that the G orbit of a nondegenerate subalgebra A containing a primitive idempotent is determined by the isometry type of Eo (2) and Ey (2). Since isometry type is not altered by y the claim holds for all nondegenerate subalgebras containing a primitive idempotent. There is only one orbit of 68 nondegenerate subalgebras containing no primitive idempotents. These are described in lemma 5.2 and are the totally dark three dimensional subalgebras. Now as f(x,x,x)7 = O iff f(x,x,x) = 0 we see that + maps totally dark subalgebras into totally dark subalgebras , completing the argument for the nondegenerate orbits. Recall from section 6 that the amber and tangerine subalgebras are defined in terms of the invariant u(R(A)). Recall if x is a singular point of id®, then u(x) is the Q- radical of xA. Clearly we have H(xy) = p(x)y and the claim follows for amber, tangerine and scarlet subalgebras. Corollary 9.5: If A is a good subalgebra then Nr(A)/NQ(A) ~I/G. Proof: By proposition 9.4 the I and the G orbit of A coincide. So the claim follows from a Frattini argument. Lemma 9.6: If D is a 27- or twisted 27-decomposition, then so is Dy , where vyer\G, Proof: First we assume that 9 is a 27-decomposition. Then By is a decomposition of J into 27 distinct dark weight spaces of N’, where N is the 27 group giving rise to 9. So C,(N’) = <id> and hence by lemma 7.10.4 the weight spaces of N’ form a 27- decomposition of J. Now let D- be a twisted 27-decomposition. Then we extend the field so that 9* jg a 27-decomposition of JK, Now by the above gk, is also a 27-decomposition of gk and the claim follows. Definition: A subgroup L of I is brilliantif L stabilizes a member of Y; where i € {1,2,3,5,6,9,10,12}. Lemma 9.7: Let L_ be a brilliant subgroup of [. Then L_ stabilizes a good 69 subalgebra. Proof; When L. stabilizes a singular subspace S, then L_ stabilizes u4(S) and the claim follows from Proposition 6.4. When L_ stabilizes a point or L_ stabilizes a member of V10: Vi9 or Yo; then the proofs of section 6 carry over verbatim. SUBALGEBRA THEOREM FORT: If L is a subgroup of [ and _ L stabilizes a proper nontrivial subalgebra, then L stabilizes a good subalgebra. Proof: If L_ stabilizes a nondegenerate subalgebra, the proof is identical to that of 5.14. If L_ stabilizes a brilliant subalgebra then by 4.9.4 and [As2] Thml L is a brilliant subgroup of E¢(F) extended by diagonal and field automorphisms. So then L is a brilliant subgroup of [ and the claim follows from lemma 9.7. STRUCTURE THEOREM FORT: Let L bea subgroup of [ and assume that F*(LNG) is not a simple group. Assume also that LNG is a closed subgroup of G. Then one of the following holds: 1. L stabilizes a proper nontrivial subalgebra. 2. L stabilizes a 27-decomposition. 3. L_ stabilizes a twisted 27-decomposition. Proof: F*(L MG) is normal in L. If the unipotent radical of F*(LNG) # 0, then by Borel Tits L_ is contained in a maximal parabolic subgroup of [ and hence L stabilizes an amber or maximal tangerine subalgebra. If the unipotent radical of F*(LQNG) = 0 and F*(LNG) contains a characteristic abelian semisimple subgroup N, then either L stabilizes C g(N?) which is a proper nontrivial L - invariant subalgebra of Jor |N| = 3° and N_ is an exotic 27-subgroup. In this case L stabilizes a 27- or twisted 27-decomposition of J by lemma 9.6. 70 Now if F*(LNG) is a direct product of nonabelian simple groups, theorem 8.1 still holds when G_ is replaced by [ (see [As4] ), i.e., we know that L is either brilliant, stabilizes a member of Us or Ug, or stabilizes a 3-decomposition. Now lemma 9.7 handles the first case. The proof of proposition 8.2 goes through verbatim forcing an L invariant subalgebra, handling the second case. If L stabilizes a 3-decomposition then L_ stabilizes the space P generated by the projections of id onto the summands of the 3-decomposition. Then by lemma 8.3, either Cr ag(P) # 0 and hence C y(C; ,Q(P)) is a proper L -invariant subalgebra or LAG contains nontrivial normal solvable subgroups. In the latter case the claim follows from the arguments given in the first paragraph of this proof. SECTION 10: MORE ABOUT SUBALGEBRAS In this section we assume that L is a simple group. Let F denote the algebraic closure of F. Definitions: Let A,B < J. Then S(A,B) := < ab:aéA and b€B> and s* (A,B) i= <a#tb: ae Aand bE B>. Lemma 10.1: Let A,B,C be L-submodules of J. The following are true. 1. S(A,B) and S¥ (A,B) are L- submodules of J. 2. If A=B then S(A,A) = S$2(A) and S¥(A,A) = s¥2(a). 3. If A,B are subalgebras then S(A,B) < A+B iff A+B _ is a subalgebra of Jiff S*(A,B) < A+B. 4. s* (A,B) and S(A,B) are homomorphic images of A@B. Moreover S2(A) and gt? (A) are homomorphic images of the symmetric square of the module A. 71 5. If A@B is asemisimple L-modules then so are S*(A,B) and s¥2( 4B). 6. If A and B are subalgebras and either s*(A,B) or S(A,B) has a one dimensional composition factor which is not <id> and A@B is a semisimple L-module, then dim(C ,(L)) > 2, and hence Np(L) is contained in a subalgebra stabilizer. 7. S(A+B,C) = $(A,C) + S(B,C) and S*(A+B,C) = $*(A,C) + $*(B,C). 8. 1fs*(a), s*2(B) , S*#(A,B) < id+A+B,then id+A+B is an L-invariant subalgebra. 9. If S2(A), $2(B) and S(A,B) < id+A+B, then id+A+B is an L-invariant subalgebra. Proof: Part 1 is clear because both # and . are I invariant multiplications. Part 2 is obvious. Parts 3, 8 and 9 are also clear given lemma 2.6. For Part 4 notice that the map ®: A x Br s* (A,B) resp. S(A,B) defined by ®(a,b) = a#b resp ab is bilinear and L-equivariant. Hence, by the universality of the tensor product, © must factor though A@B as an L-map. Using the commutativity of J we see that 4: AxA a s#2(a) resp. $2(A) factors through A@A/ I where I is the ideal generated by the commutators. This shows part 4. Parts 5 and 6 follow form Part 4. Part 7 is clear because the multiplications (. and #) on J are bilinear. Definition: We call an L-module small if its F-dimension is less than or equal to 26. Lemma 10.2: Let L<G. If {A,; : i€{1,....m}} are the small L-modules and for all ije {1,...,.m}, A;@ A; is a semisimple L-module then the socle of J is an L-invariant subalgebra. Moreover the socle of J is a semisimple L - module. 72 Proof: By assumption L is a subgroup of G. Let U = & wi be the socle of J; notice that the set of U,’s is a subset of the A,’s. Now 52) =z S(U;,U5) by part 7 of lemma 10.1. By part 5 of lemma 10.1 and our assumptions on the Us we conclude that S(U;,U;) is a semisimple L-module contained in J. So S(U;,U5) is contained in U for all ij € {1,....r}. Hence s2(U) Cc U and from Lemma 4.5 we can conclude that U is a subalgebra of J. Now the claim follows because U is semisimple. Lemma 10.3: If F is algebraically closed and L < G, then id®@ does not have a composition factor of dimension 24 or 25. Proof: Assume otherwise. Then as J is a selfdual L-module, id® must contain a one or two dimensional L-module U. Because L_ is assumed to be a simple group, U must be a trivial L-module and hence L centralizes a point of J. But id® regarded as a Cq(x)-module, x€ idO, does not not have composition factors of dimension greater than 16 (see orbit on points lemma and subalgebra theorem). So as L < Cq(x) for some x € id®, idO regarded as an L-module cannot have a composition factor of dimension greater than 16. Lemma 10.4: Let U_ be an L-submodule of idO. If s#2(y) < <id>, then U is singular. Proof: By lemma 3.4 it suffices to show that every u€U is singular. Now recall from lemma 4.6 that Ue =S*?(U) < <id>. So U@>id+ = id@ > U and hence, by definition, U is brilliant. If u € U_ is nonsingular then by lemma 4.2 u is G- conjugate to Xy+xX5 or x1 +ax¢. In the first case lemmas 4.3 and 4.7 show that u#u € idO contradicting our assumption. 73 In the second case u#u is singular. This can’t happen because <id> does not contain singular points. So every point in U is singular, hence the claim. Lemma 10.5: Let U be an L-submodule of idQ@. Then s¥2(y) < idO iff U is Q- singular. Proof: Assume s#2(y) < idO. We must show that U is Q-singular. By Lemma 3.3.1 and 2 it suffices to show that for all ueU we have P;,(u) = 0. So it suffices to show that id € U®. Now by lemma 4.6 we have idO@ > s#2(y) = Uet. So Ue > id@+ = id and the first half of the claim follows. Now if U is Q-singular, then for all usywEU we have 0 = Q(u+v) = Q(u)+Q(v) + 2(u,v) = -2f(id,u,v) = - (u#v,id); the last two equal signs follow from lemma 3.3 and the definition of # respectively. So the second half of the claim holds as u,v were arbitrarily chosen. Lemma 10.6: If U is the unique L-submodule of id@_ in its quasiequivalence class, then either U has an L invariant complement in idO or U is Q-singular and ide/ut ~ U* (the dual module of U). In particular if dim(U) > 14 or U is the unique dim(U) dimensional composition factor of the L-module id®, then U has an L invariant complement. In any event ut is always L invariant. Proof: These are well known facts about self dual modules. Recall that idO is self dual as Q is a nondegenerate G-invariant quadratic form on id9. Lemma 10.7: Let U_ be a three dimensional irreducible L submodule of J. Suppose that dim(S*2(U)) > 3, U_ is brilliant and contains no singular points and that s#2(y) is not brilliant. Then s¥2(y) is a member of Ug. Proof: Let {u;: 1 <i<3} be a basis for U. First we observe that s¥2(y) is generated 74 by B:= { (uta)? : 1<i<j<3 } and that by lemma 4.4 and our hypothesis on U we have that ut? must be singular for all 0 4 u € U. Next we observe that % contains a special triple, otherwise <B> = s#2(y) is brilliant. Next we find a basis f = {s::1<i<3} of U such that { si?) is a special triple. This can be achieved as follows. Pick three points in % which form a special triple, say (xi? }. Now if {x,} is a basis of U we are done. So suppose otherwise. Pick y evu\ <x;: 1<i<3> then for alla € F* { X19Xq.Xgtay} is a basis for U. Now (xi? xF? (xg-+ay)* 2) is a polynomial in a of degree 2 with nonzero constant term. #2 oF? (xg-tay)*”) # 0. So we So as |F| >3 there exists an a #0 such that f(x have the desired basis for a proper choice of a. Next we want to show that s. i#s; € 3? AN a for all i# j. For this it suffices to prove that: (*) (3, #5,)#sfP = 0 for if j. Now because the # - square of a singular point is zero and the #-square of an element of U is singular the following holds: 0 = ((st a," >)F? = (sF? + 20 (s:H#8:) + 07 fh”)? = 2a 2(s,#s;)#57"°) + a(s* 2 4 r? + 2 cay? )+ 2 a*(s:#s,)#sit” ] for all aeéF. Now (*) follows from the following easy fact: If {v; : 1<i<3} are elements of a vectorspace over a field of characteristic # 2 and for all O4a € F we have 0 = vy +a Vo + a2 V3 and F contains more than 4 nonzero elements, then for all i v= y#? _ a2st? 1 0. As U_ is brilliant and contains no singular points 0 # (as; +s; + prait® + af s#8; and singular. Now [As3] 4.3.1 states that if {V¥1,V9,¥3} is a special triple, and x = Vyt Vot w is singular, where we vy ANvoA. Then Pya(w) = -1, and in particular w # 0. Soif a,f8 # 0, then we can apply [As3] 4.3.1 we to get OF 8,55. So (81 +89+83)"" is singular and not contained in csi st? #2, Then by 75 [As3] 4.4.1 and 4.5.2 we have that { st? st? sF? (s; +89 +83)" 7} is a special 4 tuple. So s#2(y) satisfies Aschbacher’s definition of ‘Ug in Section 1 of [As3]. Hence the claim. Lemma 10.8: Let U_ be an irreducible L-submodule of J. If U* is not a nontrivial homomorphic image of Sym2(U), then U is brilliant. Furthermore, if U is self dual then U is brilliant if U is not a submodule or homomorphic image of Sym2(U). Proof: First recall from [As5] 2.1.1 that the space of symmetric bilinear forms on U is isomorphic to the dual of Sym2(U). The map W: U + Symmetric Bilinear forms of U given by uW = f(u,-,-) is an L equivariant map. Now W is either injective or trivial as W is L equivariant and U_ is irreducible. So *: Sym?(U) r+ U* is either surjective or trivial. By our assumption we force ¥* to be trivial and hence W is trivial. Thus f is trivialon U and hence U is brilliant. In the next series of lemmas we will investigate the normalizers of simple subgroups that stabilize subalgebras. Lemma 10.9: Let L bea simple subgroup of NG(A) where A is a good subalgebra and suppose that C y(L) = <id>. Then Aé Ug Vg,Ug,115,13,V, and L is isomorphic to a subgroup of PSLo(F), SL3(F), SU3(F), Spg(F), SL3(F), Spg(F) respectively. Proof: By assumption on L, Cq(A) N L = {e}. So an isomorphic image of L is contained in NG(A)/Cg(A). The claim follows from the subalgebra theorem in section 6 by inspection. Lemma 10.10: If L stabilizes a subalgebra of type Ug, then Np(L) _ stabilizes a subalgebra. 76 Proof: Recall from lemma 5.13 that NG(A) ~ PSO3(F) x Go(F), and that Go(F) centralizes A. Let L, be the projection of L into the i’th factor of Ng(A). First we investigate how J can break up as an L,@ Lg - module. Now At is the tensor product of the seven dimensional irreducible Go(F)-module with the three dimensional irreducible PSLy(F) module M(2A,) (see [As3] 5.7). Moreover as a Go(F) module A+ can be decomposed into U L Ua 1 Ud’, where a€N(,(Go(F)), such that s¥ (ual) is a primitive idempotent contained in A. Now by [As7] Ly can act either irreducibly on U, in which case L ~ PSL(F9), Fo<F, and U~ M(6),), or Lo fixes a Q-nondegenerate point x of U and U ~ <x> © W\V as an Lo-module where W,V are irreducible three dimensional Lo-modules. We denote the two Lo actions by Li i€{1,2} respectively . Now if L < L, @L3 then as an L-module J ~ A © <xL,;>@ V@M(3)\W@M(3), where M(3) denotes some three dimensional irreducible L-module. As L_ is isomorphic to a simple subgroup of L,, L_ is isomorphic to Alts or PSLO(F 9), where F is a subfield of F. Now when L ~ PSLo(F9) then V®M(3) is either irreducible or isomorphic to M(4A,)®M(2A5)®M(0). When L = Alt, then V@M(3) isomorphic to M(5)®M(3)®M(1) or M(5)@M(4), where M(i) denotes an irreducible Alts module of dimension i. Now either Cy(L) # <id> or <xL > is the unique three dimensional irreducible L submodule and is hence an N/(L) invariant submodule of J. Now s¥(<xL,>) = A as x#x € Ais a primitive idempotent and as an L-module Ax M(0)®W,Wa five dimensional irreducible L-module. So then A is also Ny(L) invariant. Now suppose L < L, @Ld. Then L~ PSLo(F 9) and asan L- module J~ A ® M(2A,)@M(6A,)° or A® M(4\,)®M(6A,)®M(8),) unless p=7, a case we treat separately. In case At is irreducible A is the unique L-invariant module isomorphic to A. So then Ais N/p(L) -invariant. In case A+ is not irreducible, M(6A,) is Np(L)- 77 invariant. Now by lemmas 10.8 and 13.1.3 M(6A,) is a brilliant subspace of J. If M(6A,) contains singular points, then by [As2] Thml Nj,(M(6Aj,)) is a brilliant subgroup of I, and hence the result follows from 9.7. In case M(6\,) does not contain singular points, the high weight vector x wrt some Cartan subgroup of L of M(6),) must be brilliant and nonsingular. Hence, by Lemma 4.3, its # square is nonzero. So 0 # x#x is a high weight vector in J of high weight 12\,. Sothen idO contains a nontrivial homomorphic image of M(12,,) or M(A,)°@ M(A,)/M(8A,) (= W(12A,) the Weyl module) or M(2A,)®M(0)/M(8\,) depending on whether p#11, p=11 and |F/4#11, |F]=11 respectively, or M(A,)9; a contradiction to the decomposition of J that we assumed at the beginning of this paragraph. Now assume p=7, L < L,® Ld and At is not irreducible. Then at~ M(2A,)@M(6A,) as an L-module and hence J ~ A ® M(4.,)/M(A,®)@M(A,)/M(4A) ® M(6A,) (ie, J contains a nonsplit indecomposable submodule) or A@M(4A,)/M(2A,)®M(0)/M(4A,) @M(6A,) when |F|#7 , |F|=7 respectively. So in the first case A is Np(L) invariant. In the second case we don’t know whether J is completely reducible or contains the projective indecomposable M(4A,)/M(2\,)®M(0)/M(4,,). In case of complete reducibility, C y(L) is a proper nontrivial N,(L) invariant subalgebra see lemma 9.3. In the other case A is Np(L) invariant. Lemma 10.11: If L stabilizes a subalgebra of type V9,Ug or Vi5> then Ny(L) stabilizes a proper nontrivial subalgebra. Proof: Extend the field F to F if necessary to contain a cube root of unity. Then the subalgebra stabilizer has a nontrivial center. Now let o be an automorphism of T whose fixed points are T. Now let . = <I,o>. Observe that 04 Cat), and hence F*(N=(L)) is not a simple group. So by the structure theorem N.(L) fixes a G r 78 subalgebra A of J. Next we observe that 7€C-(Np(L)). Thus A is o@Ny,(L) invariant so by 25.7.2 of [As9] there exists an FNp(L) module A such that dim p(A) = dim;(A). Moreover A. is the set of fixed points in A of a, and hence A is a subalgebra by lemma 9.2.1. This is the claim. Lemma 10.12: If L stabilizes a subalgebra of type VY, then Ny(L) stabilizes a proper nontrivial subalgebra. Proof: WLOG let U = <X1.XQ5Xg> be an L invariant member of V3. Then a straightforward computation shows: id < <id,\U> < < idJUA> < vat < ut <s is an L-invariant flag of subspaces of dimensions 1,4,10,18,24 resp. Moreover, it’s easy to check that the first four spaces in the flag are indeed subalgebras. If all the Nj,(L) conjugates of U are contained in vat, then Ny(L) _ stabilizes the subalgebra of uA+t which is generated by the Ny(L) conjugates of U. We assume for the moment that L stabilizes a G-conjugate U of U such that U is a complement to utin J. Then { x4taq, X5+a5, Xgtag: a; € ut} is a basis for U, for some proper choice of a.’ By corollary 6.5.3 we have s¥(u,ut) = UA. Let Ww = <x X56 oXB>> then we compute that s¥(u,w) = <Xo 40% Q5 9X1 99% Q4 2X35 X14 0% 369% 15% 06> is a subalgebra of type To. Another calculation using 6.5.3 shows that s*(u,) < vat, and that s¥(u,0) has trivial Q- radical. Also, we can compute that UA is the Q-radical of uA+. So uAt = UA®e s*(u,U) as an L-module. Similarly, we obtain that tat =Ta © s¥(u,U) as an L-module. Another calculation shows that UANUA = 0. So s*(u,U) = UAtn tat, and is hence an L-invariant nondegenerate subalgebra. So L stabilizes a member of Tyg or Ug. So now lemma 10.11 applies and hence Ny(L) stabilizes some proper nontrivial subalgebra . 79 Now let V denote the subspace of J which is generated by the N/(L) conjugates of U. By the previous paragraph we may assume that every N/(L) conjugate of U lies in U+. Thus as V_ is a direct sum of conjugates of U, we conclude that V_ is totally Q-singular as U<Rad(V). So as V is a proper N,(L)-invariant subspace of J, we will show that either V is a subalgebra or else that there exists an L-submodule U as in the previous paragraph. If V_ is brilliant, then Ny(L) is a brilliant subgroup of Eg, by [As2] Thm], as V is generated by singular points. If V is a direct sum of 2 members of Y, then V is brilliant and contains singular points. So V is a direct sum of 3 or 4 Nvy( L) conjugates. The latter case is impossible as then id@ ~ V\M(0)@M(0)\V* has eight composition factors Galois conjugate or dual to U and two composition factors of dimension 1. Then by lemma 12.2 an element of order three in L_ has trace 2: a contradiction to lemma 12.1. So V= Up © U,@ Up with U= Ug and U; N/p(L) conjugate to U. Now as U; < u,+ for all ij we have, by corollary 6.5.3, that s*(U;,U;) < U;ANU,A for all ij. Now as s#(y.a) = U; and U;NU; = 0 we get that s**(s*(u;,U,)) = 0. So by lemma 3.4 s*(U,,U,) is a singular subspace of UA Nn U;4. As_ V is not brilliant, we may assume that U; is not contained in U;A and hence s*(U;U;) # 0, and U; 9 s*(U,,U;) = 0. As s*(U,,U;) is L-invariant and we assumed that Cigo(L) = 0 it follows that dim(S*(U,,U,)) > 3. So now U, © S*(U;,U;) is at least six dimensional and singular, as s*(U;,U;) is a singular subspace of U;A, and hence a member of ¥,. So s*(U;,U;) is the intersection of two members of Y, and is hence a three dimensional amber subspace by [As1] 9.9, i.e., a member of Y3. Now V is not brilliant so there exists a vector vE€V such v = ugtu,+u9 with u;€U; and T(v)40. Thus OFT(v)= 6f(ug,u,,u9) = 3 ( ug, uy#uUg). So as s#(U,,Uy) is a three dimensional irreducible L-module, it is a complement to Up- in 80 J. Moreover, s*(U,,U9) is G conjugate to U and L-invariant. So the argument in the second paragraph of this proof applies and we are done. Lemma 10.13: If L stabilizes a subalgebra of type Yg, then N/(L) stabilizes a subalgebra. Proof: Let U be a maximal tangerine subspace of id@. Then by corollary 6.5.4 ux is a subalgebra of J and s¥(u,u+) = U. Now if every Np(L) conjugate of U is contained in ut, then the subalgebra generated by the Ny(L) conjugates of U is proper and N_(L) invariant. So assume that U is an Ny(L) conjugate of U not contained in u+. Then by [As1] 9.9 1-3 one of the following holds: i. UNUT is one dimensional. ii. UN T is an amber plane. ii, Ut NT =0. If i. holds Ciae(L) # 0 and the claim holds. If ii. holds the claim follows from lemma 10.12. If iii, holds J =U ® (utnAt+) @®U and (utntt) is an L invariant nondegenerate subalgebra of dimension 15. By inspecting the conjugacy classes of nondegenerate subalgebras, we see that (utnt+) is a V1, type subalgebra. Now lemma 10.11 proves the claim. We summarize lemmas 10.10 to 10.13 in: Proposition 10.14: If L stabilizes a subalgebra, then N,(L) stabilizes a subalgebra. Proof: If C(L) # <id> then lemma 9.3 gives the claim. If L stabilizes a subalgebra, then L also stabilizes a good subalgebra. In case C ,(L) = <id> Lemma 10.10 gives the possible choices of good subalgebras that L can stabilize. Lemmas 10.11 to 10.16 81 deal with each of the possibilities listed in lemma 10.10. Corollary 10.15: If L_ stabilizes a brilliant subspace which is generated by singular points, then N/(L) stabilizes a subalgebra. Proof: By [As2] Thm1 L is a brilliant subgroup of Eg(F). Now by proposition 6.9L stabilizes a subalgebra. So the claim follows from proposition 10.14. Corollary 10.16: If K is an extension field of F and L stabilizes a subalgebra of JK then Np(L) stabilizes a subalgebra of J. Proof: Let o be the automorphism of TK whose fixed points are [T. Let [I = <PK gs, Then o € C(Np(L)). Now by proposition 10.14 N,(L) stabilizes a subalgebra AK. so oxNy(L) stabilizes AK. So there exists an FN/(L) invariant submodule A of J. by [As9] 25.7.2. A = CqK(?) so A is a subalgebra by lemma 9.2.1. Lemma 10.17: If the minimal faithful representation of L over F has dimension > 14, then L is not a subgroup of G unless L has an irreducible F-representation of dimension 26. Proof: Let p be any 26 dimensional F-representation of L into G. Therefore Lp centralizes a nondegenerate quadratic form, as G does. We observe first that L can only have one nontrivial composition factor. If this composition factor does not have dimension 26, L centralizes a nontrivial subspace of id® by lemma 10.6. In this case L_ centralizes a point # id. The point centralizers are universal groups of Lie type that have nontrivial representations of degree less than 9. So because L is simple, L must also have a nontrivial representation of degree less or equal to 9: a contradiction to our hypothesis. The claim follows. 82 At this point we know that any closed simple subgroup L < G < G_ such that Ny(L) is maximal in I does not stabilize a proper nontrivial subalgebra of J. We will use this fact frequently in showing that certain subgroups of G do not give rise to maximal subgroups. 83 CHAPTER II: ALMOST SIMPLE SUBGROUPS OF LIE-TYPE DEFINED OVER FIELDS WHOSE CHARACTERISTIC EQUALS char(F) Throughout this chapter L will denote a simple adjoint group of Lie type of char(F) = p (p=0 allowed). We regard J as an L-module and id@ and <id> as_ L- submodules of J. We assume throughout this chapter that F is a splitting field for L and for us an L- module is always an FL-module unless otherwise stated. The arguments in this chapter will be of a more module and representation theoretic nature. SECTION 11: INITIAL REDUCTIONS Lemma 11.1: If L is a simple adjoint group of Lie type in characteristic p , then L embeds into G only if one of the following holds: 1. L is of type 3D,. 2. L is of type Go. 3. Lis of type A; or 2A., i€{1,2}. 4. Lis of type F 4. Proof: If L is an orthogonal group PQ, (for us it is irrelevant what the type of L is), then L_ contains a subgroup gn-l, Altyp, when n is odd, and a subgroup of type 20-2. Alt 4 when n is even. To see this let {bi:l< i < n} be an orthonormal basis of an odd dimensional orthogonal F-space, V. The group generated by the reflections Tp, across the hyperplanes orthogonal to b; is an elementary abelian 2-group of 2-rank n. Its normalizer in O(V) contains the permutation matrices. Hence the claim for n odd. When n is even any orthogonal F-space will contain a nondegenerate hyperplane and the same construction works. So when n>5 L contains an elementary abelian 2 84 group E and 5 divides |A;(E)|. So in this case L is not a subgroup of G by lemma 7.11.4. So now we have eliminated all groups containing PQ, n> 5. This leaves us with the groups listed in the statement of the lemma and the groups of type C, i€¢{3,4,5} and 2Ee. The minimal dimensional faithful representations of the groups not listed in the statement of the lemma have dimension > 14 and no 26 dimensional irreducible representations (see [As6]), so these are out by lemma 10.17. Lemma 11.2: If L is of type Go, then L centralizes a proper nontrivial subalgebra unless char(F) = 7 and L acts irreducibly on idO. Proof: In characteristic 7 L possesses an irreducible 26 dimensional representation. For now we will exclude this case from our considerations and treat it separately in Section 14. The other small L-modules have dimensions 1,7, and 14 and have corresponding high weights 0,1,,A)_ respectively, see [As6]. Moreover, dim(M(2A,)) = 27 if p#7 and 26 if p = 7. So either id@ contains seven dimensional submodules or else Cra@(L) # QO and L centralizes a proper nontrivial subalgebra. Now using [G1] we see that Sym?(M(A,)) = M(0) ® M(2A,) if p#7 and is M(0)/M(2A,)/M(0) when p= 7. Soif U_ is a seven dimensional irreducible L submodule of idO then dim(S*2(U)) < 1. So either Ciae@(L) ~ 0 or lemma 10.4 applies and U_ is singular. The latter is impossible as id@ contains no seven dimensional singular subspaces. This is the claim. Lemma 11.3: If L is of type 3D,, then L stabilizes a subalgebra. Proof: As F is a splitting field for L, we see from [As6] that the only small L modules have dimension 8. From [Jol] we know that H}(L,V) = 0 for every 8 85 dimensional L module . So dim(C;4@(L)) > 2 and the claim follows from lemma 7.1 . SECTION 12: GROUPS OF LIE TYPE Ay AND 2A5 In this Section we assume that L is of type Ay or 2 Ao. Here we will start to use our knowledge about the possible character values of involutions and elements of order three in G acting on id©@. We record this in a lemma. Lemma 12.1: Let géG be an involution, then triyo(g) € {-6,2}. If heG is an element of order 3, then tr|yo(h) € {-1,8}. Proof: The first part follows from [As3] 6.5 and an easy calculation. The second part follows from [As3] 8.2 and an easy calculation. In this section we will prove that L stabilizes a subalgebra. Combined with corollary 10.16 this will show that an almost simple group of type Ay oF 2A5 is never maximal in T. From [As6] we get the set of small irreducible modules. These will be listed in the table below. Let g be the involution that acts as the diagonal matrix (- 1,-1,1) on the natural L-module. Let h be the element of order three that acts on the natural L-module as the diagonal matrix (w,w"4,1). Lemma 12.2: The following table lists all small nontrivial irreducible FL-modules up to conjugacy and duality: M(A,) M(2A,) M(A,+A9) M(A,)@M(A,)? -M(3A,) M(2A,+A9) M(4Aq) Dim(M) 3 6 8 9 10 15 15 tr(g) -1 2 0 1 -2 tr(h) 0 0 “1 0 1 86 M(5A,) M(2Ay+2A9) M(34q +p) M(Aq)P@M(AYHAQ)-- M(Ay)@M(2A,)* dim(M) 21 27 (19) 24 (18) 24 18 The character values of g and h are the same on the dual modules and Galois conjugates. The numbers in parentheses denote the value in case p=5 , and 6 denotes a nontrivial field automorphism. Proof: That the modules listed in the table together with their duals are the complete set of submodules is a consequence of [As6]. The character values are easily computed as follows: The module M(A,) is the natural module so the traces of g,h are easily computed. Now tr(x)lA gB = tr(x)|, tr(x)|p, allowing us to compute the traces on M(A,)@M(A,)°. It is well known that M(A,)@M(Aj) = M(A+A9) © M(0) so the traces on M(A,+A9) are easily obtained. Now in [Sel] it is shown that M(cA,) is the space of homogeneous polynomials in 3 variables of degree c. From this one can easily get the traces for c = 2 or 3. Lemma 12.3: If id@ contains a submodule isomorphic to M(2A,), then Ny(L) is not maximal in I. Proof: We first recall from [As6] that Sym?(M(2A,)) ~ M(4\,) ® M(2A,) and that M(2A,)@M(2\,) ~ M(4\,) ® M(2A,) ®@ M(2A\,+Ay). Further observe that M(4A,) and M(2A,+A9) are not self dual modules. Thus neither of these modules can be submodules of id@ as id@ is self dual and hence can not contain nonself dual modules of dimension greater than 13. Now let U ~ M(2A,) be a submodule of id@ and let W= s#2(y), By the above W~ U. If W< U_ then id+U is a subalgebra by lemma 10.1 and hence Ny(L) is not maximal by corollary 10.16. By lemma 10.1 the only alternative is WOU 87 = O0Oand W ~ U. Now either s*(U,w) or s#2(w) is not contained in id @ U @ W else id®U®W is an L invariant subalgebra and Ny(L) is not maximal. So there exists a submodule V of s*(u,w) or s¥2(w) and V ~ U_ which is not contained in id ® U ® W. Similarly if id@U®@WEV is not a subalgebra, there exists a submodule T ~ U_ which is not contained in id@U@®WO@V. Now dim((id@U@W@VOT)*+) = 2 and hence dim(C;49(L)) = 2. Thus in this last case N/(L) stabilizes a subalgebra , completing the proof. Lemma 12.4: If id® contains a submodule isomorphic to M(A,), then Ny(L) is not maximal in [. Proof: Let Ux M(A,) be a submodule of idO. As Sym*?(M(A,)) ~ M(2A,) we have s#2(y) is either trivial or isomorphic to M(2A,). In the first case U is singular and we are done by corollary 10.16. The second case is handled by lemma 12.3. Lemma 12.5: If id® contains a submodule isomorphic to M(3A,), then Ny(L) is not maximal in I. Proof: Let U ~ M(3A,) be a submodule of id®@. First we observe that U_ is Q singular as_ U is not self dual. So id@/ ut ~ M(3A5) and ut/ U_ is six dimensional. Now checking traces of g and h shows that the only composition series of uty, U_ that does not violate lemma 12.1 is the series consisting of trivial composition factors. Now in [As6] it is proved that Sym2(U) has neither ten nor one dimensional composition factors. From this we conclude that gt U) =0 and hence U is singular. This contradicts that the maximal dimension of a singular subspace is 6. Soin fact U can never be a submodule of idO, establishing the claim. 88 Lemma 12.6: The following are true: 1. Sym?(M(A,+Ag)) © M(2A,+2A9) @ M(A,+Ay) © M(O) when p# 5. 2. Sym?(M(A,+A9)) has composition factors M(2A,+29), M(0), and M(A,+A5) occurring with multiplicity 2 when p=5. 3. Sym(M(A,)@M(A1)°) ~ M(2A,)@M(2A,)° @ M(Ag)@M(Ay)°. Proof: Part 3 follows readily from the well known formula Sym2(A@B) ~ Sym*(A)@Sym2(B) a) A?(A) @A2(B), see [As5] section 2.. Part 1 and 2 can be deduced from [As6] once we observe that M(0) and M(2(A,;+A9)) are submodules of Sym?(M(A,+A9)) and that Sym?(M(A,-+A9)) is a self dual module. Lemma 12.7: If id@ contains a submodule isomorphic to M(A,)®M(A,)°, then Np(L) is not maximal in PI. Proof: Let U=M(A,)@M(A,)° be a submodule of id@ and let W= s#2(y). If id® U_ is a subalgebra the claim follows. So assume otherwise. Then dim(W) = 9 and UN W =0. So dim((U@W)+) = 8. Now only the choice (U@W)+ ~ M(A,+A9) does not lead to a contradiction of lemma 12.1. Now we see using lemma 12.6.1 and 2 that s*2((u@w)-) < id® (Uaw)+ and hence L stabilizes a subalgebra. We are done by corollary 10.16. Lemma 12.8: If id®@ contains a submodule isomorphic to M(A,+ Ag), then Np(L) is not maximal. Proof: Let U ~ M(A,+A9) be a submodule of id@ and let W = s#2(y), If W< U @ id then U@ id is an L invariant subalgebra and we are done by corollary 10.16. If M(0) < W and M(0) #id, then C:44(L) # 0 and again we are done. So we may assume that W~U and WNU = 0. Then (weu)t is a ten 89 dimensional submodule of id®. Now by the previous lemmas (wnu)+ can only contain a submodule isomorphic to M(A,+Ag9). This forces three composition factors isomorphic to U and two trivial composition factors. Now [Jol] shows that H}(L,U) =0. Thus Crae(L) % 0. This gives the claim. Proposition 12.9: If L is of type Ag or 2A then Ny(L) is not maximal in [. Proof: L has no irreducible 26 dimensional representations so L stabilizes a submodule U of id. So either U or U+ has dimension less than or equal to 13. So WLOG U is isomorphic to one of the modules of dimension less than 13 listed in table 13.1. Now the lemmas 12.3, 4, 5, 7 and 8 handle each of the cases. SECTION 13: GROUPS OF LIE TYPE A, Let L = PSLO(F). Let g,h € SLo(F) be the diagonal matrices (i,-i) and (ww!) where i, w are primitive fourth and third roots of unity of F, respectively. Lemma 13.1: 1. The small irreducible L-modules are: M(2a\,) , 2a <pand 2a < 26 M(ad,)@M(bA,)° , 641,2| (a+b) and (at1)(bt1) < 26 M(aA,)@M(bA,)°@M(cA,)%, « # 1,6, 2| (atbte) and (at1)(b+1)(ct+1) < 26 M(A,)®M(A,)°@M(A,)"@M(Aq)9, a # 1,8 and f # 5,a,1. 2. Tr(B)Iy(2aa,) = CD* » T@)laba,) = ° Tr()IM(ar,) = a+1 mod 3 and Tr(B)IM(aa,) é€ {-1,0,1} 3. Sym?(M(ad,)) = @M((2a- 4i)d,) i € {0,..a/2]} if 2a < p. If a< p< 2a define U, ~ M(2kA,) for O< k < p-a-lor k = (p-1)/2 and k 1 90 define Uy, ~ M( 2 ( p-k-1) 4,) / Ay / M(2 (p-k-1) 4,) for (p-1)/2<k <a. If |F| # p then define A, ~ M((2k-p)A;) @ M(A,)® and_ if |F|=p then define A, ~ M( (2k-p+1)A,) © M( (2k-p-1)A,). Then Sym?(M(aa,)) © ® Uaok k € {0,...,[a/2]} if a <p < 2a. Moreover U, is indecomposable if |F| # p. 4. Let U bean L submodule of the L module V. Then: Sym*(V) = Sym?(V/U) / ( (V/U)@U) / Sym>(U). 5. Let A,B be L modules. Then: Sym2(A@B) ~ Sym2(A)@Sym2(B) ® A2(A) @A2(B). Proof: Parts 1 and 3 can be found in {[As6]. Part 4 is the content of [As5] 2.2.1. Part 2 is a simple calculation using the fact that M(aA,) can be identified with space of homogeneous polynomials in two variables of degree a. Moreover we use that Tr(x)lyow = Tr(x)|y Tr(x)|w- Lemma 13.2: If Ny(L) is maximal then J contains no submodule isomorphic to a Galois conjugate of M(2.,). Proof: Let U ~ M(2A,) be an L-submodule of J. Then s#2(y) is a homomorphic image of M(4A,) ®@ M(0). So U is brilliant by lemma 10.8. So U does not contain singular points else Lis brilliant and hence N,(L) nonmaximal. If dim(S*°(U)) is 0 or 1 then either U_ is singular and hence N/(L) is nonmaximal or C,y9(L) # 0 and again N,(L) is nonmaximal. If s¥2(y) is brilliant then L_ is brilliant as s¥(u) contains singular points. So we can assume that dim(S*2(U)) > 5, s¥2(y) is not brilliant; i.e., the hypotheses of lemma 10.7 are satisfied. So s#2(y) is a member of ‘Ug and as an L-module is isomorphic to M(0) ® M(4,). Now if id € s#2(y) then by lemma 5.10 s#2(y) is an L invariant subalgebra so we are done by corollary 10.16. If id¢ s#2(y) then C,39(L) #0 and hence N/(L) is not maximal. 91 Lemma 13.3: If Ny(L) is maximal then no Galois conjugate of M(A,)@M(A,)° is isomorphic to a submodule of J. Proof: Let U be an L - submodule of J isomorphic to M(A,)@M(A,)°. Recall from 13.1 that Sym?(U) ~ M(2A,)@M(2A,)° @ M(0) and that Sym?(M(2A,)@M(2A,)°) ~ M(4A,)@M(421)°@ M(44) © M(41)@ M(0) © M(2A,)@M(2A,)°. Let S$ = s#2(u) nid@ and T = $*%(s). Now if dim(S) < 1 then either U is singular or Crge@(L) # 0. So then N/(L) is not maximal. So we may assume that S ~ M(2A,)®M(2A,)°. Now observe that T is not contained in id+S ; else id+S is a ten dimensional subalgebra of J. So T contains a submodule W_ such that WNU = WNS = 0 and W is isomorphic to S or M(4),). If WS then dim((U@S@W)+) = 4 and so in view of lemma 13.2 either Ci ae@(L) #0 or (Uesew)t is isomorphic to a Galois conjugate of U . The first case leads to a nonmaximal Ny(L). The second case leads to a contradiction to lemma 12.1 as the trace of a three element is 2. If W~M(4.,) then s#2(w) is not contained in W-+id and dim(S*7(w)) > 5. If p47 s¥2(w) contains a submodule V ~ W and VOW = VNU = VNS = 0. Thus dim((U@S@WeV)+) = 3 and we are done by lemma 13.2. If p=7 and s¥2(w) does not contain a submodule V~ W and WNV = 0, then s#2(w) contains a submodule R ~ M(4,)/M(A,)°@M(A,)/W. Now dim(U@®S@R) = 27: a contradiction as dim(id@) = 26. Lemma 13.4: If Np(L) is maximal then J does not contain an L submodule that is Galois conjugate to M(6\,), M(10A,), M(A,)@M(3A})° or M(A,)°@M(5A,), unless L ~ PSL2(7). Proof: Let U < J bea Galois conjugate of one of the irreducible L- modules in the 92 statement of the lemma. We observe first that U_ is brilliant as U is not a homomorphic image of Sym2(U). So in view of corollary 10.15 we assume that U does not contain singular points. So if x € U is a highest weight vector of U = wrt some Cartan subgroup of L of weight A, then 0 4 x#x is a highest weight vector of weight 2\ of <x#xL> < S*?(u). if U # M(6A) then dim(S*7(U)) > 21 and UN<x#xL> = 0, a contradiction as dim(id@) = 26. So for the rest of the proof U ~ M(6A,). Then let S = <x#x L >. Now dim(S) = 13 and S < id@ unless char(F) =7 and S > <id>. Now (U@S)+N idO isa six respectively seven dimensional L submodule of id. So in view of the two previous lemmas id@ contains a five dim. irreducible submodule T or in case char(F) = 7 one of dimension 7. In the latter case the trace of an involution of L is -2 contradicting lemma 12.1. In case dim(T) = 5 J contains a 25 dimensional submodule, namely U@®S@T. Now the orthogonal complement of this space is a two dimensional L module. Thus dim(C ,(L)) 2 2 and Ny(L) is not maximal. Lemma 13.5: If U is an L-submodule Galois conjugate to M(44,) then Ny(L) is not maximal, unless L ~ PSLo(7). Proof: We consider first the case p#7 and |F| #5. Let T = s¥2(y) nidd&. T contains either an S~ M(4\,) # U or an M(8A,), else id + U_ is a subalgebra or Crag(L) # 0 and N,(L) is not maximal. So then at least one of s#?(s) or s¥(s.u) contains a submodule W # S,U and W ~ M(aA,) a€{4,8,}. Now 0% dim( (U@S@W)+nide)) < 6, unless U ~ S ~ W. In the first case idO has either a three dimensional L-submodule contradicting lemma 13.2 or the involution geéL has tr(g) = 1 on every composition factor of id@ and id@ has at least five compfactors so that tr; g@(8) > 5 contradicting lemma 12.1. In the second case s*(w,x ) XE {U,S,W} must contain another submodule R distinct from U,S5,W and isomorphic to 93 M(aA,) a€{4,8}, else id+U+S+W is a subalgebra. Now in both cases (U@S@®WOR)* is of dimension < 6. So as before either id® contains a three dimensional irreducible or the trace of an involution of L is > 5 contradicting lemma 12.1. Now assume |F| = 5. Then L = Alts. It is well known that the irreducible FL modules fall into 2 blocks, B= {M(0),M(2A, )} and By = {M(4A, )}. So if idO@ has composition factors isomorphic to M(0) or M(2\,), then id@ contains a submodule isomorphic to M(0) or M(2\,). Observe that dim(M(4,)) does not divide dim(id@) and hence id@ contains a submodule isomorphic to M(0) or M(2,,). Now either lemma 13.2 or corollary 10.16 applies. Now let char(F) = 7 and |F|#7. As before let T = s#2(y) N id®. Suppose for the moment that T = M(4),)/ M(A1)@M(A,)°/ U. We compute that Sym?(T) = (T@M(0))/M(4A,)@A/(T@M(0)) where A = (M(A,)@M(A,)°)/M(4A,), by using lemma 13.1.4. Also by definition of T we have T< s¥2 (7), Now let Y := M(44,)@A then: Y = (M(5A,)@M(Aq)°® M(3A,)@M(Aq)°)/ (T © M(0) ® M(6A,)@M(2A,)) So as s¥2 (7) is a homomorphic image of T, we violate either lemma 13.2 or 13.4, or id@®T is a subalgebra in which case we violate proposition 10.14. So from now on we may assume that whenever W,V ~ M(4),) are submodules of id®, then s¥(w , V) does not contain a module isomorphic to T. So then we can proceed as in the case char(F) # 5,7. So now we assume that idO contains no submodules isomorphic to T. Thus we can now proceed as in the case p # 5,7. The proof is now complete. At this point we observe that N/(L) maximal in [ forces that every L submodule of id® is at least nine dimensional, unless L ~ PSL2(7). 94 Lemma 13.6: If S# T are irreducible L submodules of id@ of dimension > 9, then (SeT)t N id® is a submodule of dimension at most 8. If Np(L) is maximal then idO = SOT. Proof: Clear in view of lemmas 13.2 , 3 , 4 and 5. Proposition 13.7: If Np(L) is maximal , then either char(F) > 17 and idO = M(8\,)®M(16A,), or char(F) = 13 |F| # 13 and id@ = M(8),) / M(A,)°@M(3A,) / M(8A,), or char(F) = |F| = 13 and id® = M(8A,)\M(2A,)®M(4A,)\M(8A,), or L = PSLo(7) and idO = M(6\,)®M(4A,)@(M(4A,)\M(0)®M(2A,)\M(4A, )). Proof: We treat the case PSLo(7) last. So for now we assume L # PSLo(7). {2 We observe first that L can not act irreducibly on id® as then by 13.1 idO M(25,,) and so the trace of an involution is +1 hence contradicting 12.1. So assume for the moment that id@ ~ S @ T with dim(S),dim(T) > 9. If Sx M(12,,) =~ T then an element of order 3 in L_ has trace +2 by 13.1 which contradicts 12.1. If Sz M(2A,)@M(2A,)° then by 14.1 Sym?(S) has no 17 dimensional homomorphic image. So this case is out by the above lemma or because S + id isa subalgebra. Also S does not have dimension 12 because by 13.1 L has no irreducible 14 dimensional representations. If idO~M(8A,) ® M(16A,) and p>17, then no contradiction to earlier lemmas arises. Moreover, in this case one may assume that s#2(s) > T and g¥2(7) > S. Now we assume that S = Soc(id@) is irreducible of dimension > 9. In case dim(S) = 13 S*x idO/S ~ M(12A,) and the element of order 3 in L trace argument eliminates this case. If dim(S) = 12 then idO / st ~ s* and dim (s+/s) =: 2. Now the trace of an 95 element of order 3 in Lis, by 13.1.2, zero on any 12 dimensional irreducible L module. Soon id@ the element has trace +2, contradicting 12.1. The case S ~ M(2A,)@M(2A,)° is out because Sym2(S) is semisimple so either S is a subalgebra or idO contains another irreducible L submodule distinct from S. The first possibility is out by corollary 10.16 and the second contradicts our assumption S = Soc(id®). So now assume that S ~ M(8\,) then id®@ = M(8A,)/ U / M(8Aj,) indecomposable and s#?( M(8A,)) contains T/ M(8A,) indecomposable with 0 4 T < U. Then by 13.1 char(F) = 13 and id® is as claimed. Now we consider the case L = PSLo(7). In this case lemma 12.1 and the usual considerations eliminate all possibilities other than the the one listed in the statement of the lemma. We have now exhausted all possibilities, and the claim follows. SECTION 14: THE EXISTENCE AND UNIQUENESS OF CERTAIN A,’s AND Go’s Theorem 14.1: If char(F) = 7 then G_ contains a unique conjugacy class of subgroups isomorphic to Go(F) and acting irreducibly on id®. Moreover idO ~ M(2\,) as a Go(F) module. Proof: This is the content of theorems 1.c and 2.c of Testerman [T2]. Theorem 14.2: If char(F) > 17 then G contains a conjugacy class of PSLo(F) *s such that idO ~ M(8A,) ® M(16\,) as a PSLo(F) module. If char(F) = 13 and |F| 4 13, then G contains a class of PSLo(F) ’s such that id@ ~ M(8A,)/ M(A,)? 96 ® M(3\,) / M(8A,) as a PSLo(F) module. Proof: Testerman constructed this PSLo(F) in [T3]. Now we want to establish the uniqueness up to conjugacy in G of the PSLo(F)’s when p > 17. Notice that we built into our assumption that Nyp(PSLo(F)) is maximal. Otherwise PSL.o(F) stabilizes a subalgebra and hence can not act as we assumed. Lemma 14.3: Let f € sy3(v) (the set of set of symmetric trilinear forms on V), L < O(V,f) with L perfect, C the conjugacy class of L under GL(V), and B the set of <b> € sy3(v) with b similar tof. The number of orbits of A(V,f) on O(V,f) N C_ is equal to the number of orbits of Neuvy)) on B. Here A(V,f) denotes the subgroup of GL(V) preserving f up to a scalar. Proof: This is 2.9 of [As5]. Lemma 14.4: We keep the notation of 14.3. If f is the trilinear form of section 3, Nouv) has one orbit on B, and Cy,,(L) = id, then G has one orbit on GNC. Proof; Let L~M,N € GNC. Then by lemma 14.3 there exists h € A(V,f) such that M® = N. Now as Cy(M) = C\(N) = id, it follows that h € Rix Nacv,p{<id>)- Now R= Cav,n(@) ®G as Agiv)(9) = G. So as Cav.) does not fuse G conjugacy classes, we can adjust h so that it lies in G. The claim follows. For the remainder of this section let L ~ PSLo(F). Until further notice we assume 97 that id@ ~ M(8A,) © M(16\,) as an L- module. Lemma 14.5: dim( Hompy( M(ad,) , Sym?(M(bA})) )) = 1. if a € {0,8,16} and be {8,16} orif a=b=0. Proof: A simple computation using 13.1. Definitions: Let fij denote the nontrivial L-invariant trilinear map W : M(iA,) x M(jA,) x M(jA,) > F defined by W(x,y, y’) = (y,y') (x(ar)) where a isa generator of Hom py (M(iA,),Sym2(M(jA,))) and fT: Sym?(M(jA,)) ~ {symmetric bilinear forms of M(jA,)} is an FL - isomorphism. When i=j let f, denote the nontrivial L- invariant trilinear form defined on M(iA,), defined as above. Definition: Let U,W _ be L-submodules of id@ isomorphic to M(8A,) respectively M(16A,). Lemma 14.6: Let f be the trilinear form of section 3. Then f= arf, + agfg tar6fig +23,168,16 + 916,8%16,8+#1,8"1,8 +#1,16%1,16° where the a’s lie in F. Moreover 2416) 2g 16°71,16 ~ O. Proof: We observe that f(id,x,y) = 0 for all x € U and ye W using the fact that U = W+ nid@ and that - f(id,-- lige is the associated bilinear form of Qhae (see lemma 3.4). Now the first part is self evident from lemma 14.5. We see that 4116 #0 as w1tnw <=. As noted in the proof of proposition 13.7wW <s*?(y) = vet. So id @ U = We? > UO, and similarly We < UL = W @ id. Now let x € U bea high weight vector of U of weight 8A,. Then x#x is a high weight vector of W of weight 16A,. Now x is brilliant as 8A,(h) € F* has order # 1 or 3 for some h in the Cartan subgroup of L with respect to which the 98 weights are defined. So by lemma 4.4 x#x is either singular or zero. In the latter case we contradict W < s¥2(y) so W contains singular points. So W is not brilliant otherwise L is brilliant by [As2] Thm1 and hence L stabilizes a subalgebra. This shows a16 #0. If ag 16 = 0 then 0 = f(y,y,x) = (y#y,x) forall x € U. Thus y#y € UL = id ® W for all ye W. But then W_ is an L-invariant subalgebra a contradiction. Let F denote the algebraic closure of F, J = JF, T= LF, andf =f". Then if f is unique so is f. So replacing F by F, we may assume F is algebraically closed. Also f =ajf, + agig + arefig + 48,168.16 + *16,8' 16,8+ 91,8f1,8 + @1,16f 1,16 and f., fis are preserved by L,so L_ preserves f. Thus replacing L by L we may . assume that F = F is algebraically closed. Also we observe that Cera ()) ~ (F*)3 is generated by the scalar actions on id, U and W respectively. Notation: We fix a Cartan subgroup T of L <G and we denote the weight vectors of U and W with respect to T by {X5; : -4 <i <4} respectively {Y9; :-8 <i <8}. Let 2z be the generator of <id>. Let Ss; denote the weight space corresponding to the weight iA}. Note that we may assume that T < Hp < H where is the Cartan subgroup of E,(F) defined in section 3. Lemma 14.7: Let A bean F- algebra, D < Aut(A). Assume that D contains only semisimple elements and let S; denote the weight space of D of weight d;- Let SiS; denote the space generated by the products. Then SiS; < Si}? where Ai 4 ;(8) is defined as A;(g)4;(8) for all gé L. 99 Proof: Let s-€S;, s5€5; and géL. Then 815; g = A;(g)A;(8) 8)5; = As 4;(8) si; and the claim follows. One consequence of lemma 14.7 is that So is a subalgebra. Moreover, as Q_ is nondegenerate on Sy and F is algebraically closed, So is a member of Us by lemma 5.12.1. Observe also that J is the orthogonal direct sum of the spaces S,@S_;. So as Q is nondegenerate on J the space S; ® Ss; is Q-nondegenerate. Lemma 14.8: The following are true: 1. @ S; i = 0 mod 8 is a subalgebra of type Yo. i 2. @ S; i = 0 mod 6 is a subalgebra of type To. i 3. @ S; i = 0 mod 4 is a subalgebra of type V 15° i Proof: We note first that in all three cases © S; is a Q-nondegenerate subspace containing id and So: The fact that @S; is a subalgebra is a consequence of lemma 14.7. Let x € So be a primitive idempotent. Then as So is a member of Ug we have E, @S (x) # 0. So, by lemma 5.8, the subalgebras have the claimed types. wei Lemma 14.9: Let {V1,V9,V3} be the three pairwise orthogonal primitive idempotents spanning Sp. Then we can find s,,8; € S; such that Wi i= vg nN VoA = < 16 :Y_ 16954 98.456 98_6:8998.9> Wo = Vy N vgA = < ¥195¥_198g98_95119-¥-19°% 25 -2> Wg = VAN vgA = <8 45 455355 9:5 65 6: 149_14> - Note that each s; ,8; is Q-singular and hence singular by lemma 3.6.2. 100 Proof: We first observe that by lemma 14.7 Y;#Y; = 0 when |i] > 8 and hence these y; are singular by lemma 3.4.1. Now we also observe that (¥}¥3) =-(1/2)f(id,y;,y_;) # 0 by lemmas 14.6 and 14.5. Soif i> 8 then y; + yj brilliant and Q-nonsingular hence squares to 2yi#Y_; which by lemmas 4.2 and 4.3 is a multiple of a primitive idempotent. Moreover, observe that Yi#y; E So by lemma 14.7, so we may assume that Y:#Y_; is a multiple of one of the V5: So without loss of generality we may assume that ¥16°¥-16 © Ww, . So now as R:= © Sg: is a member of Vg we have dim(R NM W,) = 2 ( see lemma 5.6). As RN Wy = < yy@,Y_16> it follows that (Wy W3) NR= Sg@ S_g. So let <s;g> = Wo MR and <S4g>= W3NR. As T < H every weight vector of H is a weight vector of T. As vj and y, |i] > 8 are singular; they are also weight vectors of H. So let Y = {Yj0Vj 8,54: li] > 8} bea basis of weight vectors of H. Define a line to be a special triple contained in Y. Let Y be the set of points and define incidence by inclusion. Then it was observed in [As1] that the geometry defined in this way can be identified with the building of 26 (2) and is hence a generalized quadrangle. Each point is contained in exactly 5 lines and 2 points can lie on, at most, one line. Now we observe that for w;€ S; {(w;, Ws wi) #0 only if itj+k = 0. So the lines containing y,g must be of the form {y16> Y.16 4 21 Bo ilsis 3} or {¥1g > Y.1g6° Vz} or {¥4g>8.g>3.g}. Now weclaim that y 14 ,y_,9 and Y-10 are not contained in W;- Suppose otherwise; then [V1 516 ¥_;] is a triangle in the building of 6 (2), a contradiction as, by definition, generalized quadrangles do not contain contain triangles. Similarly we argue that Yi4> Y 19 and Y1Q are not contained in W,. So WLOG let Y4i4 € W3- Then again we can argue as above that y 412° Y419 are not contained in W3- Leaving as the only alternative y +12) 101 Now as R:= @ S6; is a member of Y9 we argue as in the second paragraph of this proof that <846> = W,OR and that <3 46> =W30R., Now if R:= @ S4; then R is a member of V15 by lemma 14.8. Also we have shown that RN Woy = < Y412 S43 > and that dim(RNW, ) = 4 (see lemma 5.6). So S4® S4< (W,®W3) 9 R and it follows that <844> < ROW, and <S 44> < RN Ws. Finally observe that we have now shown that W3 is as claimed. Thus S9@S_» is contained in We Wo and so Wi, Wo are also as claimed. The lemma follows. Lemma 14-10: f16(¥_16.¥19:¥@)> fg 16(¥-16:¥19>%g)s f6,86*%6Y.12)> fg(% 45x_45%g), *y (4% 40%4)> f6(¥16:¥.19¥.4)) fg 16(¥16¥-199%.4)> 1 16(@¥p¥;) with i 4 0 are all not equal to zero. Proof: Let h(,) = [2,2] and g(t) =[} | then rh(A) = Y tr; for all i and Then zm g(t) = +3 be, where b_ . denotes the binomial coefficient m over j=o MJ J m,j Jj, see [As 6]. Moreover the same formula holds for wm in place of Zm- Let r= : °| then by [As6] 2m r = (-1)™ Zm and the same formula holds for Wm- Using the invariance of the forms involved we see that (s;,8;,5)) = i+j+k f(s;h(A),s;h(A),8, h(A) y= VT f(s; ,85,9,,) for all \ € F and si€ S; s;€5; 8, € S,. So (8; 85,5, ) # O only if i+j+k = 0. Now suppose that for a fixed i £16 (¥j1¥j-¥y) = 0 for all j,k such that itj+k = 0. Then y; € y;A for all j. Thus fig a nontrivial radical. Since L acts irreducibly on U, this implies that fig = 0; a contradiction to lemma 14.5. Now assume that fie(Yig> Y¥_16°Y) = 0 then: 102 0 = f16(29:216:29) = f16(298(t),21 6 8(t).z98(t)) 25 -j-i So f16(¥16¥-14¥-2) = (-9/16) fig(Yig> Y¥.16°Y9) = 0. So we conclude that fig(Y16¥-14Y-2) =O as figie> Y¥.16°Y9) = 0. Now using that fj) ¢(y¥16,¥.14:¥.4) = 0 we get that £16(¥16¥-19°¥-4) = (- 10/15)(-9/16) fig(Y16> Y.16Y9) = 0. Continuing this way we get that f16(%16-9-16419-i) =a fig(Yig Y.16-Y9) = 0 for all i >0. Thus Y16 is contained in the radical of f16 a contradiction. The only way out is that £16(¥16> Y-16%9) # 0 and hence f16(Y16-¥-16421°9-21) #0 for all 0 <i < 8. Now we apply r to get 0 # fig( ¥167) Y_197 9-67) = fig(¥_16.Y19-¥6): All the other claims are proved in the same fashion. Proposition 14.11: There is a unique class of subgroups of G_ which are isomorphic to PSLo(F) such that J ~ M(0) ® M(8A,) ® M(16\,) as a PSLo(F) module. Proof: In view of lemma 14.6 and the fact that Cory) = (F*)3 we may choose 416= 9g.16 = 91,16 = 1. If we can now show that the remaining four a’s of lemma 14.6 are uniquely determined, we will have shown that L_ stabilizes a unique Eg - form and hence by lemma 14.4 we will have proved the claim. We observe from lemma 14.9 that for -8 <i<0 <s. > = Ker( (¥16-Y_16-17 - Ns.) and similarly for 0>i1>8 <s,>= Ker( f(Y_16> Y16-i* Ng.)- As figl¥_16> Yi90%¥6) # 9 # fg. 16(¥-16°710°%6) we compute that sg = Axgtayg) where a= f(y_16,¥19%8) / (Y.16:%19%6) # 9- Now recall from lemma 14.9 that 8g is singular. So we have: 103 0 = f(56,56,Y. 19) = aia g A*fia go (KasX ) + 2 r2e f(x )- rd? a? £ ) = 916,8 “ '16,8 *6%6-Y-12 6°%6°7-12 ¥6%6°Y-12): So a gg is determined as A # 0 and by lemma 14.10 fy g(XgoXg.19) # 0. Similarly we have: s4 = Xx, + @ y4) where 0 4 a@ = fg 16(¥16-¥-12*-4)/fie(Y16¥-12¥-4) and 0= f(s_4,8_45Xg) = ag dF Qar? f da? = agh"fg(x_49X_40Xg) + 20d” f(x gsy_4oxg) + AMO” f(Y_4s¥_goXg): So ag is determined as fg(x_4,x_4.xg) # 0 # A. Now we also have: 0 = f(s458_4.V9) = fA(xg+ 8 ¥4) , HOX_gt+oy_4), 142 + Tq Xq + 73YQ) = a) 971 AH fy g(X 4X4) z) + determined quantities. Now App #0 as sy #0 F 8 4: Moreover 047, as 0 = {Y 16¥_16°V2) = 73fi6(¥16¥-16%0) + 1fi16(%16¥-162) 29d fy igl¥ie¥-167) # 9 # f16(Y16¥-16°%0): So a1 8 is determined as fy g(X4>%_4> z) #0. Finally we use that 0 = f(V9,V9,2) = a7 f,(2,2,2) + determined quantities to determine ay: So aj, is determined as f,(z,z,z) # 0 and the proof is complete. Proposition 14.12: Let Fo be a subfield of F. Then G contains a unique conjugacy class of subgroups isomorphic to F4(FQ). Proof: Let L~ F4(Fo). Then it is shown in [As5] that L stabilizes a unique Eg - form so the claim follows from lemma 14.4. We now want to establish the uniqueness of the PSLo(F), char(F) = 13, described in theorem 14.2. Our methods will be similar to those used in the proof of proposition 14.11 which were previously used in [As5] in a similar situation. For the remainder of 104 this section we will assume that char(F) = 13, [F| # 13, L ~ PSLO(F), and J ~ M(0) © M(8A,)\ M(3A,)@M(A,)°\ M(8A,) ~ M(0) @ W(16A,)\M(8A,)_ where W(16A,) denotes the Weyl module of highest weight 16A,. This last isomorphism is in [As6]. Now we establish some notation. Let Y = Soc(id®@), W = W(16A,)/ Y and Z= id6/W(16A,). We fix a Cartan subgroup H, of L. Then let { y; } be a basis of weight vectors of Y, {w;: i= 16,14,12,10,-10,-12,-14,-16} be a basis of weight vectors spanning W and {z;} be a basis of weight vectors spanning Z. By S; we will denote the weight space of weight i. For aé€ F and be F%, let Wb € Endpy (Y\W\Z) be the element defined by Yi%ab = by; Wis, = bw, » Bay = bz. + ay;. Let fy € Sym3(Y\W\Z) be such that f; #0 on Y. We define recursively fray i= fie - f,. Similarly, let g, € L(M(0),Y\W\Z,Y\W\Z) such that g,(m,w,v) # 0 for some mé€M(0), w, v € W, and let S47 = 81 Be Lemma 14.13: . Sym3(Y\W\Z) is spanned by {f, ,fo,f3,£4}- pay i) . {o. 5 } is transitive on <f,,fg> \ <fy> modulo <fg,f4>. ~ Lp (M(0), Y\W\Z,Y\W\Z) is spanned by {&189}- 4, f= f+ agfs + a,t, + g, + boo + ch, where h spans Sym?(M(0)), where a; ,b;,c EF. Proof: We prove part 3 first. Observe that 8} and &) are linearly independent. Next we observe that Ly, (M(0), Y\W\Z,Y\W\Z) ~ Homp, (Y\W\Z,Y\W\Z) which is clearly two dimensional, and part 3 follows. Part 2 is easy to see. Given parts 1,2,3 we can see part 4 as follows. First, Sym®(M(0)@Y\W\Z) is spanned by {f,,8,h}. If f is an Eg form, then fly# 0 otherwise fly\w =0 and (Y\W)9 > 105 Y\W and so by lemma 4.5 s#2(y\w) < Yand Y\W © M(0) is a subalgebra. But then by the subalgebra theorem Y\W/\Z is not indecomposable contradicting our choice of L. So we may assume f= fy + other terms. By part 2 we may also set ag = 0. In view of the fact that f,,(-,-) is nondegenerate and that rad(o iq) #0, we may further assume that f = f}+ &, + other terms. So part 4 follows. Now we prove part 1. First we want to see that {f, ,f5,f5,£4} are linearly independent. To see this we make the following easy to check observations: a. f,(w,v,u) = 0 for all wiv,u € W(16\,), but f,(z,w,w)#40 for some zéZ and we W(16,). b. fg(x,w,w) = 0 for all wEW(16\,) and x € Y\W\Z, but Rad(fg) # 0. c. Rad(f,) = W(16))). To complete the proof of part 1 we need to check that dim(Sym?(Y\W\Z)) < 4. First we want to observe that dim(Sym3(Y\W)) = dim(Hompy (Y\W.Sym2(Y\W))) = 1. We can see this using 13.1.4 to compute Sym2(Y\W) ~ Sym2(Y)\Y @W\Sym2(W). We can then compute Sym2(W) using formula 13.1.5 and then observe that Hom(Y\W,Sym2(W)) = 0. Furthermore we can compute that Hom(Y\W,Y@W) = 0 and that dim(Hom(Y\W,Sym2(Y))) = 1 establish our observation. So now it suffices to show that the space of trilinear forms J which vanish on Y\W is three dimensional. We consider the FL-map D: J + Homp,(Z, (Y\W)@(Y\W)) defined by D(g)(z) = g(z,-,-). Let % = Ker(D), then Dim(T/%) < 1 = dim(Homp, (Z, (Y\W)@(Y\W))). Now define E: % + Hompy, (Z, (Y\W)@Z) by E(g)(z) = g(z,_,_). As dim(Homp, (Z, (Y\W)@Z)) = 1 dim(%/Ker(E)) < 1. Finally we observe that Ker(E) ~ Hompy, (Z,Z@Z), which is one dimensional, as can be computed using lemma 13.1. The claim follows. 106 As in the case char(F) > 17, we can work over an algebraically closed field. For let F denote the algebraic closure of F. Then the fg, and h are PSLo(F) invariant maps. Lemma 14.14: The following are true: 1. @S; i=0 mod 8 is a subalgebra of type T9- i 2. @S- i=0 mod 6 is a subalgebra of type V9: i 3. @ S; i = 0 mod 4 is a subalgebra of type V 15: i Proof: The proof of lemma 14.8 carries over verbatim. Lemma 14.15: Let {V1,V9,V3} be three pairwise orthogonal primitive idempotents spanning Sy . Then we can find 8,3, € S; such that Wi = vg nN Vo = < W165W_ 16984 98_498g8_g98958_9> Wo = van vga = < Wy9W_1998g:8_g¥19_195 98 _9> W3 = v,4 ia) v34 =< 5455 _4:5 955 _955 653 _6:¥149¥_14>- Moreover the s:,8; are singular. Proof: The proof of lemma 14.9 carries over verbatim. Lemma 14.16: The following quantities are not equal to zero: £3(Wy65W_y9¥_4), £10 16>¥_10°¥-6)> £10 19).127.4)> £1(W 19.1976)» f3(24:24¥_g)> f4(24:24:2.g)> £(id,z9,z9), h(id,id,id). Proof: The proof is similar to that of lemma 14.10. The idea is to show that if one of the quantities is zero, then the form f, resp. (g;,h) is trivial, contradicting the definition of f; resp. (g;,h). 107 Proposition 14.17: Let char(F) = 13, |F| # 13. Then G contains a unique conjugacy class of subgroups’ isomorphic to PSLo(F) and idO ~ M(8A,)\M(3A,)@M(A,)°\M(8A}) as an FPSLo(F)- module. Proof: The proof is analogous to that of proposition 14.11. First we observe that <s.g> = Ker(fy(wigw 19g .)- So as fy(wi6¥19-9.6) #9 # £016. 107.6) we can write sg = Y¢ + @ 2, where a = - f)(Wy¢g,W_iqy.g) / (1619.26): For the same reasons we can write 84=Y4+ B24. As s.¢ is singular by lemma 14.15 we have: 0o= f(84,84,Y_g) = Bragts(24,24,9_g) + determined quantities. So as £5(24,24,Y_g) %0 we have determined ag. Also we have: 0O= f(84584,2_g) = Bragt, (2424529) + determined quantities. So as £4 (24,24 .2_g) # 0 we have determined a4. Now v; = aid + BYg + 729 where a,f,y are expressible in determined quantities. We choose i such that 740 4a. Then as Vi is singular we have: 0 = f(v;,v;,29) = 2 aybo8o(id,z9,29) + determined quantities. Also as &o(id,29,29) #0 we have determined bo. Finally, we use 0 = f(vj,vj,id) = cah(id,id,id) + determined quantities to determine c. This.completes our proof. 108 CHAPTER III: CROSS CHARACTERISTICALLY EMBEDDED SIMPLE SUBGROUPS, ALTERNATING AND SPORADIC SIMPLE SUBGROUPS In this chapter let L denote a finite simple group. We will assume that the defining characteristic of the group L is not equal to char(F) = p. For the rest of this chapter F will denote the algebraic closure of F. SECTION 15: INITIAL REDUCTIONS Proposition 15.1: The structure of the Weyl group of F4(F) is 2° ( Alt,@Alt,) .2 : 2. Proof: This is well known. One way of seeing this is as follows. The Weyl group of Eg(F) is O¢(2). The Weyl group of F,(F) is the stabilizer of a singular line in the building of the Wey] group of E¢(F), the structure of which is well known. Lemma 15.2: The orders of elements of the Wey! group of F4(F) are : 1,2,3,4,6,8,12. Proof: Well known. Lemma 15.3: If L< G and_ L is of Lie type of characteristic r #4 2,3,p, then L ~ PSLo(q) and q€{5,7,13,17,25}. Proof: From lemma 7.8 we know that the Sylow - r- subgroup of G is abelian. So L ~ PSLo(q) as all other simple groups of Lie type of char = r have nonabelian Sylow- r-subgroups. Now the Borel group of PSLo5(q) is a Frobenius group of order q(q-1)/2. Again by lemma 7.8 the Frobenius complement has to be a section of the Weyl group of G. So (q-1)/2 has to be the order of an element of the Weyl group of G. The claim now 109 follows from lemma 15.2. Lemma 15.4: If L < G_ is of Lie type of characteristic 3, then L ~ PSLo(9), PSL,(27), PSL4(3), PSU3(3), or 7Go(3). Proof: Recall from lemma 7.10.3 that a Sylow-3-group R of G contains an abelian normal subgroup A, such that R/A is elementary abelian of order 9, and that m3(G) < 4. This restricts the type of L to one of the following: PSLo(q) and q< 81, PSU3(q) and q €{3,9}, PSL3(q) 4 € {3,9}, PQ,(3) ~ PSp4(3), 27Go(3), Go(3) and PSU4(3) = PN%(3). Now the argument in lemma 11.1 shows that PQ,(3) and P&(3) contain a subgroup isomorphic to 94, Alts ; so lemma 7.11.4 eliminates these two possibilities. The Borel subgroup of PSL5(81) is Frobenius of order 81:40. By 7.10.1 the group 3* is contained in a Cartan subgroup of GF. So by 7.10.2 Ag(34) is a section of the Wey! group of G. Now 15.2 eliminates PSL» (81) as a possibility as 40 does not divide the order of the Weyl group of G. The group PSLo(9) contains a subgroup 3*:PSLo(9). But by lemma 7.10 34 is contained in a Cartan subgroup of G and AG(3*) is a section of the Weyl group of G. As PSLo(9) contains elements of order 5 lemma 15.3 eliminates PSL4(9) as a possiblity. The group U3(9) has no faithful F-representation of degree less than 72 (see [Lal]) so it can’t embed into G. The only small faithful irreducible F-representations of Go(3) have degree 14 (see Atlas and Parkers Character Tables), so Go(3) is out by lemma 10.17. Lemma 15.5: If L < G is of Lie type of characteristic 2, then L is isomorphic to one of the following: PSLo(q) q < 8, PSp4(2), PSL3(2), Go(2)', 3D,(2) or PSL,(2). Proof: Recall from lemma 7.10 that mo(G) = 5, so in general q< 32. Now recall 110 from lemma 7.10 that Ag(E) is a 2,3,7 - group when E is an elementary abelian 2 group. Soif L has Lie rank 1, then q < 8. So L ~ PSLo(4), PSLo(8), PSU3(4), PSU4(8), or Suz(8). Now the small nontrivial irreducible F-representations of Suz(8) have degree >14 and hence 10.17 disposes of this case. Now by [Lal] we see that the smallest nontrivial irreducible F-representation of PSU3(8) has degree at least 56; so this case is out. The case PSU3(4) is out because the trace of an element of order 3 in PSU.s(4) on any 26 dimensional F-representation is either 2 or 14; a contradiction to lemma 12.1 (see Atlas and Parkers character tables ). Now suppose that the Lie rank of L is 2. Then L contains subgroups of Lie rank 1 so in general q < 8. As mo(G) = 5 this leaves only the possibilities listed in the statement of the lemma and the groups PSL3(4), PSU4(2), PSU5(2), and 2F 4(2)!. The last four possibilities are out by lemma 7.11.4 as the groups contain subgroups isomorphic to 24:PSL5(4) ) 2*:PSL5(4) , 3x PSU,(2) and 24; 5 respectively. Now if L has Lie rank greater than or equal to 3, then as L contains a subgroup of Lie rank 2, we see q=2. Now as mo(G) = 5 the only possibility is L ~ PSL4(2). Now the claim is established. The following result can also be found in [K12]. Proposition 15.6: Let L be a sporadic finite simple group. Then L is a subgroup of G iff p=11 and L~ J, or Mi: Proof: It is well known that PSL3(4) < Mo9:Mo3.Moq4,McL,HS,Co.3,Co.2,Co.1. So these groups are out by lemma 15.5. It is also known (see Atlas) that the 3-ranks of the following groups, Suz, Ly, Th are at least 5. It is also known (see Atlas) that the 2-ranks of the groups J4, He, Figo, Figg, Fig4, Ru, HN, B, M_ sare at least 6. So 111 these groups are out by lemmas 7.10 and 7.11. The group Jz contains a subgroup PSL.(16), so it is out by proposition 15.5. The group O’N has subgroups PSL(11) and PSL2(31). So O’N is out by corollary 15.3. The small nontrivial irreducible FJ» -modules have dimensions 14 and 21 (see Atlas and Parkers tables). So J. is out by lemma 10.17. The groups Jy, Mi and M9 contain PSLo(11), so by lemma 16.3 these groups can embed into G only if p= 11. Now J, was originally constructed as a subgroup of Go(11) (see [Jal]). The centralizer in G of a subalgebra of type Ug is Go(F), by subalgebra theorem, which certainly contains Go(p). So J, is a subgroup of G iff p=11. Now consider M,,. From the modular character table we see that Mj, is a subgroup of Qg(11). Moreover because M,, has trivial Schur multiplier, we see that M11 is also a subgroup of Sping(11). Thus Mi is a subgroup of G, because it is a subgroup of the stabilizer of a primitive idempotent. Recall from proposition 2.7.5 that the stabilizer of a primitive idempotent is Sping(F). So Mj, isa subgroup of G iff p=11. Now it remains to show that Myo is not a subgroup of G when p=11. To see this we observe that the small nontrivial 11-modular irreducibles of M19 have dimensions 11 and 16, and that M19 contains an involution whose trace is 3 resp 0 on these irreducibles. So we see that this involution has trace 10 on any nontrivial 26- dimensional FL-representation. So Mj 9 is out by lemma 12.1. Proposition 15.7: Let L ~ Altn, n> 7. Then L is not a subgroup of G unless n=7 and p=5. Moreover if p=5 dim(C,(Alt7)) = 3 and hence Ny(Alt7) is never maximal in TP. Proof: We will show first that when p35, then Alt is not a subgroup of G. Then we will show that when p = 5 Alt is a subgroup of G_ only if Altz 112 centralizes a three dimensional subalgebra. Finally, we will show that when p=5 Altg is not a subgroup of G , completing the proof. The following part of the Altz ordinary character table is an excerpt from the Atlas: Xy XQ x3 X4 Xn og Xz -Xg 1A 1 6 10 10 14 14 15 21 2A 1 2 -2 -2 2 2 -1 1 3A 1 3 1 1 2 -1 3 -3 3B 1 0 1 1 -1 2 0 0 For now we will assume that p# 5,7. The character y of any 26 dimensional AltoF - module is a sum of irreducible characters. Now we will survey the possible y and observe that either (2A) ¢ {-6,2} or y(3A), x(3B) ¢ {-1,8}, showing by lemma 12.1 that the module affording y is not a G module. if x= x gt Oxy, X7t Xi + Xy 1 = 3,4, x7 + XQ + 5Xqs X{F2Xy 1=5,6 , Xj + Xq + 6x, i=5,6, x; +2x9+4x, J=3,4, xj+ Xqt10x, j=3,4, Xj + 16x, j = 3,4, or axgtbx,| Ga+b =26, then y(2A) ¢ {-6,2} Ify = xi + Xj+2X4 i=5,6 j=3,4 , Xj + x, + XQ j,k = 3,4, then y(3A) or x(3B) ¢ {-1,8}. If XGtXj, F6Xy and j,k € {3,4}, then Alt, must stabilize a point. This is impossible because point centralizers do not allow the proposed Alt series as a refinement. Now assume p=7. The following table is an excerpt from the 7-modular character table (see Parker): 113 Xy XQ XZ X4 X5 =X6 1A 1 5 10 14 14 21 2A 1 1 -2 2 2 1 3A 1 2 1 2 -1 -3 3B 1 -1 1 -1 2 0 The only 26 dimensional characters that are feasible by Lemma 12.1 are: Xgtxo and 2xg+ 6x. In the first case observe that idO is the direct sum of the irreducibles affording ygand Xq and that dim(Sym?(xo)) = 15. So s#(y.) < <id , ¥9> and hence <id,x9> is a proper nontrivial subalgebra of J. Now <id,xo> is nondegenerate and hence it contains a primitive idempotent x by Springer’s lemma. If E,(x) al <id,x9> = 0 then lemma 5.5 applies forcing an Altz invariant subalgebra <id,x> < <id,x9>: a contradiction. If E,(x)N<id,x9> # 0 then <id,y9> must be a member of Ug by lemmas 5.9.5 and 5.12.2. In this case we see from lemma 5.13 Altz < NQ(<id,x9>)/Cg(<id,x9>) = SO,(F): a contradiction as Alt7 has no nontrivial three dimensional 7-modular representation. In the other case regard J as an Altg module. As 7 does not divide the order of Altg, J must be semisimple when regarded as an Altg module and hence dim(C ,(Alt¢) 27. The trivial Alt, module induced up to Altz is of shape X4/Xo/Xx4- So when x € CiyQ(Altg), then <x Alt7> isa homomorphic image of X4/Xo/X4- So <x Alt7> =X, as we assumed that idO = 2xX3+6x, as an Alt module. So we see that C glAltz) is a seven dimensional nondegenerate subalgebra of J. So by Springer’s Lemma Altz < C.,(x) for some primitive idempotent x € C ,(Alt7). Now C,(x) has only one composition factor of dimension greater than 10. We assumed that Alt has two ten dimensional composition factors: a contradiction. 114 Now assume that p=5. We will show first that Altg is not a subgroup of G. To see this we observe that Altg contains an involution whose trace on every small module is at least 1 (see the 5-modular character table). Now in all but one case a 26 dimensional Altg representation has more than 2 composition factors. In this case the trace of this involution is more than 2, and hence the claim follows from lemma 12.1. The remaining case is the case 13/13, where the trace on the other involution is 10; again lemma 12.1 yields the claim. It remains to show that in case p=5, Alt? must centralize a subalgebra of dimension 3. Here is an excerpt from the 5-modular character table of Altz: xq x2 X3 X4 X5 X6 X7 1A 1 6 8 10 10 13 15 2A 1 2 0 -2 -2 1 -1 3A 1 3 -1 1 1 -2 3 3B 1 0 -1 1 1 1 0 As above, the only 26 dimensional representation not contradicting lemma 12.1 has character 2x, + 3x3. Now we restrict this character to PSLo(7). We observe that CgQ(PSL2(7)) is at least two dimensional as 5 does not divide the order of PSLo(7). An easy calculation using the Atlas and Parkers character tables shows that the trivial PSLo(7) module induced up to Altz affords the character 2x,+ xg. Thus if id@ affords the character 2x,;+3xg as an Alt7 module, then dim(C; 49(Alt7)) = 2 as for any x € C4@(PSLo(7)) <xAlt7> is a homomorphic image of 2x;+Xx¢ and hence a trivial Altz module. The proposition is finally proved. 115 SECTION 16: ESTABLISHING NONMAXIMALITY OF CERTAIN SIMPLE SUBGROUPS OF G The results of the previous section leave the following groups for our consideration: List 16.1 Alts = PSLo(5) = PSLo(4) , Altg = PSp4(2) ~ PSLo(9) , PSL9(7) ~ PSL3(2) ~ PSU3(2)' , PSLo(8) = 7Go(3)', PSLo(q) q €{13,17,25,27} PSL(3), PSU3(3) = Go(2)', 3D,(2), J, when p=11,My, when p= 11 Remark: From now on we will denote the ordinary irreducible characters of our simple groups as in the Atlas [Con}]. Lemma 16.2: Let p=11. Then Ny(J,) is never maximal in I. Proof: From the 11-modular character table and lemma 12.1 we can see that the J,- module id@ has three composition factors of dimension 7 and five composition factors of dimension 1. By lemma 7.1 it is enough to show that J, centralizes a point of id. Assume for now that this is not the case. Then let A be a seven dimensional J,- submodule of idO. Using the fact that J; contains PSLo(11), we will show in a moment that the symmetric square of A is <1> @ B_ where B is the unique 27 dimensional 11-modular irreducible of Jj. So from lemma 10.1.4 we observe that dim(S*2(A)) <1. If s¥2(A) = 0, then by lemma 3.4.2 A is singular. This is impossible because the maximal singular subspaces of J are five and six dimensional. If s#2(a) < <id>, then lemma 10.4 applies and hence A is singular: a contradiction. 116 So we conclude that s¥(4) is a point of J distinct from <id>. Now the claim follows as we have shown that CigQ(I DF 0. Now we will show that the symmetric square of the 11-modular irreducible is as claimed. We establish first the 11-modular irreducible is M(6A,) as a PSLo(11) module. To see this we note that the character value an element of order 6 on the 7 dimensional J, module is -1. The character value of the element of order 6 on the PSLo(11) modules M(2A,) and M(4A,) is 2 and 1 respectively, leaving M(6A,) as the only possible choice. Now recall from chapter II that Sym(M(6.4)) = M(81)/M(2,)®M(0)/M(8\,) © M(4\;) © M(0) as a PSLo(11) module. Now Sym?(M(6A,)) as a J, module can have only 1,7,14 or 27 dimensional composition factors. The trace of an involution respectively an element of order 3 on Sym?(M(6A,)) is 4 respectively 1. Now it is easy to check using the excerpt of the 11 modular character table of Jy given below, that the traces can only add up right when the composition factors are as claimed. To see that the module splits we recall that J, is a subgroup of Go(11) and that the symmetric square of the 7 dimensional Go(11) module is M(2A,) @ M(0). Here is the excerpt of the 11-modular character table: 1A 2A 3A 1 1 1 7 -1 1 14 = -2 -1 27 3 0 117 Lemma 16.3: Let p=11 . Then Ny(M,,) is contained in the stabilizer in [ of a primitive idempotent. Proof: By lemma 12.1 the only feasible 26 dimensional characters for M1, are the ones whose composition series have composition factors of degrees 1,16,9 or 19 102, not necessarily in that order. The first case leads to semisimple module as id® is a self dual module. So in this case Mi centralizes a nondegenerate two dimensional subalgebra. So in this case, by lemma 5.1, M1, embeds into the stabilizer of a primitive idempotent. In the second case we regard id@ as an Mj module to see that dim(C; 19(Mj9)) = 6 as idO is a semisimple Mio module. Now the trivial Mig module induces up to 1\9\1 as an M,, module mod 11. So as in the Alt, case we see that dim(C; 49(M1,)) = 6 and C y(M14) is a nondegenerate subalgebra containing primitive idempotents. Now C 4(M 41) is not isomorphic to some Jp for reasons of dimension. So by proposition 5.9 and lemma 5.5 M,, must centralize a primitive idempotent. But then J can’t have two ten dimensional composition factors as an Mi module see prop. 2.7.3. This completes the proof. Lemma 16.4: Np(Alt,) is never maximal in I. Proof: We observe that the case p=5 was handeled in section 13 as Alts ~ PSLo(5). So p #5 and idO is a_ semisimple Alt; module. So if Ny(Alts) is maximal then idQ has no trivial composition factors. Now if idO has an irreducible three dimensional Alt. submodule U then, by lemma 10.8, U is brilliant. If U contains singular points then lemma 10.15 applies. If U does not contain singular points but s#2(y) is brilliant, then as s#2(u) contains singular points, lemma 10.15 applies. So U satisfies the hypothesis of lemma 10.7 and hence s¥(u) is a member of Ug and Alt, centralizes a point of st (U). If this point is not id, we are done because 118 then Alte centralizes a point of id®. If the point centralized is id, then by lemma 5.10, s¥2(u) is a subalgebra and proposition 10.16 applies. So now we assume that id® is the direct sum of four and five dimensional Alt, modules. Thus id® has to be the sum of four four dimensional and two five dimensional Alt, modules. This contradicts lemma 12.1 as the trace of the element of order 3 is 4x1 + 2x(-1) = 2 ¢ {-1,8}. This completes the proof. Lemma 16.5: When p#7, then Np(PSLo(7)) is never maximal in I. Proof: As we assume p#7 id® is a semisimple PSLo(7) module, so if Np(PSLo(7)) is to be maximal, then Ci 4@(PSLo(7)) = 0. Checking the character table of PSL.(7) we see that the only 26 dimensional representations which do not violate lemma 12.1 and have no trivial submodules are afforded by characters of the form axgtbxg + X6 where a+b = 6, deg(x9)= deg(x3) = 3 and deg(xg) = 8. Now a simple calculation using the PSL5(7) character table shows that Sym?(x5)=Sym2(x3)= X4> the unique irreducible character of degree 6. This shows that when idO affords a character of the form axo+bx3 + Xg, then s#2(y) = 0 for every three dimensional submodule U. So U is singular by lemma 4.8 and hence PSLo(7) stabilizes a subalgebra. So then N,(PSL2(7)) stabilizes a subalgebra by corollary 10.16. Lemma 16.6: Np(PSU3(3)) is not maximal in I when p#7. When p=7 L has an irreducible 26 dimensional representation, and this representation of L is realized in G via the Go(7) described in section 14. Proof: Let L ~ PSU3(3). First we treat the case when p # 7. In this case every L - module is semisimple. If L centralizes a point, the statement clearly holds. So then the only character which does not violate lemma 12.1 and contains no trivial characters is X9 + Xq + Xg (in Atlas notation) where deg(x.) = 6 and deg(xg) = 14. We 119 compute easily using the character table of PSU3(3) that Sym?(x9) = x7 (in Atlas notation) and deg(x7) = 21. So, by lemma 10.1.4, s¥(a) = 0, where A is a submodule affording x9. So A_ is singular by lemma 3.4.2. So <id,A> is a PSU3(3) invariant subalgebra and we are done by corollary 10.16. Now we consider the case p=7: The Borel group B of PSU,(3) has order coprime to 7. The trivial B module induced up to PSU3(3) mod 7 is of the form 1\26\1. Therefore if id@ affords a PSU9(3) character with trivial constituents, then C49 (B) 0, and hence C:4Q@(PSU3(3)) # 0. So then the arguments above go through in case PSU.(3) does not act irreducibly on idO, as the character y7 stays irreducible when reduced mod 7. We recall from [As7] that Aut(PSU3(3)) is an irreducible subgroup of Go(7) acting on its seven dimensional irreducible module. Now PSU3(3) contains a PSLo(7) such that the 7 modular seven dimensional irreducible is M(6A,) as a PSL.(7) module (this can be seen from the ordinary and modular character tables of PSL»(7) and PSU,(3)). Now from lemma 13.1.3 we know that: Sym?(M(6A,)) = M(6\,) © M(0)/M(4.,)/M(0) © M(4A,)/(M(2A,)®M(0))/M(4))) (is the sum of two indecomposables and an irreducible). So now we can see with the aid of the 7-modular character table that Sym?(M(6,)) as a PSU.(3) module is of the form 1/26/1. Now as a Go(7) module Sym?(M(6A,)) is also of the form 1\26\1. So we conclude that the 26 dimensional 7 modular irreducible of Go(7) stays irreducible upon restriction to Aut(PSU,(3)). So Aut(PSU,(3)) embeds irreducibly into F4(7) via the Go(7) of section 14. Lemma 16.7: Let L ~ PSL2(3). Then L is an irreducible subgroup of G. All embeddings of L are equivalent, in GLo7(F). Proof: The irreducible embedding of L can be seen in the stabilizer of a 27- or twisted 120 27-decomposition of J. The usual argument using the character tables and lemma 12.1 shows the second part. Lemma 16.8: Let L ~ 3D,(2) . If L<G then L acts irreducibly on idO. Proof: By [Lal] L has no representations of dimension <13. So the claim follows from 10.17. Lemma 16.9: Let L ~ PSL5(25). Then Ny(L) is maximal only if idO affords Xa Moreover G contains such a subgroup when F is a splitting field for x? +1= 0. Proof: The usual argument using lemma 12.1 shows that Ny(L) is maximal only if id® affords X11 OF X43: Now the Borel subgroup of L is of the form 52: 12. By lemma 7.8 the element of order 12 must be conjugate to an element in the Weyl group of F, and hence also to an element in the Weyl group of Eg. Now the Weyl group of Eg acts as 1+ et + 20- on J. So from the atlas [Con1] we see that the element of order 12 of the Borel subgroup of L_ has to have trace -1 or 2 on idO. But X13(12a) = X43(12b) = 1 eliminating x13 and leaving only x,, as a possibility. Now 2F (2)! is an irreducible subgroup of E,(F) iff F is a splitting field for x24 1. Now. it is known that L is a subgroup of 2F,(2)'. Now the claim follows from character restriction. Lemma 16.10: Let L ~ PSLo(27). Then N,(L) is maximal only if idO affords x; i € {6,7,8}. Proof: Clear from 12.1 and [Con1]. 121 Lemma 16.11: Let L ~ PSLo(8). Then Ny(L) is maximal in [ only if p # 7 and id® affords the L-module character Xg + Xitx; where ijé{7,8,9}. Proof: The usual argument when p # 7. Now let p=7 and let B ~ 93.7 be the Borel subgroup of L. Then recall form the proof of lemma 7.11.4 that the 2° is unique up to conjugacy in G and that C (23) is a member of Ug: Now as observed in [As5] 19.6 the trivial 23 module induced up to L is ofthe form 1\8\i. So in this case either Ciqe@(L) #0 or id~ 8 @ 8\1 @ 8\1. In the last case let U = 8@8@8. Then U+ isa three dimensional submodule of J which must be trivial as an L module as L contains no nontrivial representations over F of dimension < 7. Lemma 16.12: Let L = PSLa(13). Then Ny(L) is maximal only if p # 7 and idO affords the character Xg t+ % i€{4,5,6} or p= 7 and id® affords 12 + 14g. When p=7 and 13 is a square in F, then the embedding of L via the irreducible Go(F) affords 12 + 14a. Proof: The usual argument. Lemma 16.13: Let L ~ PSL9(17). Then N,(L) is maximal only if idO affords the character x7 + x; i€ {2,3}. Proof: The usual argument. Lemma 16.14: Let L = Alte. Then Ny(L) is a maximal subgroup of [ only if p > 5 and id@ affords the character 2xg + x; ie {4,5}. Proof: When p> 5 the usual argument using 12.1, Crg@(L) = 0 and [Con]1] gives the claim. So let p = 5 then using lemma 12.1 we find that the degrees of the composition factors of id@ must be 102, 18 or 8° 12. In the first case we let M 122 be the normalizer of a Sylow 3 subgroup of L; in this case 5 does not divide |M| and so dim(C;49(M)) > 6. Then the character of the trivial representation of M induced up to Lin characteristic 5 is 1\8\1. So if L has composition factors 107,19, then dim(C;49(L)) = 6,and L stabilizes a subalgebra. Now we assume the other case. First we observe that idO must be uniserial 8\1\8\1\8 otherwise id® has a submodule of dimension 10. As dim(H1(L,8)) = 1 any such ten dimensional submodule contains a trivial submodule and hence so does id®. On the other hand as dim(C; 49(M)) 2 2 we have that either C,y9(L) # 0 or id® contains two distinct submodules U, W ~ 8\1. This contradicts uniseriality and the claim follows. SECTION 17: PROOF OF THE SIMPLE SUBGROUP THEOREM In the last section we restricted the isomorphism type of a simple subgroup giving rise to a maximal subgroup of I to the following: PSLo(q) q€{8,9,13,17,25,27}, PSU4(3), PSL4(3), °D4(2) Moreover, with the exception 3D, (2), we have identified the cross-characteristic embeddings of L into G which could make Ny(L) into a maximal subgroup of [; recall also that we dealt with the non cross-characteristic case in chapter II. However, in the cases 3D,(2), PSLo(q) a € {8,9,13,17,27} we did not decide if such embeddings of L into G_ existed. The next few lemmas deal with the existence question. Lemma 17.1: If p # 7, 13, then PSL9(13) does not embed into G via a character specified in lemma 16.12. Proof: See [Co4]. 123 Recall that G = F,(F) where F is an algebraically closed field. Lemma 17.2: If p #17, then PSLo(17) embeds into G via the character XgtXxg: Proof: See [Co4]. Lemma 17.3: If p # 7, then PSLo(8) embeds into G via the character XgtXx7+Xg- Proof: See [Co4]. Lemma 17.4: Let p # 7, then PSL.(27) embeds into G via the character X7: Proof: See [Co4]. Lemma 17.5: 3D, (2) embeds into G_ via the character Xg: The embedding is unique up to G conjugacy when p # 7. Proof: See [Co4] and [Col]. Now we sumarize the results. Let M_ be a maximal subgroup of [ such that M MG _ is a closed proper subgroup of G. If F*(MNG) is not a simple nonabelian group, then the Structure theorem of Section 9 asserts that M_ either stabilizes a good subalgebra, a 27- decomposition, or a twisted 27-decomposition. Moreover, we know the structure of the stabilizer of each. Now suppose that F*(MNMG) is a simple nonabelian group. Then the results of section 10 show that F*(MNG) does not stabilize a subalgebra. If F*(MNG) is of Lie type of characteristic p, then the results of chapter II restrict F*(MNMG) to one of { Go(F) p = 7, PSLo(F) p > 13 or |F| = 7, F4(Fg) Fo < F } and moreover 124 none of these groups stabilizes a good subalgebra. If F*(MNG) is not of Lie type of characteristic p, then the results of Sections 15 and 16 restrict the type one of PSL,(q) q€{8,9,13,17,25,27}, PSU3(3), PSL3(3), 9D4(2). With the exception of PSL5(9) we have exhibited suitable embeddings into G in section 16 and in this section. However, the only known embeddings for PSL3(3), PSU3(3) p=7, PSLo(7) p=7, and PSL5(13) p=7 are contained in proper subgroups of G. So we have proved the necessary conditions for maximality as stated in the simple subgroup theorem. When do we have sufficient conditions for maximality? We mention here, without proof, the following facts: The stabilizers of good subalgebras are maximal. If F is finite the stabilizers of 27- or twisted 27-decompositions are maximal iff F is a prime field. If F is not a prime field then the stabilizers are contained in a conjugate of F4(Fq) , where Fo is the prime field of F. If F is finite F4(F,) is maximal iff [F:F,] is a prime. If F is finite and p # 7, then 3D,(2) is maximal iff F is a prime field. Similar conditions will hold for PSLj(q) q € {8,9,17,25,27}. References: [Al] Albert A.A.: A theory of power associative commutative algebras ; Trans. Amer. Math. Soc. 69 (1950) 503-527. 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