OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Xxxx XXXX, Pages 000 000 S 0002-9947XX)0000-0 OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II BEN FORD ABSTRACT. Determining the subgroup structure of algebraic groups over an algebraically closed field K of arbitrary characteristic) often requires an understanding of those instances when a group Y and a closed subgroup G both act irreducibly on some module V, which is rational for G and Y. In this paper and [4], we give a classification of all such triples G Y V when G is a non-connected algebraic group with simple identity component X, V is an irreducible G-module with restricted X-high weights), and Y is a simple algebraic group of classical type over K sitting strictly between X and SL V. CONTENTS. Introduction 2. Q X -Levels and Embeddings of Parabolics 3 3. THE CASE X A m 5 4. THE CASE D m 20 5. THE CASE X E 6 32 6. THE CASE X D 4, G : X 3 34 References 42. INTRODUCTION E. B. Dynkin in 957 [3, 2] classified the maximal closed connected subgroups of simple algebraic groups when the underlying algebraically closed) field has characteristic 0. Seitz [9, 0]) and Testerman [3]) completed the same program in positive characteristic in the 980 s. Their analyses for the classical group cases were based primarily on a striking result: If G is a simple algebraic group and ϕ : G SL V is a tensor indecomposable irreducible rational representation, then with specified exceptions the image of G is maximal among closed connected subgroups of one of the classical groups SL V, Sp V, or SO V. What is most striking is the brevity of the list of exceptions. From a slightly different perspective, the question these authors answered was: Given a closed, connected subgroup G of SL V for some vector space V, with G acting irreducibly on V, find all possibilities for closed, connected overgroups Y of G in SL V. This question of irreducible overgroups appears in other contexts as well, sometimes for non-connected subgroups. Here and in [4] we present results for some such non-connected subgroups; namely, those with simple identity components. The overall program is to classify all possible triples GY V with G and Y both closed subgroups of SL V acting 99 Mathematics Subject Classification. 20G05. Supported in part by the NSF and the NSA. c997 American Mathematical Society

2 BEN FORD irreducibly on V, G Aut Y, Y SL V SO V, or Sp V, and Y a simple group of classical type the corresponding question for Y of exceptional type is also open). We give complete results for the case when G is not connected but has simple identity component X, and the T Y -high weight and T X -high weights of V are restricted. Specifically, the papers are concerned with the proof of Theorem. Theorem. Let G be a non-connected algebraic group, over a field K of arbitrary characteristic p 0, with simple identity component X. Let V be an irreducible KG-module with restricted X-high weights). Let Y be a simple algebraic group of classical type such that X Y SL V and G Aut Y. Then V Y is irreducible with restricted high weight if and only if Y SO V, Y Sp V, or X Y V appears in Table page 4) or Table 2 page 42). If G has simple identity component X, then G Aut X. Since we require that G X, we therefore may restrict our attention to X of type A m, D m, or E 6. We assume henceforth that Y is simply connected, and that X and Y act on W, the natural module for Y. The analysis is different depending on whether X acts reducibly or irreducibly on W. We settled the reducible case in [4], and we consider the irreducible case here. Also, we will assume here that the involutory graph automorphism of X, if it is in G, also acts on W though it need not be in Y ), as we dealt with the case when it does not act on W in the final section of [4]. If V X is irreducible, then we are in the case examined by Seitz in [9], with the additional condition that X have an outer automorphism which acts on V. We examine Table of that paper, and find that we have such a situation in the examples there labelled I 4, I 5, I 6 for n 3, II, S, S 8 in S 8 we could take G X t, G X s, or G X s t, where t s are outer automorphisms of X of order 2 and 3 respectively), and MR 4. These examples are collected in Table, and henceforth we shall assume that V X is reducible... Notation and Conventions. All structures are assumed to be constructed over the same algebraically closed field K, of characteristic p 0. Throughout, X will denote a simple algebraic group over K admitting an outer automorphism so X is of type A m, D m, or E 6 ). A fixed standard graph automorphism of order 2 will be denoted by t, and if X has an outer automorphism of order 3 i.e. if X D 4 ), we will fix one and denote it by s. Thus G is X t except possibly when X D 4, in which case we also consider G X s and G X s t. For any reductive group H we consider, with fixed maximal torus T, Σ H will denote the roots of H relative to T. If γ Σ H, we let h γ : K T be the one-parameter subgroup of T such that α h γ x x α γ for any α Σ H and x K. We let B X be a fixed t-stable Borel subgroup of X, containing a fixed t-stable maximal torus T X. Define sets of simple roots β β 2 β m Π X"! Σ X and fundamental dominant weights δ # δ m with respect to T X and B X, but with the opposite of the standard convention: The set of positive roots Σ$% X is defined by B X U X T X where U U& α for α Σ$ X. Then for J! Π X, P X is the opposite of the standard parabolic X corresponding to J. We assume the δ i are numbered so that δ i corresponds to β i for every i. The group Y will be a simple algebraic group over K of classical type and rank n A n, B n, C n or D n ), such that X Y and G Aut Y. Let α α 2 α n Π Y be a set of simple roots of Y, and λ i the set of fundamental dominant weights such that λ i corresponds to α i. Notation and conventions similar to those used for X are used for parabolic subgroups of Y.

+ OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 3 For a group H acting on a module M, M l H will denote the l-fold commutator of H with M. The K-vector space V is assumed to be a restricted irreducible Y-module with high weight λ a i λ i, such that V is irreducible as a G-module but not as an X-module see the comment at the end of the previous subsection). We assume that the T X -high weights of V are restricted as well. So if G X t, then V X V ' V 2, where each of V, V 2 is a restricted irreducible X-module. The natural module for Y will be denoted by W. We assume that W is irreducible as an X-module, and δ will denote its T X -high weight. As in [4], we will always assume that Y is the smallest of SL W, SO W, Sp W containing X. Finally, we assume that G acts on W, as the case when it does not was considered in [4] We label Dynkin diagrams for the groups we will be dealing with as follows, and we always number fundamental roots and fundamental dominant weights to agree with this labelling: 2 A l : ))) l* l 2 B l : ))) l* l 2 C l : ))) l* l D l :,,, 2 ))) l* 2 - l* -- 3 4 5 6 2 E 6 : l We will sometimes use the standard partial order on weights: ν. µ if and only if ν* µ is a sum of positive roots. 2. Q X -LEVELS AND EMBEDDINGS OF PARABOLICS In this section we introduce important facts about the commutator series of a module of a simple algebraic group. Lemma 2.. If H is a simple algebraic group whose root system has only one root length, then restricted irreducible H-modules are tensor indecomposable in particular, restricted irreducible X-modules are tensor indecomposable). Proof. This is part of.6 of [9]. Lemma 2.2. Let M be an irreducible restricted H-module with high weight γ for some simple algebraic group H. Let P be a proper parabolic subgroup of H, with P QL a Levi decomposition. Then M/0 M Q is irreducible for L and for L2 3 L L, with T L4 -high weight γ T L4. Proof. This is.7 and 2. of [9]. Let H, M, γ, and P be as in the last lemma. Let ε i be the set of fundamental roots of H.

4 BEN FORD Definition 2.3. Let µ be a weight of M, say µ γ* c i ε i, with each c i 0. The Q-level of µ is c j, where the sum ranges over those j for which ε j Π H5* Π L6. The Q-level l of M is the sum of weight spaces for weights having Q-level l and is denoted M l. Lemma 2.4. H, M, and P as above. If H is simply laced or if p+ 2 p+ 3 for H G 2 ), then. M l 7 Q ' M µ, the sum taken over those weights µ having Q-level at least l. 2. M l 8/0 Q M l$ Q 79 M l 3. dim M l 8/0 Q M l$ Q :; s) dim M l& Q :/< M l : Q, where s is the number of positive roots β such that U& β Q and β ε i= β for some ε i Π H>* Π L, with β? 0 or a sum of roots in Π L. 4. dim M l 8/0 Q M l$ Q :@ dim Q ) dim M l& Q 8/0 M l : Q. Proof. This is 2.3 of [9]. We will write M l Q for the quotient M Q l& 8/0 M Q l. Lemma 2.5. Let H A l ; let c be an integer such that 0 c p; and let γ # γ l be the fundamental dominant weights for H. The irreducible module M having high weight cγ or cγ l has all weight spaces of dimension ; in particular, dim M A l= c!/ l!c!. Proof. This is.4 of [9]. We will occasionally use the Weyl character formula for dimensions of Weyl modules. Finally, it was shown in [4] that when X acts irreducibly on W, we may assume W is in fact restricted as an X-module: Lemma 2.6. If X acts irreducibly on W, then as an X-module, W has a restricted high weight. Now let P X be a parabolic subgroup of X, and P X Q X L X a Levi decomposition with T X L X if P X is t-stable, choose L X to also be t-stable). Now X acts irreducibly on W with high weight δ, which is restricted by the Lemma above. We gave in [4] the following construction of a parabolic subgroup P Y of Y with P Y Q Y L Y a Levi decomposition) such that P X P Y, Q X Q Y, L X L Y. Let Z Z L XB. Lemma 2.7. The stabilizer in Y of the commutator series + W W Q X + W Q X Q X + ))) + 0 P Y and Q X Q Y R u P Y. C Y Z is a Levi factor of P Y containing L X. is a parabolic subgroup P Y of Y satisfying the following:. P X 2. L Y 3. If T Y is a maximal torus of Y containing T X, then T Y L Y. Proof. This is 2.7 of [4]. We give more information about this embedding for particular groups X and parabolic subgroups P X below and in subsequent sections. For the next Lemma, we assume that t G where t is the fixed outer automorphism of X) and V X V ' V 2, with V V 2 irreducible X-modules. This Lemma was proved in [4, 2.8 and 2.9]: Lemma 2.8. If P X is a t-stable parabolic subgroup of X and P X is embedded in a parabolic subgroup P Y of Y as above, then. P Y is likewise t-stable; 2. V/< V Q Y < V/0 V Q X 0 V /< V Q X ' V 2/0 V 2 Q X as L X -modules.

OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 5 Let P Y L Y Q Y be a parabolic subgroup of Y. For each γ Π Y5* Π LY, we define a certain normal subgroup K γ Y of P Y, as in [9, page 44]: Let Σ γ Y denote the set of roots in Σ Y having γ-coefficient* and zero coefficient for other roots in Π Y>* Π LY. Then let K γ Y be the product of those T Y -root subgroups U β for β Σ& YC* Σ& LY* Σ γ Y. From the commutator relations it follows that K γ Y is normal in P Y and we let Q γ Y Q Y/ K γ Y. This construction also applies to a parabolic subgroup P X of X. In particular, if P X is a maximal parabolic subgroup corresponding to α Π X, then set Q α X Q X/ K α, where K α is the product of those T X -root subgroups corresponding to roots having α-coefficient strictly less than*. The Lemma below will be used heavily in dimension arguments. Lemma 2.9. If P X Q X L X is a maximal parabolic subgroup of X corresponding to α Π X, and P X is embedded in a parabolic subgroup P Y of Y as in Lemma 2.7, then dim 2 V Q Y@ dim QX ) dim α V Q X. Proof. This is part of Proposition 2.4 in [9]. Lemma 2.0. If P X Q X L X is a maximal parabolic subgroup corresponding to α Π X, then: α A. K Q X Q X. with* 2. Q α X is an irreducible LX -module α as its T -high weight. L4X Proof. See 3.2 in [9] remembering that X is of type A m, D m, or E 6 ). Again assume t G. Let P X be a parabolic subgroup of X not necessarily t-stable) containing the fixed t-stable Borel subgroup B X. Embed P X in a parabolic subgroup P Y of Y via the above construction. Write LY L D ))) D L r, a direct product of simple groups. By Lemma 2.2, LY acts irreducibly on V Q YE V/< V Q Y. Then V Q YE F V )))F V r where for each i, V i is an irreducible module for L i. The embedding L X L Y gives an embedding of LX D ))) D into L L r, and via the projections LX L i, any L i -module, in particular V i, can be regarded as a module for LX. Since Q X Q Y, we have V Q X 2G V Q Y and hence V/0 V Q Y is a quotient of V/< V Q V /0 V Q X ' V 2/0 V 2 Q X, with each of these summands irreducible LX -modules. Since X H LX V/< V LY Y V/0 V, this implies that either Q is irreducible for LX Y I V/< V, or Q Q V /0 V Q X ' V 2/0 V 2 Q X. Lemma 2.8 tells us that the latter happens when P X is t-stable. X I The following was proved in [4, 2.]: Lemma 2.. If V, P X L X Q X, P Y L Y Q Y, and L i are as above with P X t-stable, then only one L i acts nontrivially on V/< V Q Y. 3. THE CASE X A m As always, let X Y be simple algebraic groups over an algebraically closed field K of characteristic p 0 or a prime), with X admitting an involutory graph automorphism t which also acts on Y, and Y of classical type. Let β i Π Y ) be the set of α i fundamental roots for X Y); δ i λ i ) the corresponding fundamental dominant weights for X Y). The fixed t-stable Borel subgroup B X of X contains a t-stable maximal torus T X. Let V V λ be a restricted irreducible t-stable Y -module with high weight λ a i λ i, such that V XJ tk is irreducible, but V X V ' V 2, with V V 2 restricted irreducible X-modules. We denote the T X -high weight of V by b δ = b 2 δ 2= ))) = b m δ m, so the T X - high weight of V 2 is b m δ = b m& δ 2= ))) = b δ m The natural module for Y is W. The main result of the section is:

Q Q 6 BEN FORD Theorem 3.. If X acts irreducibly on W with high weight δ, t acts on W, and X is of type A m, then p L / 2 3 5 7, X A 3, Y D 0, δ 2δ 2, and the high weights of V X V Y are as in U 5 of Table 2. Notice that since t acts on W, the high weight δ d δ = ))) = d m δ m of W X must be symmetric, i.e. d d m, d 2 d m&, etc. But then by a result of Steinberg [, page 226]), X fixes a nondegenerate bilinear form on W; the form is orthogonal if p 2. The strategy we use to rule out most possibilities for the high weight δ is to show that the construction outlined in Lemma 2.7) of a parabolic subgroup of Y containing the fixed t-stable) Borel subgroup of X gives a contradiction in all but a few cases. After giving the Lemma which we usually use to produce the contradiction, we will treat the A 2 and A 3 cases first, followed by the general argument. 3.. Some Facts About P Y. We use the construction given in Lemma 2.7 of a parabolic subgroup P Y Y containing the fixed t-stable Borel subgroup B X. Namely, P Y is taken to be the stabilizer in Y of the flag in W given by U X -levels. We want to use Lemma 3.2 below to produce a contradiction in most cases; we will show that LY has a factor of type A only under strong conditions. Before proceeding with the general proof, we need some facts about the flag in W of which P Y is the stabilizer. 2$EMMMN iw δ& e β & e 2 β 2&>OOO, the sum taken over e j 0. Each Recall that for i 0, W i e$ e space W i is T X t -stable, and if u U& α, then uw δ& e β & e 2 β 2&5OOO! W mp 0 δ& e β & e 2 β 2& MMM& mα So B X t stabilizes each factor W ip m ir / W ip m$ ir By Lemma 2.4, W i9 W UX 8/0 W i$ i UX. Let l be minimal with respect to W l$ UX < 0, and notice that l is then the level of the low weight * δ. If Y Sp W or Y SO W, with the form denoted by "S, then we noted in [4, proof of 2.7] that u v@ 0 for u W i, v W j unless i= j l. Thus the W i for i+ l/ 2, along with a maximal totally singular subspace of W lt 2 if l is even), span a maximal totally singular subspace of W. Let w$ be an T X -high weight vector of W. Then W 0 3 w$, W l 3 w 0 w$, and B X is contained in the full stabilizer P Y of the flag W W ip 0 i W ip i ))) W ip l i U w 0 w$ V 0 Let P Y L Y Q Y be a Levi decomposition of P Y ; then if u U X w W m, we have uw* w ip m$ W i, so U X Q Y. We have T X L Y C Y Z for Z Z L XB. Choose a basis for each W i with the basis for W lt 2 chosen maximally hyperbolic note that W lt 29 XW ip lt 2W iy / W i lt 2W iy is the only possible non-singular quotient in the flag); the union of these bases is a basis for W. With respect to this basis, LY consists of block matrices, each block corresponding to W i for some i. On the other hand, each W i for 0 i l/ 2 corresponds to a connected component of the Dynkin diagram for L Y. dim W i 2 for some i l/ 2, or dim W lt 2V 4. So the only possibilities for an A to appear as one of the simple factors of LY are when

LY OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 7 To show dim W i@ m, it suffices to find m T X -weights of W which occur in W i. By the result in [2], weights which appear in characteristic 0 also appear in characteristic p. This is the approach we use to obtain contradictions for most embeddings XZ Y. For a T X -weight ω of W, ω= t ω ω* w 0 ω is a sum of roots, and we let l ω be the height of ω= ω t in the root lattice the number of summands when we express ω= ω t as a sum of fundamental roots). So for l as above, l l δ where δ is the T X -high weight of W. The constructions in this section also apply to the embedding of an arbitrary parabolic subgroup P X Q X L X of X in a parabolic subgroup P Y of Y, using Q X -levels in place of the U X -levels. The weights appearing in W i then are those of the form δ* e β ))) * e m β m where the sum of the e j for β j Π XE* Π LX is i. Again by Lemma 2.4, * W i9 W QX :/< W i$ i QX. One more fact: If P X is a t-stable parabolic subgroup of X including P X B X ), then eachw i is clearly t-stable since then W δ& e β & e 2 β 2& MMM& e m β m is sent by t to W δ& e m β & e m[ β 2& MMM& e So P Y is t-stable. But then V Q Y is T X t -stable, hence V Q Y? \ V Q X V Q X 2] V Q since Q X Q Y, and we get equality because V/0 V Q X is an irreducible T X t -module and Y T X L Y ). So if P X is t-stable, then V/0 V Q Y is a sum of two irreducible LX -modules. The following Lemma provides the basis for the proof of the section s main result; it will also be used throughout the paper. Lemma 3.2. If P Y is a t-stable parabolic subgroup of Y such that B X P Y, U X Q Y, T X where P Y Q Y L Y B X U X T X are the Levi decompositions), then at least one of the simple factors of LY has type A ; and if this factor corresponds to α j, then a j. In addition, a i 0 for α i Π LY, i j. Proof. This is Lemma 5.2 of [4]. 3.2. The Cases X A 2 and X A 3. We use Lemma 3.2 heavily. As always, δ d δ d 2 δ 2= ))) is the T X -high weight of W. Let β be a nonnegative sum of fundamental roots = of X, of height j in the root lattice. Note that if δ* β is a dominant weight such that the X-module with high weight δ* β has m weights at level i, then W has m weights at level j= i. So in our attempt to prove that there cannot be a U X -level of dimension 2 in W, we will proceed by induction on the high weight δ. X A 2 : Since δ is symmetric, δ aδ = aδ 2 for some a+ 0. Here we will always have dim W V 2, since the only two weight spaces in level are δ* β and δ* β 2, both of dimension ; we will deal with level after we discuss levels 2 and higher. In evaluating the numbers of weights at these levels, we will first use an induction to deal with the case a 4, and then deal with a 3 and a 2. Assume a 4. Then l l δ 6, so we must check to level 8 we must list three weights at every level 2-7, and 5 at level 8). We have the weights in the table below: Level Weights 2 δ* 2β, δ* β * β 2, δ* 2β 2 3 δ* 3β, δ* 2β * β 2, δ* 3β 2 4 δ* 4β, δ* 3β * β 2, δ* 4β 2 5 δ* 4β * β 2, δ* 3β * 2β 2, δ* 2β * 3β 2 6 δ* 4β * 2β 2, δ* 3β * 3β 2, δ* 2β 2* 4β 2 7 δ* 5β * β 2, δ* 4β * 3β 2, δ* 3β * 4β 2 8 δ* 6β * 2β 2, δ* 5β * 3β 2, δ* 4β * 4β 2, δ* 3β * 5β 2, δ* 2β * 6β 2. So LY has no factors of type A except possibly the factor L corresponding to W. Assume a+ 4. Then δ* β * β 2 ^ a* δ = a* δ 2 is dominant, and by induction δ has enough weights at all levels except possibly at levels, 2 and 3. At level 2, we have β m ).

a a a 8 BEN FORD δ* 2β, δ* β * β 2, and δ* 2β 2 ; at level 3, δ* 3β, δ* 2β * β 2, and δ* 3β 2. So again, L is the only possible A -factor of LY. Assume a 3. Then l 2; we must check dimensions to level 6. In levels 2-5, we have enough weights as above. So we must show that W 6 has dimension at least 5. The weights at level 6 are δ* 4β * 2β 2, δ* 3β * 3β 2, and δ* 2β * 4β 2. If p 7, then dim W δ& 3β & 3β 2" 3, so dim W 6" 5. So unless p 7, here again we have only the L A possibility. If a 2, then l 8 and we must check dimensions to level 4. For level 2, we have enough weights as above. At level 3, we have δ* 2β * β 2 and δ* β * 2β 2 ; if p 5, each has dimension 2, so dim W 3 + 3. At level 4, the weights are δ* 3β * β 2, δ* 2β * 2β 2, and δ* β * 3β 2. If p 5, then δ* 2β * 2β 2 has dimension 3, so dim W 4_ 5. As above, unless p 5, we have only the L A possibility. From the construction of P Y we can see that in any case covered above including a p A 2 5 3 7 ), there is only one node in the Dynkin diagram between the L i, since there are no U X -levels of dimension other thanw$. Also notice that from the a 2 and a 3 cases above we know the embeddings in the cases a p 3 2 5`# 3 7 we simply compute the dimensions of the levels): If a 2 p 5, then dim WV 9, so Y has type B 9 and P Y is the parabolic subgroup of Y corresponding to the indicated nodes: a a + a If a 3 p 7 then dim W 37, Y is of type B 8, and P Y corresponds to: + So the possibilities are: ) a 2, with the first simple factor of LY corresponding to ))) ; 2) a 2 p 5; 3) a 3 p 7; 4) a. By Lemma 3.2, in the marking for the high weight of V on the Dynkin diagram for Y there is only one non-zero label on the nodes representing L Y, and this non-zero label must be a on a node corresponding to an A factor of LY ; call this node γ. By the comment above, all nodes in the Dynkin diagram are either in or adjacent 2 to Π LY except possibly in case 4). Our aim is to show that all nodes except γ have marking 0. As in the introduction, let V Q Y@ b V Q Y :/< V Q Y Q Y 2 ; similarly define V U X and Vi 2 U X for i 2 2. Recall that λ is the T Y -high weight of V. By Lemma 2.4, the weights in V Q Y are those of the form λ* βc T LY where β Σ$ Y and there exists ε Π Y * Π L Y such that β* ε Σ L Y. Then by Lemma 2.4, we have dim 2 V Q dim 2 V U X@ 4. YV First consider the case a. If p 3, then dim W; 8 and Y D 4 ; if p 3, then dim W; 7 and Y B 3. In both cases, constructing P Y as usual, we have α 2 Π LY, so the only possibility for an A factor of L Y to occur is Π LYd ^ α 2 as all other nodes adjoin α 2 ). For p 2 3, if another node has a non-zero label, say α 3 all are equivalent by symmetry for this argument), then in V Q Y we have the high weights λ* α 3 T, giving L4Y a composition factor of dimension 3; and λ* α 2* α and λ* α 2* α 4 each giving one of dimension. But this contradicts dim 2 V Q Y_ 4. So λ λ 2, giving an example of ) above, which we deal with below. If p 3 and a a 3 0, then λ* α T, L4Y λ* α 3 T L4Y each give a composition factor of dimension 3 in V 2 Q Y, again a contradiction. Finally, irreducible B 3 -modules with high weights eλ = λ 2 or λ 2= eλ 3 are all too large to be the sum of two restricted irreducible A 2 -modules unless e 0 by counting weights that appear

g g g OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 9 in the Weyl module ofe B [8]). So again λ λ 2, which 3 we deal with below. eλ = λ 2 ore B 3 λ 2= eλ 3 ; all these weights appear in V by For cases ) 3), assume there is an ε Π Y which adjoins γ and has non-zero marking m. If ε is not 2 an end node, then it also adjoins another factor L l of LY by our comment above that there are never two adjacent nodes outside Π LY ; then λ* ε is an LY -high weight in V Q Y, and the LY -high weight module of this high weight has dimension 6 λ* εf T L4Y 2λ j= λ k where λ j is the fundamental dominant weight corresponding to γ and λ k is the node of L l which adjoins ε the LY -module with this high weight has dimension at least 2) 3 6). This contradicts dim 2 V Q Y@ 4. If ε is an end node, it cannot be the short root in a B n that root is in Π LY because we m 0 saw that dim W lt 2; ))) 3 in all cases). Then we have the picture: ε γ µ. We of dimension 3 and λ* γ* µc T of L4Y dimension 2. Again, have high weights λ* εf T L4Y By Lemma 3.2, ε / Π LY this contradicts dim 2 V Q Y; 4. So all nodes adjoining γ have 0 label. Assume there is an ε Π Y which does not adjoin γ and has non-zero marking m. ; and ε adjoins Π LY since every fundamental root is either in Π LY or adjoins it. If γ is not an end node, we have the pictures different pieces of the Dynkin diagram for Y): Here λ* α* γ T L4Y 0 0 m 0 ))) ))) ))) ))) α γ β µ ε ν gives dimension 2; and λ* ε gives dimension 2. Again, this is a contradiction. The node 2 is a high weight in V Q Y, giving dimension ; λ* γ* β T L4Y γ can be an end node only in the cases a p; h 2 5` 3 7 these are the only cases in which an A factor of LY corresponds to an end node of the Dynkin diagram for Y ), with γ 0 0 ))) the short root of a B n + ; then the picture is: µ δ γ. Here λ* γ* δ T gives dimension L4Y 4; and λ* ε T L4Y gives, again a contradiction. So all nodes other than γ must have label 0. For both of the above non-end node cases, similar arguments hold if there is a double bond in one of the relevant pieces of the Dynkin diagram for Y. So γ is the only node in the Dynkin diagram with a non-zero label. We need only show that the few remaining possibilities do not lead to examples. Case ): V has high weight λ 2. Then V 9 hi 2 W by [7, II.2.5]. Regard W as an X-module. Remember that δ aδ = aδ 2 is the T X -high weight for W. Let v W δ be a maximal vector in W; 0 v 2 W δ& β 0 v 3 W δ& β 2. Then v j v 2 and v j v 3 are X-maximal vectors in V, so KX v j v 2 ' KX v j v 3C V. We now consider the dimension of KX v j v. The vector v j v 2 has weight 2δ* β 2a* 2 δ = 2a= δ 2. So dim KX v j v 2k dim Weyl modules 2 2a* 2a= 2 4a= S 3= 2* 8a 6a 3a*. Also, dim KX v j v 3S dim KX v j v 2 since t interchanges them), so dim V@ 3= 2* 6a 2a 6a* 2. On the other hand, dim WS 2= 3a 3a= this is the number of weights that appear in the Weyl module with the same T X -high weight as W ; all these weights appear in W by Lemma.2), and dim VC dim i WC W diml Wm 2 3a 2 Y W 2$ 3a$ 2 Y A 2= 3a 3a= 2= 3a

s trr } rr 0 BEN FORD 3a/ 2 A 4= 3= 2= 9a 8a 2a 3a/ 4= 3= 2= 2. So 9a 8a 2a 3a 2 3= 2* 6a 2a 6a* 2. But this has no solutions in positive integers. So this case is ruled out. The only cases not ruled out by the above are: Case 2): a p5 n 2 5 here remembery B 9 ). If λ ))) λ 7 then the picture is g + g and λ* γ* ε L Y λ* γ* µ L Y each give dimension 3. So λ λ 7. If λ λ 9, then dim VS 9 2 52. But the dimension of any irreducible A 2 -module with high weight cδ = dδ 2 c d 5 c d) is at most 90. So V is too large to be the sum of two restricted irreducible modules for X. Case 3): a pv h 3 7 here Y B 8 ) with λ λ 8. Here dim VV 2 8. But c d 7 c d; again, V is too large to be the sum of two restricted irreducible X-modules. So X is not of type A 2. X A 3 : We use a similar induction. Let δ aδ = bδ 2= aδ 3 be the T X -high weight of W. First we eliminate the case b 0 with the A factor of LY referred to in Lemma 3.2) corresponding to α 2. Assume δ aδ = aδ 3 and λ λ 2= ))). Let P X L X Q X be the maximal parabolic subgroup of X corresponding to β 3 Π X, and embed P X in a parabolic subgroup P Y of Y via the construction in Lemma 2.7. Now L the simple factor of LY corresponding to Q X -level 0 of W) is of type A 2 if a, and type A l with l 5 if a+ ; and the root system of L contains α. We determine the rank of L by counting weights that appear in Q X -level 0 of W, using Suprunenko s result [2] that weights which appear in the Weyl module also appear in the irreducible module.) Assume V/< V Q Y is reducible as an L X -module; that is, V/0 V Q Y 7 L X V /0 V Q X ' V 2/< V 2 Q X o Then Z Z L Y since L Y C Y Z. So Z induces scalars on V/0 V Q Y an irreducible L Y -module). vvuwc 0 0 0 xcyy 0 c 0 0 Z qprr 0 0 c 0 0 0 0 c& z c K2{rr 3 ~ h β c h β2 2 c h β3 3S c c K 2b 3b Now h β c h β2 2 c h β3 3 c acts as multiplication by c $ 2$ b 3 on a high weight vector v V, and as multiplication by c 3$ 2$ b on a high weight vector v 2 V 2. But by our assumption, v and v 2 both have nonzero images in V/< V Q Y ; because b = 2b 2= 3b 3 b 3= 2b 2= 3b, we have a contradiction b b 3 because V 9 X V 2 ). So V/< V Q Y is an irreducible L X -module, and thus an irreducible LX Assume V/< V Q Y < X V /< V Q X. Now we are in the situation studied in [9]: V/0 V Q Y is an irreducible module for LX of type A 2 ) and for L. As there are no examples matching this setup in [9, Table ], we know that either the embedding LX Z L is an isomorphism, or V/0 V Q Y is the natural module for L. Both possibilities are excluded if a+, as then L is of type A l with l 5 so L 9 LX 2 ), and a so V/< V Q Y is not the natural module for L ). On the other hand, if a then necessarily LX 9 L as W/0 W Q Y then has LX -high weight δ ). 2b 3b 0 0 0 0 -module by Lemma 2.2. ε γ µ,

OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II So V/< V Q Y must have LX -high weight a δ = δ 2, since it has L -high weight a λ = λ 2 and the embedding is an isomorphism. But V /< V Q X also has high weight b δ = b 2 δ 2. So a b and b 2. Now the argument above can be repeated with the maximal parabolic subgroup of X corresponding to β instead of β 3, with the conclusion that b 3 a. But then b b 3, a contradiction. So if δ aδ = aδ 3, then a 2 0 Every weight of the form aδ = bδ 2= aδ 3 except δ = δ 3, δ 2, and 2δ 2 has one of 2δ = 2δ 3 or δ = δ 2= δ 3 as a subdominant weight. It is easy to check, as in the A 2 case, that the modules with these latter two high weights have enough weights at every level, so we can proceed by induction: If b 2 and a+ 2, then by induction δ* β * β 2* β 3 a* δ = bδ 2= a* δ 3 has enough weights at all levels; we need to check δ-levels 2-5. As before, there are enough weights in each of these levels, so by induction, L is the only possible factor of LY of type A. If b 2, then by induction δ* β 2 has enough weights at all levels, and we need check only δ-levels and 2. If a 0, then dim W, so L is trivial; if a+ 0, then dim W 3, so L is of type A 2. Again, there are enough weights at level 2. So there are no possible A factors of LY here. So the possibilities which have not been ruled out are: δ δ = δ 3, δ δ 2, and δ 2δ 2. If δ δ 2, then X SO W; Y, so there are no examples here. If δ δ = δ 3, we can check the dimensions of the weight spaces and find that dim W5 5 if p 2, and P Y is the +. parabolic subgroup of Y corresponding to the subset α 2 α 4 α 5 α 7 of Π Y, as indicated on this picture of the Dynkin diagram for Y B 7 : If p 2, then dim W> 4, and in [9, page 273] it is shown that X stabilizes a quadratic form on W. So in this case Y D 7 and counting dimensions of U X -levels, we see that the parabolic subgroup P Y corresponds to,,, - -- the stabilizer of the 2- dimensional level 3 is SO 2, which is a torus). So Lemma 3.5 rules out this case. If δ 2δ 2, again we can check dimensions of all the weight spaces to find 20 p 2 3 dim W@ ; dim W> 9 p 3, and P Y is indicated by the circled nodes of the Dynkin diagram for Y : or,,, - -- p 2 3 ; + p 3` Notice we need not consider p 2 by Lemma 2.6. So the cases we must deal with are: W X Y and embedding of B X P Y 0 ) + p 2)

2 BEN FORD 0 2 0 2),,, - -- p 2 3 + p 3 In the marking for V on Π Y, there is a on one of the nodes α j corresponding to an A factor of LY. There can be no other nonzero marking on any of the indicated nodes, since dim V/0 V Q Y :@ 2. Recall that j 2 by Lemma 3.5; thus, as we see from the pictures above, j n is the only possibility or j n* in case 2, but we may assume j n by symmetry). We claim that α n is the only node with a nonzero label. Claim 3.3. In any of the above cases λ λ n. Proof. Since U X Q Y, we have V Q Y Q Y < n V U X U X. Then because V Q Y 2 ^ V U X, we have dim V Q Y :/< V 2 Q Y Q Y k dim V U X 8/0 V U X U X :" 6 by Lemma 2.4. The weights that appear in V Q Yd A V Q Y :/< V Q Y Q Y are those of the form λ* β, where if β e i α i, then the sum of the e i for those α i Π Y>* Π LY is. Let us consider the above cases. Remember λ a i λ i is the T Y -high weight for V: ) The node α 7 has label a 7 ). Consider the possibilities for another nonzero label: If a 0, then λ* 2 α is a high weight in V Q Y, giving a composition factor of dimension 4. Another high weight is λ* α 6* α 7, giving dimension 6 since p 2. But above we noted that dim 2 V Q Y; 6. So a 0. If a 3 0, then λ* α 3 gives dim 2 V Q Y; 2; so a 3 0. If a 6 0, then λ* α 6 gives dimension 2 since p 2. So a 6 0. 2) If p 2 3, we are in the case Y D 0, with a 0. If a 8 0, then λ* α 8 gives dimension 8 in V Q Y. If a 5 0, then λ* α 5 gives dimension 8. If a 2 0, then λ* α 2 gives dimension 6 and λ* α 8* α 9 gives dimension 6. If a 0, then λ* α 8* α 9 gives dimension 6 and λ* α dimension 2. So in fact a i 0 for all i 9. If p 3, then Y B 9 and a 9. If a 8 0, then λ* α 8 gives dimension 6 and λ* α α 9 gives dimension ; so a 8 0. If a 5 0, λ* α 5 gives dimension 8; if a 2 0, l* 8* α 2 gives dimension 6 and λ* α 8* α 9 gives dimension 6. If a 0, then λ* α 8* α 9 gives dimension 6 and λ* α dimension 2. So a i 0 for all i 9. Lemma 3.4. If λ λ n i.e. if we are in one of the remaining cases), then as an X-module, W has high weight 2δ 2 and V is as in the statement of Theorem 3.. Proof. We still have to check cases ) and 2) on page with λ λ n : ) Assume W has T X -high weight δ = δ 3 and p 2. Consider the embedding in a parabolic subgroup P Y Y of the parabolic subgroup P X X corresponding to β β 2! Π X. Checking the dimensions at different Q X -levels as before, we see that for any characteristic, P Y corresponds to α α 2 α 4 α 5 α 6 α 7. Since λ λ n dim V/0 V Q Y ; 6 V/0 V Q Y is isomorphic to a spin module for the simple factor of LY of type B 4 corresponding to α 4 α 5 α 6 α 7 ). The quotient V/0 V Q Y is also an irreducible LX -module again by considering the action of Z Z L X B on the two T X -high weight vectors of V); but LX A 2, which has no irreducible representations of dimension 6 in any characteristic. So we have no examples here. 2) Assume W X has high weight 2δ 2. If p 3, then dim WE 9 and Y B 9. Using P X corresponding to β β 2 as above, we get an embedding of P X in the parabolic subgroup P Y corresponding to Π YE*ƒ α 6. Since λ λ n, dim V/< V Q Y ; dim spin B 3; 8. But V/0 V Q Y is an irreducible LX -module by the action of Z again), and LX A 2, which has no 8-dimensional irreducible representations in characteristic 3.

OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 3 If p 3 then take P X as above; again P Y corresponds to Π Y * α 6 and dim V/0 V Q dim spin D 4 8 dim V i/< V i Q X for i or 2, say i. So b b 2. Y : Now let P X correspond to β β 3. This P X is t-stable, so V/0 V Q Y < V/0 V Q X 0 V /0 V Q X ' V 2/0 V 2 Q X The embedding gives P Y corresponding to Π Y5* α α 5. So here, dim V/0 V Q Y : a 6 dim V/< V Q 2dim V X : /< V Q 2 b X = b 3= E 4 b 3= ` So b 3 3. Now dim V D0 λ n 2 9 in any characteristic; dim V 3δ = δ 2= δ 3E 256 2 8 3 when p+ 7 or p 0. So V X 3 ', V Y spin D 0 is a possibility for p 2 3 5 7. In this case, consider again the embedding of B X in the parabolic subgroup P Y corresponding to,,, -. We define some temporary notation: Let λ j A λ * α * ))) * α jf T X --. By the construction of the embedding, we know that λ T X δ 2δ λ A λ * α f T 2 X δ* β 2 λ 2 λ 3 λ 4 δ* β * β 2 δ* β 2* β 3 δ* 2β 2 λ 5 λ 6 λ 7 δ* β * 2β 2 δ* 2β 2* β 3 δ* β * β 2* β 3 λ 8 λ 9 λ 0 δ* 2β 2* 2β 3 δ* 2β * 2β 2 δ* β * 2β 2* β 3 with λ * α * ))) * α 8c T X X*ˆ λ * α * ))) * α 0c T X since λ * α * ))) * α * 8 λ 8= λ 9= λ 0 U*ˆ λ * α * ))) * α 0 ). This gives enough information to determine the possibilities for α i T X for i 0. We can write the T Y -high weight for V, λ 0, in terms of the α i, and we find that with any of the possible choices made above, λ 0 T X λ 0* α 0 T X 3 3δ = δ 2= δ 3 δ = δ 2= 3δ 3. So V contains A 3 -submodules of these 2 high weights; since their dimensions add to dim V, we have the case stated in the theorem. This completes the proof for X A 3. 3.3. When Lemma 3.2 doesn t help. Using our standard construction of P Y Lemma 2.7), the obvious situation in which the Lemma 3.2 is of no help is when δ aδ i= bδ j, i.e. when U X -level 0 has dimension 2. In this case L LY is of type A, corresponding to α 2 Π Y. Remember that δ must be symmetric, so that in fact the following is what we will need. Lemma 3.5. The situation δ aδ i= aδ m& i$ i m/ 2), a 2 does not give any examples in the Main Theorem if m+ 3. Proof. With the given δ and with P Y as in Lemma 3.2, we have α 2 Π LY since level in the construction of P Y has dimension 2. So Lemma 3.2 tells us that a 2 is the only nonzero coefficient on Π LY.

4 BEN FORD Assume i+. Consider the parabolic subgroup P X of X corresponding to Π X* β β m ))), as in the following picture:. So LX m& is of type A 2, and if we embed P X Z P Y by the usual construction, then level 0 that is, W/0 W Q X ) of the flag which results is totally singular and has dimension 7 the smallest W/0 W Q X could be is when LX is of type A 2 with a label of on each node; this has dimension 7 in characteristic 3 and dimension 8 in other characteristics); thus L the simple factor of LY corresponding to the quotient W/< W Q X in the flag) is of type A l for some l 6. By Lemma 2., only one L j acts nontrivially on V/< V Q Y. So V/0 V Q Y is a nontrivial there is at least one nonzero label on L in the marking for the high weight of V/0 V Q Y, namely, on the second node), irreducible L -module, and as an LX V/0 V -module, Q Y is the sum of two irreducibles, V /0 V Q X and V 2/0 V 2 Q X. The natural module for L is isomorphic to W/0 W Q Y and is irreducible as an LX -module. So unless V /0 V Q X 9 V 2/0 V 2 Q X, we are in the situation we consider in this section and by induction no examples arise the single case which arises below for X A 3 does not arise inductively because δ does not have the form we are considering here). Let the T X -high weight of V be b δ = b 2 δ 2= ))) = b m δ m so the high weight of V 2 is b m δ = b m& δ 2= ). Assume b 2 b m& b 3 b m& 2 i.e. V /0 V Q X 9 V 2/< V 2 Q X ). This implies b b m since V 9 V 2. Now take P X to be another parabolic subgroup of X, corresponding to Π X * β m, and embed P XZ P Y via the same construction. Again we have W/0 W Q Y 7 W/< W Q X irreducible for LX and for L. We show that in this case V/< V Q Y is irreducible as an L X -module in contrast to the situation when the parabolic subgroup of X is t-stable, which forces V/0 V Q Y to be the sum of two irreducibles for L X ): Let Z Z L XB. Since L Y C Y Z, we have Z Z L Y, so Z induces scalars on V/< V Q Y since L Y acts irreducibly on this module). But if V/< V Q Y is reducible as an L X -module, then V/< V Q Y 0 V /0 V Q X ' V 2/0 V 2 Q X, and we show that Z acts differently on these two L X -modules: Z diag a a a& mf a K h β a h β2 27 a h βm mc a a K The two X-modules V, V 2 have high weight labelling b b 2 b ))) m& b m b and m b m& b ))) 2 b so h β a h β2 27 a h βm m a acts as multiplication by a $ b 2b 2$MMM$ mb m on a high weight vector v V and as multiplication by a m$ b 2b m[ $MMM$ mb on a high weight vector v 2 V 2 ; these two exponents are not equal. Since v j has a nonzero image in V j/0 V j Q X, this shows that only one of the V j/0 V j Q Y can have a nonzero image in V/0 V Q Y. So V/0 V Q Y is irreducible as an L X -module. Assume V is the summand which projects nontrivially to V/0 V Q Y so V 2!^ V Q Y ). As no irreducible restricted representations of A m& are tensor V/0 V decomposable by Lemma 2., L is the only simple factor of LY Y to act nontrivially on Q. Note that the rank of L is 5, since this rank is one less than the dimension of W/0 W Q Y : With δ of the form we are assuming, the A m& -high weight of W/0 W Q Y is of the form aδ i= aδ m& i$ c T. L4X The Weyl module for A m& with this high weight has at least 6 weights, and these weights all appear in the irreducible module W/0 W Q Y by the result in [2]. So now we are inductively in the situation examined in [9]: V/< V Q Y is an irreducible module for both LX and L, and it is not the natural module for L since in the labelling

OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 5 for the L -high weight of V/0 V Q Y, there is a on the second node of Π L ). Also, LX 9 L because the natural module for L, W/0 W Q Y, has high weight δ T and thus has dimension larger than m, which is the dimension of the natural module for LX. So any examples here would appear in Table of [9]; examining that table, we see that in fact there are no examples. This completes the case i+. So we need to consider only the case i, i.e. δ aδ = aδ m, with λ λ 2= ))). Let P X L X Q X be the maximal parabolic subgroup of X corresponding to β j Π X and embed P X in a parabolic subgroup P Y L Y Q Y of Y via the usual construction. Notice that L the simple factor of LY corresponding to the Q X-level 0 of W ) is of type A l, with l+ 3 unless m 4 we have taken care of the cases m 2 3 in 3.2). We wish to show that for at least one choice of j, V/< V Q Y is irreducible as an L X -module. We will again use the action of Z Z L XB on T X -high weight vectors in V. Z diag Š al m& j$ m al m& j$ Œ m al m& j$ m j Š a& j a& `Œ j# a& ; a K j m& j$ h β al m& j$ mh β2 2l a m& j$ m7 h β j jl a m& j$ mh β jž jl a m& jmh β jž 2 jl a m& j& m7 h βm j_ a a K If V/< V Q Y is reducible as an L X -module, then as above, Z must act as multiplication by the same scalar on a high weight vector v V as on a high weight vector v 2 V 2, and we get the equation: m* j= b ) = 2 m* j= b 2= ))) = j m* j= b j m* j b j= j$ = j m* j* b j$ 2= ))) = j) ) b m U m* j= b m= 2 m* j= b m& = ))) = j m* j= b j m* j m& j$ = b m& j= j m* j* b m& j& = ))) = j) ) b If we assume that V/0 V Q Y is reducible as an L X -module for every j, then we have b m b 2 b m&. For example, the a system of equations which together imply b equations for j m and j m* are: b = 2b 2= )))= m* b m& = mb m b m= 2b m& = ))) = mb 2b = 4b 2= )))= 2 m* b m& = m* b m 2b m= 4b m& = ))) = m* b Twice the first equation minus the second gives m= b m m= b. Knowing b b m, the equations for j m and j m* 2 give b 2 b m& ; continuing in this manner we obtain b l b m& for every l. But this is impossible, as it would imply that V is reducible for X t l$. So for some j j m, V/< V Q Y is irreducible as an LX -module, where L X, Q Y are as above. But then again we are in the situation examined in [9]: V/0 V Q Y is an irreducible module for LX A j& D A m& j and for L. So one of the following occurs:. The embedding L X Z L is an isomorphism. 2. V/< V Q Y is the natural module for L which happens only for m 3 4, since L has rank 4 in other cases we consider and the L -high weight of V/0 V Q Y has a nonzero label on the node α 2 ). 3. V/< V Q Y appears in Table of [9]. We deal with 3 first. Of the appearances of the inclusion A j& D A m& j A l in Table of [9], only one case I 7 there) gives the correct restriction of the natural module for A l to the L4X

6 BEN FORD subgroup. So the possible picture here is LX A m& ; L Al m of W X 2δ = 2δ m ): 2$ m& 2m T 2 ; the high weight V /< V Q X 0 2 0 b ))) m and V/0 V Q Y 7 0 0 0 ))) Now we look at yet another parabolic subgroup of X: Let P X be generated by B X and the root subgroups for β and β m ; embed P X in a parabolic subgroup P Y of Y as usual. By V/< V Lemma 2., L is the only simple factor of LY Y to act nontrivially on Q. Notice dim W/< W Q X 9 since W X has T X -high weight 2δ = 2δ m ), so L has rank 8. Thus 3 2) b m= C dim V /0 V Q X : = dim V 2/0 V 2 Q X :C dim V/< V Q X :C dim V/< V Q 9 W 2Y 36, which tells us b m 5. This tells us that the only two T Y :C L4X -high weights of V/0 V Q Y are 2δ 5δ m and 5δ 2δ m. But in fact V/0 V Q Y <9 Ui 2 W/0 W Q Y : p 2 because of the 2 appearing in the picture above and ofi our assumption that the T X -high weights of V are restricted), and two T -high weights L4X 2 W/0 W Q Y : are 4δ 2δ and 2δ 4δ m. Again we have a contradiction. m Next consider item 2. This can only occur if dim W/< W Q X " 4 since there is a on the second node in the marking for the high weight of V/0 V Q Y on L ); and this occurs only in the cases we excluded X A 2 A 3 ) and the case X A 4, with j 4 or, equivalently, j ), δ δ = δ 4. But this in fact gives an instance of item it implies that W/< W Q Y has T -high weight δ L4X ). So we are left with item. Note that the equations hold for all j m, since in the consideration of item 3 above we obtained a contradiction to V/< V Q Y being irreducible for LX when j m. We have the pictures V /< V Q X 7 a a 3 a ))) m& b m V/0 V Q Y 7 a a 3 a ))) m& a m ))) So a b a 2 b 2# a m& b m&. Now let P X correspond to the indicated ))) nodes:, and embed it in a parabolic subgroup of Y as usual. Then V/0 V Q Y is the sum of two irreducible LX -modules, and the simple factor L 2 of LY ))) corresponds to the indicated nodes at the beginning of the Dynkin diagram for Y : with L 2 of rank 2m* 3. But by the construction of P Y, we know the embedding LX Z here the natural module for L 2 is Q X -level of W and restricts to LX as the sum of the two irreducible modules with high weights δ 2 T and δ L4X m& T ). As this situation gave L4X no examples in [4], we know by induction that either: ) a 3 a 4 ))) a 2m& 2 0 i.e. V/0 V Q Y 29 the natural module for L 2 ); or 2) a 2 a m& a 3 a m& 2 i.e. V/0 V Q is reducible for LX t ). Y ))) ))) ))) L 2 L 3, L 2 We noted that all the equations on page 5 hold for j m. The equation for j m* is: 2b = 4b 2= )))= 2 m* b m& = m* b m 2b m= 4b m& = ))) = m* b

OVERGROUPS OF IRREDUCIBLE LINEAR GROUPS, II 7 Now if 2) above holds, we know that b 2 b m& b 3 b m& 2 remember that a i b i for i m), which together with the above equation gives b b m, which is a contradiction. So 2) does not occur. If a 3 ))) a 2m& 2 0, i.e. ) holds, then once again we examine the equation above and see that now it reduces to 2b = 4= m* b m 2b m= 2 m* = m* b remembering a i b i for i m and a 2 b 2 ), or b m* b 2. Let P X be the parabolic subgroup of X corresponding V/< V to β β m. Then L is of type A 3, L is the only factor of LY Y acting nontrivially on Q by Lemma 2., and the picture is: V /< V Q X 7 b F b = 2 and V/< V Q Y < b 0 ))) By the Andersen-Jantzen sum formula []), the A 3 -module with the high weight pictured above is in fact the Weyl module except when b p* 2. But p+ b m b = 2 here, so the dimension of V/0 V Q Y is the dimension of the Weyl module, which is b = c b 3 b = 4/ 2. On the other hand, dim V/< V Q Y : dim V/0 V Q X 5 2dim V /0 V Q X : = 2 b = c b = 3. These two equations together imply b 0. So now all the b i are known: b 2, b m 2, and all others are 0. Since one of the coefficients is greater than, we know p 2. And since the T X -high weight of W is δ = δ m, we know dim W m m= 2 or m m= 25*. We claim that dim VV dim i W 2 this is clear if λ λ 2, as then V i 2 W since Y D n and p 2). The T Y -high weight λ of V has a λ 2 -coefficient of, and W is orthogonal. We claim that any B n or D n weight of this form has one of λ 2, λ 3, or λ n as a subdominant weight. For B n, every fundamental weight except λ n is a sum of roots, and 2λ n is a sum of roots, so any λ with a λ 2 -coefficient of has λ 2 or λ n as a subdominant weight. For D n, any fundamental weight λ k for even k n* 2 is a sum of roots; for odd k with k n* 2, λ k differs from a sum of roots by λ 3. The weight 2λ n or 2λ n& is either a sum of roots or differs from one by λ 3 ; finally, λ = λ 2 has λ 3 as a subdominant weight. So the claim holds. Say λ. λ i. Then by the result in [8], every weight which appears in the Weyl module with T Y -high weight λ appears in V. So dim VV card ω is a weight occurring in V Y λ i = card Weyl group-orbit of λ 2 ˆ? dim W 2 * dim W = dim W dim W` Now we have a chain of inequalities the second line is a computation of the dimension of the Weyl module with the specified high weight, using the Weyl dimension formula): dim V 2dim V Am λ 2= 2λ 2 m* m m m= 2 m= 3/ 4 4= 3= 2* m 4m m 6m/ 2 4= 3= 2* m 4m m 6m= 2#/ 2 m m= 25* 2 dim W 2 dim V`

8 BEN FORD which is a contradiction. So we have ruled out all possible configurations, and the proof of the Lemma is complete. 3.4. The General X A m Case. We must prove that there are no triples X Y V with X acting irreducibly on W and t acting on W, for X of type A m + with m 3. We use the same argument as in the A 2 and A 3 cases to limit the possibilities for the embeddings XZ Y, relying on Lemma 3.2. Lemma 3.5 tells us we need not worry about level in the computation of dimensions of U X -levels. As in the small cases, the method we use to generate weights that appear in a representation is simple: If µ is a weight in the X-module M and µ β i_ a+ 0, then µ* aβ i is another weight appearing in M. Every symmetric weight for X A m except δ = δ m has either δ 2= δ m& or δl m$ m T 2 as a subdominant weight δ i= δ m& is a sum of roots for any i$ i m). It is relatively simple to find enough weights in each level for δ δ 2= δ m& : The level l δ of the low weight* δ in this case is 2 m= m* 2V 4 m*, so we must show there are at least three weights at level j for 2 j 2 m*, and at least 5 at level 2 m* ; this is easy. Unless otherwise stated, the levels we discuss are U X -levels. The case δ δl m$ mt 2 m odd) is considerably more difficult. At U X -level 2 there are only the two weights δ* βl m$ m T 2* βl m& mt 2 and δ* βl m$ mt 2* βl m$ 3m T 2; so Lemma 3.2 is of no help here. We have l δ h m= 2/ 4, so if we can show that there are at least 3 weights at each level 3 j^ m= 2/ 8, and at least 5 at level m= 2/ 8 when this is integral), we will only have level 2 to worry about. Notice that for any weight δ, the dimensions and numbers of weights of U X -levels of the X-module with high weight δ are symmetric about level l δ/ 2. In other words, dim W i dim W lδ& i and the same numbers of weights appear in these two spaces, since w 0 interchanges them. So, for instance, if V Al δ has at least 3 weights at all levels j for i j l δ/ 2, then it has at least 3 at all levels j for i j l δ* i. For m 5 7, it is easy to see that there are enough weights at levels 3 through l δ/ 2 three at every level except level 8 for m 7, in which case there are at least 5 weights). We proceed by induction on m considering the subsystem group of X of type A m& 2, corresponding to Π X>* β β m ): Assume δ δl m$ m T 2 for m 9. Notice that δ * δl m$ mt 2= δ m is a weight at level ll δ mž 2 A m[ 2m 3 m* 2/ 4, and that if δl m$ m T 2* m& c 2 in i β i is a weight at level j in the A m& 2-module with high weight δl m$ mt 2, then δl m$ m T 2* m& c in i β i is a weight 2 at level j in W for δ δl m$ m T 2. So, using the comment in the last paragraph and by induction from the A n& 2 case, W has at least three weights at levels 3 through m* 2/ 4 * 2. But m* 2/ 45* 2+ m= 2/ 8 for m+ 7, which shows that the only possibilities for a simple factor of LY of type A are the above-mentioned factor corresponding to U X -level 2, and level m = 2/ 8 when this number is integral. We need to show that there are at least 5 weights at level m= 2/ 8 when 4 m= and δ δl m$ mt 2. It is easy to write down 5 such weights for m 7, as noted above; in particular in this case there are two at this level which are symmetric with respect to the graph automorphism). So assume m and consider the A m& 4-subsystem subgroup of X, corresponding to Π X>* β β 2 β m& β m. Assume there are at least two symmetric weights at level m* 3 2/ 8 for the A m& 4-module with high weight δl m$ m T 2; as in the last paragraph each corresponds to a weight of W at level m* 3 2/ 8. Let γ be one of these