Simple Lie algebras over fields of positive characteristic. II. Classifying the absolute toral rank two case

de Gruyter Expositions in Mathematics 42 Editors V. P. Maslov, Academy of Sciences, Moscow W. D. Neumann, Columbia Uni...

Author: Helmut Strade

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de Gruyter Expositions in Mathematics 42

Editors V. P. Maslov, Academy of Sciences, Moscow W. D. Neumann, Columbia University, New York R. O. Wells, Jr., International University, Bremen

Simple Lie Algebras over Fields of Positive Characteristic II. Classifying the Absolute Toral Rank Two Case by

Helmut Strade

≥ Walter de Gruyter · Berlin · New York

Author Helmut Strade E-mail: [email protected]

Mathematics Subject Classification 2000: 17-02, 17B50, 17B20, 17B05. Key words: Simple Lie algebras, Lie algebras of positive characteristic, divided power algebras, Cartan prolongation, recognition theorems.

앝 Printed on acid-free paper which falls within the guidelines 앪 of the ANSI to ensure permanence and durability.

ISSN 0938-6572 ISBN 978-3-11-019701-3 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. 쑔 Copyright 2009 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage or retrieval system, without permission in writing from the publisher. Typeset using the author’s TEX files: Kay Dimler, Müncheberg. Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen. Cover design: Thomas Bonnie, Hamburg.

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

10 Tori in Hamiltonian and Melikian algebras . . . . . . . . . . 10.1 Determining absolute toral ranks of Hamiltonian algebras 10.2 More on H.2I .1; 2//.2/ Œp . . . . . . . . . . . . . . . . . 10.3 2-dimensional tori in H.2I 1I ˆ.//.1/ . . . . . . . . . . 10.4 Semisimple elements in H.2I 1I ˆ.1//Œp . . . . . . . . . 10.5 Melikian algebras . . . . . . . . . . . . . . . . . . . . . 10.6 Semisimple Lie algebras of absolute toral rank 1 and 2 . 10.7 Weights . . . . . . . . . . . . . . . . . . . . . . . . . .

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11 1-sections . . . . . . . . . . . . . . . . . . . . . . . 11.1 Lie algebras of absolute toral rank 1 . . . . . 11.2 1-sections . . . . . . . . . . . . . . . . . . . 11.3 Representations of dimension < p 2 . . . . . . 11.4 More on H.2I 1/.2/ . . . . . . . . . . . . . . 11.5 Low dimensional representations of H.2I 1/.2/

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121 121 127 131 138 150

12 Sandwich elements and rigid tori 12.1 Deriving identities . . . . . 12.2 Sandwich elements . . . . 12.3 Rigid roots . . . . . . . . . 12.4 Rigid tori . . . . . . . . . 12.5 Trigonalizability . . . . . . 13 Towards graded algebras . . . . 13.1 The pentagon . . . . . . . 13.2 An upper bound . . . . . . 13.3 Filtrations . . . . . . . . . 13.4 More on Hamiltonian roots 13.5 Switching tori . . . . . . .

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242 242 247 253 278 285

vi

Contents

13.6 13.7

Good triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 On the existence of good tori and good triples . . . . . . . . . . . . 303

14 The toral rank 2 case . . . . . 14.1 No root is exceptional . . 14.2 S is not of Cartan type . 14.3 Graded counterexamples

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313 313 326 354

15 Supplements to Volume 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 378 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

Introduction

This three volume monograph on “Simple Lie Algebras over Fields of Positive Characteristic” presents major methods on modular Lie algebras, all the examples of simple Lie algebras over algebraically closed fields of characteristic p 5 and the complete proof of the Classification Theorem mentioned in the introduction of Volume 1. The first volume contains the methods, examples and a first classification result. It turned out during the work on the reproduction of the classification proof that one has to pay for a reasonable completeness by extending the text considerably. So the whole work is now planned as a three volume monograph. This second volume contains the proof of the Classification Theorem for simple Lie algebras of absolute toral rank 2. We have already mentioned details outlining the proof of the Classification Theorem in the introduction of the first volume. Therefore we will just recall very briefly some strategy in order to place the content of this volume into the whole picture. Already in the early work on simple Lie algebras over the P complex numbers people determined, as a general procedure, 1-sections L.˛/ WD i 2Fp LP i˛ .H / with respect to a toral CSA H , described their representations in the spaces i 2Fp Lˇ Ci˛ .H /, and determined 2-sections. The breakthrough paper [B-W88] made this procedure work for modular Lie algebras as well (if the characteristic is bigger than 7). It turned out, however, that the many more examples and the richness of their structures made things much more involved. Imagine that in the classical case only sl.2/ ˚ H \ ker ˛ occurs as the 1-section L.˛/, while in the modular case the classical algebra sl.2/, the smallest Witt algebra W .1I 1/ and the smallest Hamiltonian algebra H.2I 1/.2/ have absolute toral rank 1 (by Corollaries 7.5.2 and 7.5.9). Hence each such algebra is a 1-section of itself. There are nonsplit radical extensions of these Cartan type Lie algebras, and it is a priori not clear which of these can occur as 1-sections of simple Lie algebras. Moreover, the representation theories of such extensions are very rich, and therefore the very details of these theories can hardly be described. Less information is, fortunately, sufficient for the Classification Theory. Namely, it is sufficient and possible to describe semisimple quotients LŒ˛ WD L.˛/= rad L.˛/ with respect to certain tori T (we have to decide which tori we take into consideration though), and it is also possible to describe the T -semisimple quotients of 2-sections in terms of simple Lie algebras of absolute toral rank 1 and 2 by Block’s Theorem Corollary 3.3.6. If one knows the simple Lie algebras of absolute toral rank not bigger than 2 one is able

2

Introduction

to describe all such semisimple quotients of 2-sections in these terms. In this second volume we will prove the following Theorem. Every simple Lie algebra over an algebraically closed field of characteristic p > 3 having absolute toral rank 2 is exactly one of the following: (a) classical of type A2 , B2 or G2 ; (b) the restricted Lie algebras W .2I 1/, S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/; the naturally graded Lie algebras W .1I 2/, H.2I .1; 2//.2/ ; H.2I 1I ˆ.//.1/ , H.2I 1I ˆ.1//; (c) the Melikian algebra M.1; 1/. Since we classified the simple Lie algebras of absolute toral rank 1 in Chapter 9 of Volume 1, the result of this second volume will provide sufficient information on the 2-sections of simple Lie algebras with respect to adequate tori. The proof of the Classification Theorem for simple Lie algebras of absolute toral rank 2 is completely different from what we have done in Chapter 9 of Volume 1 and what one has to do in Volume 3 for the general case. In fact, when writing this text I have changed some of the original items, so that even more the description in the introduction of the first volume does not correctly describe the present procedure. Let me say a few words about the sources for the proofs of this volume and the citation policy. The breakthrough paper [B-W88] gave the general procedure and provided many ideas for the solution of the absolute toral rank 2 case. In the present exposition I stressed the point of using sandwich elements and graded algebras in combination with the Block–Weisfeiler description of these. The major contribution of sandwich element methods is due to A. A. P REMET. Most of the other material can be found in the papers [P-S 97]–[P-S 01]. I will not quote these results in detail. If the reader is interested in the original sources he should look into [B-W88], [Pre 85]–[Pre 94] and [P-S 97]–[P-S 01]. Chapter 10, the first chapter of this volume, is somewhat different from the rest. In that chapter we determine which of the Cartan type and Melikian algebras have absolute toral rank 2, determine automorphism groups of these algebras, describe orbits of toral elements in the minimal p-envelope under the automorphism group, compute the centralizers of toral elements and estimate the number of weights on restricted modules. In doing this we decover a lot of details of the structure of these algebras. This already indicates a weakness of the theory: at present one needs really much information on the algebra structures to apply some sophisticated arguments. The notations in this volume and all references to Chapter 1–9 refer to the first volume. As a general assumption, F always denotes an algebraically closed field of characteristic p > 3 (while in the first volume we also included the case p D 3), and all algebras are regarded to be algebras over F .

Chapter 10

Tori in Hamiltonian and Melikian algebras

In this chapter we determine the Hamiltonian and Melikian algebras having absolute toral rank not bigger than 2. For those algebras the automorphism groups are determined and conjugacy classes of tori in the minimal p-envelopes are described. We gain structural insight in root spaces and sections with respect to 2-dimensional tori.

10.1

Determining absolute toral ranks of Hamiltonian algebras

We remind the reader to the concept of an absolute toral rank of a Lie algebra g. Namely, let g be a Lie algebra with subalgebra q and let .gŒp ; Œp; / be any penvelope of g. In general we suppress the notion of , regard g as a subalgebra of gŒp and just say that gŒp is a p-envelope of g. The p-envelope of q in gŒp is the restricted subalgebra of gŒp generated by .q/ and is denoted by qŒp . By Definition 1.2.1 the toral rank TR.q; g/ of q in g equals the maximum of dimensions of tori in qŒp C C.gŒp /=C.gŒp /. Due to [S-F, Theorem 2.4.5] then TR.q; g/ equals the maximum of dim t=t \ C.gŒp / for tori t qŒp . Accordingly, the absolute toral rank of g, denoted by TR.g/, equals the maximum of dim t=t \ C.gŒp / for tori t in gŒp . These definitions do not depend on the p-envelope chosen (Theorem 1.1.7). Suppose g is centerless and gŒp denotes a minimal p-envelope (Definition 1.1.2). By definition, gŒp is centerless as well. In the Notation 1.2.5, TR.g/ D MT.gŒp / is the maximal possible dimension of tori in gŒp . Theorem 1.1.7 proves that the minimal p-envelope is unique as a restricted Lie algebra. It is stated in Volume 1, p. 299 that every simple finite dimensional Lie algebra of Cartan type can be viewed as an algebra X.mI nI '/.1/ W .mI n/ for some X D W; S; H; K and ' 2 Autc O..m// satisfying the compatibility property (6.2.2). Throughout this volume we always will without further mentioning assume that Cartan type Lie algebras X.mI nI '/.1/ satisfy the compatibility property (6.2.2). The algebras W .m/, S.m/, H.2r/ and K.2rC1/ carry natural gradings (Volume 1, p. 186). Correspondingly, the commutative algebra O.m/ and the space of differential forms .m/ are graded. Every Cartan type Lie algebra X.mI nI '/.1/ has a uniquely determined subalgebra X.mI nI '/.1/ .0/ of minimal codimension (Theorem 4.2.6)

4

10


and a natural filtration (Definition 4.2.8). This filtration is compatible with the grading in case of graded algebras. Next we remind the reader to the deliberations of Chapters 6.3–6.5. Consider U.2rI n/ D ¹u 2 O..2r// j u 1 du 2 1 .2rI n/; u.0/ D 1º (see Volume 1, p. 302). P By Definition 6.4.1 a Hamiltonian form subordinate to O.2rI n/ is a form ! D 2r fj;i , d! D 0, det .fi;j / is invertible i;j D1 fi;j dxi ^ dxj where fi;j D in O..2r// and fi;j 2 uO.2rI n/ for some u 2 U.2rI n/. Such a form is of “first type” if one can take u D 1 and of “second type” otherwise. Due to Theorem 6.4.11 any Hamiltonian form ! subordinate to O.2rI n/ of second type is conjugate under Autc O.2rI n/ to a form 1 .!H;l / where is a permutation of indices and !H;l D .p nl / P2r / j D1 .j /xj dxj 0 , and one may take D Id if the homogeneous d exp .xl P degree 2 part of ! is j2rD1 .j /dxj ^ dxj 0 D 2!H . For a Hamiltonian form ! subordinate to O.2rI n/ we put (Volume 1, p. 337) H.2rI nI !/ WD ¹D 2 W .2rI n/ j D.!/ D 0º and observe that H.2rI nI '/ D H.2rI nI '.!H // (Volume 1, p. 337, Equation (6.5.3)) holds. Theorem 6.5.7(2) and Corollary 6.5.9 show that every simple Hamiltonian algebra satisfies H.2rI nI '/.1/ D H.2rI nI '/.2/ . Let us call Hamiltonian forms !; ! 0 2 2 .2rI n/ similar Hamiltonian forms and write ! ! 0 , if 9 ˛ 2 F ;

2 Autc O.2rI n/; 2 1 .2rI n/.3/ W

.!/ D ˛! 0 C d:

Lemma 10.1.1. Hamiltonian forms ! D '.!H / and ! 0 D ' 0 .!H / subordinate to O.2rI n/ are similar if and only if H.2rI nI '/.2/ Š H.2rI nI ' 0 /.2/ . In this case there is Q 2 Autc O.2rI n/ such that the isomorphism can be given as D 7! Q ı D ı Q 1 . Proof. (a) Suppose !; ! 0 are similar and , ˛, are as in the definition such that .!/ D ˛! 0 C d. As d is a coboundary whose homogeneous summands have degree 3, ˛! 0 C d and ˛! 0 determine the same element in the second cohomology group and the homogeneous degree 2 parts coincide. Therefore ˛! 0 D 0 .˛! 0 C d/ for some 0 2 Autc O.2rI n/ (Theorem 6.4.9). Consequently, . 0 ı /.!/ D ˛! 0 . Then H.2rI nI '/.2/ Š H.2rI nI ' 0 /.2/ by the isomorphism D 7! . 0 ı / ı D ı . 0 ı / 1. 6.5.4 shows that there are (b) Suppose H.2rI nI '/.2/ Š H.2rI nI ' 0 /.2/ . Theorem ˛ 2 F and 2 Autc O.2rI n/ such that '.!H / D ˛' 0 .!H /. Then '.!H / and ' 0 .!H / are similar. In this section we are going to determine lower bounds for the absolute toral rank of Hamiltonian algebras.

10.1

5


Lemma 10.1.2. For every simple Lie algebra of Hamiltonian type one has the identity TR.H.2rI nI '/.2/ / D TR.H.2rI nI '//. Proof. The compatibility property (6.2.2) gives H.2rI n/.2/ gr H.2rI nI '/.2/ gr H.2rI nI '/ ,! H.2rI n/ .2/

D H.2rI n/

C

2r X

.p ni /

FDH .xi

/ C FDH .x ..n// /:

i D1

Therefore ad H.2rI nI '/ ad H.2rI nI '/.2/ C Der H.2rI nI '/ .p

2/

;

where Der H.2rI nI '/ .p 2/ is the set of derivations which map H.2rI nI '/.k/ into H.2rI nI '/.kCp 2/ (for all k; we considerthe natural filtration). Since H.2rI nI '/Œp is a restricted ideal and Der H.2rI nI '/ .p 2/ is a restricted subalgebra in the restricted Lie algebra Der H.2rI nI '/, Jacobson’s formula on p-th powers yields H.2rI nI '/Œp H.2rI nI '/.2/ Œp C Der H.2rI nI '/ .p 2/ for the respective p-envelopes in Der H.2rI nI '/. Note that these p-envelopes are centerless. As Der H.2rI nI '/ .p 2/ clearly is a Œp-nilpotent algebra, one gets TR.H.2rI nI '// D MT.H.2rI nI '/Œp / D MT.H.2rI nI '/.2/ Œp / D TR.H.2rI nI '/.2/ /:

Lemma 10.1.3. n

(1) The element t WD .1 C x .p 1/ /@ 2 W .1I n/ is semisimple and its minimal n X. polynomial in End O.1I n/ is X p P i (2) Suppose D 2 H.2I n/ has minimal polynomial P .X/ D ˛i X p as an element of End O.2I n/. Then P .X/ is also the minimal p-polynomial of ad D as an element of Der H.2I n/. Proof. (1) Set f.n/;˛ .x/ WD

n 1 pX

˛ i x .i/ ;

˛ 2 Fpn :

(10.1.1)

i D0 n

Since ˛ p D ˛, one has t.f.n/;˛ / D

n 1 pX

i D1

˛ i x .i

1/

C ˛x .p

n

1/

D ˛f.n/;˛ :

(10.1.2)

6

10


Consequently, t has p n different eigenvalues on O.1I n/, and this shows that t acts n semisimply on O.1I n/ and that the minimal polynomial of t is X p X. i i (2) Let d 2 H.2I n/ be arbitrary. Since ŒD p ; d D .ad D/p .d /, one has P i that ˛i .ad D/p .d / annihilates O.2I n/. Therefore it represents the 0-element of H.2I n/. As this is true for all d , P .ad D/ is the 0-element of Der H.2I n/. P j Suppose conversely that ˇj .ad D/p is the 0-element of Der H.2I n/. Put E WD P j ˇj D p . Then E defines a module endomorphism of the H.2I n/-module O.2I n/. ..n// is the only subspace of H.2I n/.0/ O.2I n/ , which is annihilated Note that F x by H.2I n/.0/ . Then E.x ..n// / D x ..n// for some 2 F . Since @1 ; @2 are contained in H.2I n/, it is immediate that E acts as Id on O.2I n/. On the other hand, P j E.1/ D 0, and this gives D 0. As a consequence, P .X/ divides ˇj X p . Theorem 10.1.4. The following holds: (1) TR.H.2I .n1 ; n2 //.2/ / D n1 C n2 (2) TR.H.2I .n1 ; n2

/I ˆ.//.1/ /

1.

D n1 C n2 .

(3) TR.H.2I .n1 ; n2 /I ˆ.l/// D n1 C n2 . Proof. (1) There is nothing to prove if n1 D n2 D 1 (Corollary 7.5.9). So assume n2 > 1. With the notation of Equation (10.1.1) we consider the form ! WD f.n1 /;1 .x1 /f.n2

1/;1 .x2 /df.n1 /; 1 .x1 /

^ df.n2

1/; 1 .x2 /

2 2 .2I .n1 ; n2

1//:

By definition ! D 11d. x1 /^d. x2 /Cgdx1 ^dx2 D dx1 ^dx2 Cgdx1 ^dx2 for suitable g 2 O.2I .n1 ; n2 1//.1/ . Choose gQ 2 O.2I .n1 ; n2 //.2/ for which @2 .g/ Q D g. Then ! D dx1 ^ dx2 C d.gdx Q 1 / which means that ! !H . Lemma 10.1.1 yields the existence of Q 2 Autc O.2I n/ such that Q ı H.2I .n1 ; n2 /I !/ ı Q

1

D H.2I .n1 ; n2 //:

Applying Lemma 10.1.2 we intend to show that there is torus of dimension n1 Cn2 1 in the minimal p-envelope H.2I .n1 ; n2 /I !/Œp Der H.2I .n1 ; n2 /I !/. Such a torus is given by Lemma 10.1.3. Namely, the elements .p n1 1/

t1 WD .1 C x1

.p n2

/@1 ; t2 WD .1 C x2

1

1/

/@2 2 W .2I .n1 ; n2

1//

annihilate ! by Equation (10.1.2). Set tQi WD Q ı ti ı Q 1 2 H.2I .n1 ; n2 //. By n n 1 Lemma 10.1.3 t1 and t2 have minimal polynomial X p 1 X and X p 2 X in End O.2I .n1 ; n2 1//, respectively, and ad tQ1 , ad tQ2 have the same minimal p-polynomials as elements of Der H.2I .n1 ; n2 1//. Then ad tQ1 and ad tQ2 have at least p n1 and p n2 1 different eigenvalues on H.2I .n1 ; n2 1// H.2I .n1 ; n2 //, and therefore the (unique) maximal torus in the abelian restricted subalgebra of Der H.2I .n1 ; n2 // generated by F tQ1 ˚ F tQ2 has dimension n1 C n2 1. Consequently, TR.H.2I .n1 ; n2 //.2/ / D TR.H.2I .n1 ; n2 // n1 C n2

1:

10.1

7


On the other hand, since Der H.2I .n1 ; n2 //.2/ =H.2I .n1 ; n2 //.2/ Œp contains the image of F .x1 @1 C x2 @2 / as a 1-dimensional torus, one gets n1 C n2 D TR.W .2I .n1 ; n2 /// D MT.W .2I .n1 ; n2 //Œp / MT F .x1 @1 C x2 @2 / ˚ H.2I .n1 ; n2 //.2/ Œp D 1 C MT H.2I .n1 ; n2 //.2/ Œp n1 C n2 : This proves the claim. (2) Proceed as in (1) setting ! 0 WD f.n1 /;1 .x1 /f.n2 /;1 .x2 /df.n1 /; 1 .x1 / ^ df.n2 /; 1 .x2 / 2 2 .2I .n1 ; n2 //: Arguing as in (1) one obtains TR.H.2I .n1 ; n2 /I ! 0 /.2/ / n1 C n2 : Note that in the case of 2 generators the Hamiltonian form and the volume form are the same. Theorem 6.3.4 shows that ! 0 is Autc O.2I .n1 ; n2 //-conjugate to one of !H ; ˛.1 C x ..n// /!H .˛ 2 F /;

.p ni /

exp .xi

/!H D !H;i

.i D 1; 2/:

As ! 0 is a Hamiltonian form subordinate to O.2rI .n1 ; n2 // of first type, it cannot be conjugate to !H;i by Theorem 6.4.11(1). Then H.2I .n1 ; n2 /I ! 0 /.2/ is isomorphic to H.2I .n1 ; n2 //.2/ or to H.2I .n1 ; n2 /I ˆ.//.1/ (see Theorem 6.3.10). In addition, H.2I .n1 ; n2 /I ! 0 /.2/ is not isomorphic to H.2I .n1 ; n2 //.2/ , because that one has smaller absolute toral rank by the first part of this theorem. Therefore we have H.2I .n1 ; n2 /I ! 0 /.2/ Š H.2I .n1 ; n2 /I ˆ.//.1/ , whence TR.H.2I .n1 ; n2 /I ˆ.//.1/ / n1 C n2 : The reverse inclusion follows from the inequality TR.H.2I .n1 ; n2 /I ˆ.//.1/ / TR.W .2I .n1 ; n2 /// D n1 C n2 : (3) Recall that H.2I .n1 ; n2 /I ˆ.l// D H.2I .n1 ; n2 /I !H;l /. Theorem 7.2.2(7) describes the minimal p-envelope of H.2I .n1 ; n2 /I ˆ.l//.1/ . Due to that result one has F .x1 @1 C x2 @2 / ˚ H.2I .n1 ; n2 //.2/ gr H.2I .n1 ; n2 /I ˆ.l//.1/ Œp : By Proposition 1.4.5, TR.H.2I .n1 ; n2 /I ˆ.l//.1/ / D MT H.2I .n1 ; n2 /I ˆ.l//.1/ Œp

MT gr H.2I .n1 ; n2 /I ˆ.l//.1/ Œp

1 C TR.H.2I .n1 ; n2 //.2/ / D n1 C n2 :

8

10


As in (2), the reverse inequality holds. Finally, observe that H.2I .n1 ; n2 /I ˆ.l//.1/ D H.2I .n1 ; n2 /I ˆ.l// because the latter is already simple. The great variety of simple Hamiltonian algebras with more than two generators only allows partial results for the general case. Theorem 10.1.5. (1) If '.!H / is subordinate to O.2rI n/ of first type, then .2/

TR.H.2rI nI '/

/

2r X

ni

r:

i D1

(2) If '.!H / is subordinate to O.2rI n/ of second type, then TR.H.2rI nI '/.2/ /

2r X

ni

r C 1:

i D1

Proof. (1) Using the compatibility property (6.2.2) and Theorem 1.4.6 one gets TR.H.2rI nI '/.2/ / TR.H.2rI n/.2/ /: Note that H.2rI n/.2/ contains a subalgebra isomorphic to ˚riD1 H.2I .ni ; ni Cr //.2/ . Theorem 1.2.8(5) and Theorem 10.1.4(1) give TR.H.2rI nI '/.2/ / TR ˚riD1 H.2I .ni ; ni Cr //.2/ D

r X

.2/

TR H.2I .ni ; ni Cr //

D

i D1

2r X

nj

r:

j D1

(2) Theorem 6.4.11(3) shows that '.!H / is similar to 1 .!H;l / for some permutation of indices and some l. Lemma 10.1.1 then gives rise to an isomorphism H.2rI nI '/.1/ Š H.2rI n0 I !H;l /.1/ where n0 is obtained from n by a permutation of indices. Theorem 7.2.2(7) describes the minimal p-envelope of H.2rI n0 I !H;l /.1/ . Due to that result one has F

2r X

xi @i ˚ H.2rI n0 /.2/ gr H.2rI n0 I !H;l /.1/ Œp :

i D1

By Proposition 1.4.5, TR.H.2rI n0 I !H;l /.1/ / D MT H.2rI n0 I !H;l /.1/ Œp

MT gr H.2rI n0 I !H;l /.1/ Œp 1C

2r X j D1

n0j

rD

2r X j D1

nj

1 C TR.H.2rI n0 /.2/ /

r C 1:

10.1

9


As a consequence, TR.H.2rI nI '/.2/ / > 2 except if r D 1 or r D 2, n D 1 and '.!H / 2 2 .4I 1/. We are going to investigate this last case. Lemma 10.1.6. Let u; v; w 2 O.2rI n/, ˛; ˛ 0 2 F . Then the following holds: (1) .uCw/p 1 v p 1 d.uCw/^dv D up 1 v p 1 du^dv Cw p 1 v p 1 dw ^dv Cd for some 2 1 .2rI n/.2p/ . (2) .˛ C ˛ 0 up 1 v p 1 /du ^ dv D .1 C u/p d0 for some 0 2 1 .2rI n/.3/ .

1 .˛ 1=.p 1/ C ˛ 01=.p 1/ v/p 1 du ^ dv C

Proof. (1) Observe that .u C w/p

1

d.u C w/ D up

1

du C w p

dw C up 1 dw C w p ! p X2 p 1 ui w p 1 i du C i i D1 ! p X2 p 1 C ui w p 1 i dw i 1

1

du

i D1

Du Pp

p 1 1 i p i i 1 iu w

1 i D1

where g WD

p 1

.u C w/p

1 p 1

v

D .up

1

du C w p

1

dw C dg; 2 O.2rI n/.p/ (as p i 1 D . 1/i ). This gives

d.u C w/ ^ dv

du C w p

1

dw/ ^ .v p

1

dv/ C .dg/ ^ .v p

1

dv/:

gv p 1 dv.

Set WD (2) Clearly, for ˇ; ˇ 0 2 F , .ˇ C ˇ 0 v/p 1 dv D .ˇ p 1 C ˇ 0p 1 v p 1 /dv C dhˇ;ˇ 0 .v/; Pp 1 where hˇ;ˇ 0 .v/ WD iD2 pi 11 1i ˇ p i ˇ 0i 1 v i 2 O.2rI n/.2/ . Applying this twice gives .1 C u/p

1

.ˇ C ˇ 0 v/p

D .1 C u D ˇp

1

p 1

1

du ^ dv

/du C dh1;1 .u/ ^ .ˇ p

du ^ dv C du ^ ˇ 0p

C ˇ 0p

1 p 1 p 1

u

v

D .ˇ

Cˇ

C d ˇ 0p

1

C ˇ 0p

0p 1 p 1 p 1

u

1

uv p

C d h1;1 .u/.ˇ p

v

1

v

1 p

v

1

1 p 1

v

du ^ ˇ p

/dv C dhˇ;ˇ 0 .v/

1

dv C dhˇ;ˇ 0 .v/ 1 /dv C dhˇ;ˇ 0 .v/

/du ^ dv

dv C udhˇ;ˇ 0 .v/ 1

C ˇ 0p

dv C dhˇ;ˇ 0 .v/

du ^ dv C up

C dh1;1 .u/ ^ .ˇ p p 1

1 p 1

1

C ˇ 0p

1 p

v

d up

1

.ˇ p

1

v C hˇ;ˇ 0 .v//du 1 /dv C d h1;1 .u/dhˇ;ˇ 0 .v/ :

10

10


Set ˇ WD ˛ 1=.p 1/ , ˇ 0 WD ˛ 01=.p 1/ and 0 WD ˇ 0p 1 uv p 1 dv udhˇ;ˇ 0 .v/ C up 1 .ˇ p 1 v C hˇ;ˇ 0 .v//du h1;1 .u/.ˇ p 1 C ˇ 0p 1 v p 1 /dv h1;1 .u/dhˇ;ˇ 0 .v/. The first part of the preceding lemma says that the mapping .u; v/ 7! up

1 p 1

v

du ^ dv

.mod d 1 .2rI n/.3/ /

is p-semilinear in both arguments. Lemma 10.1.7. Every Hamiltonian form ! 2 2 .4I 1/ is similar to dx1 ^ x3 C dx2 ^ dx4 or to a form of type X

˛i;j .1 C xi /p

1

.1 C xj /p

1

dxi ^ dxj

1i<j 4; .i;j /¤.2;4/

C .1 C x2 /p

1

0 .˛2;4 C ˛2;4 x4 /p

Proof. Write according to Proposition 6.4.4 X p 1 p .ai;j C bi;j xi xj !D

1

1

dx2 ^ dx4 :

/dxi ^ dxj C d;

1i<j 4

where 2 1 .4I 1/.3/ . Set ai;j WD aj;i and bi;j WD bj;i if i > j and ai;i WD bi;i WD 0. As ! is a Hamiltonian form, det.ai;j / ¤ 0. (a) Suppose bi;j D 0 for all i; j . There are indices k; l for which ak;l ¤ 0. Permuting indices defines an automorphism of O.4I 1/. Therefore we may take k D 1 and l D 3. Next put u WD a1;3 x1 C a2;3 x2 C a4;3 x4 ; and define

v WD x3 C a1;31 a1;2 x2 C a1;31 a1;4 x4

2 Aut O.4I 1/ by setting .x1 / WD u;

.x2 / WD x2 ;

.x3 / WD v;

.x4 / WD x4 :

Then ! D du ^ dv C dx2 ^ dx4 C d D .dx1 ^ dx3 C dx2 ^ dx4 C d 1 .// for some 2 F . Since det.ai;j / ¤ 0, one has ¤ 0. The substitution x2 7! 1 x2 defines an automorphism of O.4I 1/. This proves the claim in the present case. (b) Suppose there are indices k; l for which ak;l bk;l ¤ 0. Proceeding as in (a) we may assume k D 1, l D 3. Let 2 F be a solution of the equation b1;3 X p 1 D a1;3 . The substitution x3 7! x3 defines an automorphism of O.4I 1/ and leads us to the case a1;3 D b1;3 . Since ! and a1;31 ! are similar, we may assume a1;3 D b1;3 D 1.

10.1

11


Choose i;j 2 F satisfying p

1;j C a1;j D 1;j C b1;j ;

j D 2; 4;

p

3;2 C a2;3 D 3;2 C b2;3 ; p

3;4 C a3;4 D

3;4 C b3;4 :

Set u WD x1 C 3;2 x2 C 3;4 x4 ;

v WD x3 C 1;2 x2 C 1;4 x4 :

The p-semilinearity in Lemma 10.1.6(1) gives the following congruence modulo p 1 p 1 x4 dx2

d 1 .4I 1/.3/ C F dx2 ^ dx4 C F x2 du ^ dv C up

1 p 1

v

^ dx4 W

du ^ dv 1 p 1 x2 /du

C .a1;2 C b1;2 up

p 1 p 1

C .a2;3 C b2;3 x2

v

1 p 1 x4 /du

^ dx4

1 p 1 x4 /dv

^ dx4

^ dx2 C .a1;4 C b1;4 up

/dx2 ^ dv C .a3;4 C b3;4 v p

dx1 ^ dx3 C 1;2 dx1 ^ dx2 C 1;4 dx1 ^ dx4 C 3;2 dx2 ^ dx3 p 1 p 1 x3 dx1

C x1

^ dx3

p

p 1 p 1 x2 dx1

^ dx2 C 1;4 x1

p

p 1 p 1 x3 dx2

^ dx3

C 1;2 x1 C 3;2 x2

^ dx4

p

p 1 p 1 x4 dx3

^ dx4

3;4 x3

p 1 p 1 x3 /dx2

^ dx3 C .a3;4 C b3;4 x3

p 1 p 1 x3 /dx1

D .1 C x1

p

p 1 p 1 x2

C a1;2 C b1;2 x1

p

p 1 p 1 x4

C a1;4 C b1;4 x1

C .1;4 C 1;4 x1

p p 1 p 1 3;2 x2 x3 p

C a2;3 C

p 1 p 1 x4

3;4 x3

p 1 p 1 x3 /dx1

p 1 p 1 x4 /dx1

^ dx4

p 1 p 1 x4 /dx3

^ dx4

^ dx3

C .1;2 C 1;2 x1

D .1 C x1

p 1 p 1 x4 dx1

^ dx2 C .a1;4 C b1;4 x1

C .a2;3 C b2;3 x2

C . 3;4

p

p 1 p 1 x2 /dx1

C .a1;2 C b1;2 x1

C .3;2 C

3;4 dx3 ^ dx4

p 1 p 1 x2 /dx1

^ dx2

p 1 p 1 x4 /dx1

^ dx4

p 1 p 1 b2;3 x2 x3 /dx2

^ dx3

p 1 p 1 x4 /dx3

C a3;4 C b3;4 x3

^ dx4

p 1 p 1 x2 /dx1

^ dx3 C .1;2 C a1;2 /.1 C x1 p 1 p 1 x4 //dx1

C .1;4 C a1;4 /.1 C x1

p 1 p 1 x3 /dx2

C .3;2 C a2;3 /.1 C x2

^ dx4

^ dx3

p 1 p 1 x4 /dx3

C . 3;4 C a3;4 /.1 C x3

^ dx4 :

^ dx2

12 Define

10


2 Aut O.4I 1/by setting .x1 / WD u;

.x2 / WD x2 ;

.x3 / WD v;

The former computation yields X p .!/ D ˛i;j .1 C xi

.x4 / WD x4 :

1 p 1 xj //dxi

^ dxj

1i<j 4; .i;j /¤.2;4/

p 1 p 1 x4 /dx2

0 C .ˇ2;4 C ˇ2;4 x2

^ dx4 C d0

where 0 2 .4I 1/.3/ . Next apply Lemma 10.1.6(2) to prove the claim in this case. (c) After these considerations we assume that (a) is not true, i.e., bk;l ¤ 0 for some indices k; l. Since det.ai;j / ¤ 0, there is an index s for which ak;s ¤ 0. If ak;l ¤ 0, then we are in the former case. So assume ak;l D 0. Permuting indices we may take k D 1, l D 3, s D 2 so that a1;2 ¤ 0, a1;3 D 0, b1;3 ¤ 0. Choose 2 F such that b1;3 C p b1;2 ¤ 0. Define 2 Aut O.4I 1/ by setting .x2 / WD x2 C x3 ;

.xi / WD xi otherwise:

Using Lemma 10.1.6(1) one gets p 1 p 1 x3 /dx1

^ dx3

p 1 p 1 x4 /dx3

^ dx4 C d Q

.!/ D ! C .a1;2 C p b1;2 x1 C .a2;4 C p b2;4 x3

with Q 2 1 .4I 1/.3/ . The coefficient of dx1 ^ dx3 in .!/ p 1 p 1 p b1;2 /x1 x3 : This brings us to the former Case (b).

d Q is a1;2 C .b1;3 C

Corollary 10.1.8. The simple Lie algebras of Hamiltonian type having absolute toral rank 2 are H.2I .1; 2//.2/ ;

H.2I 1I ˆ.//.1/ ;

H.2I 1I ˆ.1//;

H.4I 1/.2/ :

Proof. (a) Suppose L is a Lie algebra of Hamiltonian type in 2 generators, i.e., L is of the form H.2I n/.2/ , H.2I nI ˆ.//.1/ , H.2I nI ˆ.l// (Theorem 6.3.10). Since TR.H.2I 1/.2/ / D 1 (Corollary 7.5.9), Theorem 10.1.4 yields that only one of the algebras listed in the statement or H.2I .2; 1//.2/ Š H.2I .1; 2//.2/ or H.2I 1I ˆ.2// are possible. However, Theorem 6.3.10 shows that H.2I 1I ˆ.2// and H.2I 1I ˆ.1// are isomorphic. (b) Suppose L is a Lie algebra of Hamiltonian type in more than 2 generators, i.e., L Š H.2rI nI '/.2/ and r 2. Theorem 10.1.5 shows that only r D 2, n D 1 are possible, and, moreover, L Š H.4I 1I '.!H //.2/ where '.!H / is subordinate to O.4I 1/ of first type. Then '.!H / 2 2 .4I 1/. Due to Lemma 10.1.7 '.!H / is

10.2

More on H.2I .1; 2//.2/ Œp

similar either to !H and then L Š H.4I 1/.2/ , or to a form

13 P

1i <j 4; .i;j /¤.2;4/

˛i;j .1 C

p 1 0 /dx2 ^ dx4 . This latter ^ dxj C .1 C x2 2;4 C ˛2;4 x4 form is annihilated by the torus T WD F .1 C x1 /@1 ˚ F .1 C x2 /@2 ˚ F .1 C x3 /@3 . Therefore T H.2rI nI '/ and hence H.2rI nI '/.2/ has absolute toral rank > 2 (see

xi /p 1 .1 C xj /p 1 /dxi

/p 1 .˛

Lemma 10.1.2). (c) Theorem 10.1.4 and Corollary 7.5.9 show that these algebras indeed have absolute toral rank 2.

10.2

More on H.2I .1; 2//.2/ Œp

In the next three sections we will derive several structural details concerning 2-dimensional tori and sections with respect to these tori in one of the Hamiltonian algebras H.2I .1; 2//.2/ , H.2I 1I ˆ.//.1/ and H.2I 1I ˆ.1// of absolute toral rank 2. In dealing with p-envelopes of Cartan type Lie algebras one has to be a bit careful. The compatibility property (6.2.2) tells us that every simple Cartan type Lie algebra X.mI nI '/.1/ may be viewed as a subalgebra of W .mI n/, and the latter is a subalgebra of Der O..m//, where O..m// is the algebra of divided power series in m indeterminates. Considering this Lie algebra together with the natural p-mapping D 7! D p of associative p-th powers in Der O..m// gives naturally a p-envelope of X.mI nI '/.1/ . This p-envelope differs from the minimal p-envelope of X.mI nI '/.1/ . Recall that the latter can be realized as the subalgebra of Der X.mI nI '/.1/ generated by ad X.mI nI '/.1/ and associative p-th powers. To distinguish p-th powers let us write Œp for associative p-th powers in Der X.mI nI '/.1/ . Note that ad D p D Œp annihilates X.mI nI '/.1/ for every D 2 X.mI nI '/.1/ . Lemma 7.1.1(3) states that the natural maximal subalgebra X.mI n/.0/ of X.mI n/ is closed under p-powers in Der O..m//. Then, for D 2 X.mI nI '/.0/ , D p 2 X.mI nI '/.0/ and therefore ad D p D Œp 2 X.mI nI '/Œp . As X.mI nI '/Œp is centerless we may identify D p and D Œp if D 2 X.mI nI '/.0/ . So there is no ambiguity for such elements. However, for elements outside X.mI nI '/.0/ this is no longer true. Let us specify these observations to the case H.2I .1; 2//.2/ , in which we are interested in this section. Theorem 7.2.2 states that H.2I .1; 2//.2/ Œp D H.2I .1; 2//.2/ C p F @2 where we identified elements D and ad D. The interpretation of this result is Œp p p2 p the following: if D 2 H.2I .1; 2//.2/ .0/ , then D Œp D D p ; @2 D @2 ; and as @1 ; @2 Œp Œp2 centralize the algebra, @1 D @2 D 0. We intend to describe orbits of 2-dimensional tori in H.2I .1; 2//.2/ Œp under the automorphism group of H.2I .1; 2//.2/ Œp . Note that every automorphism ‰ of H.2I .1; 2//.2/ Œp induces (by restriction) an automorphism ‰ 0 of H.2I .1; 2//.2/ . Theorem 7.3.2 tells us that there is a continuous divided power automorphism of O..2// which stabilizes O.2I .1; 2// such that .dx1 ^ dx2 / 2 F dx1 ^ dx2 and

14

10


‰ 0 .D/ D ı D ı 1 . Then ‰ 0 extends to an automorphism of H.2I .1; 2//. Conversely, each automorphism of H.2I .1; 2// defines an automorphism of H.2I .1; 2//.2/ by restriction and a restricted automorphism of H.2I .1; 2//.2/ Œp Der H.2I .1; 2//.2/ by extension. We therefore will not distinguish between automorphisms of these three Lie algebras. For any finite dimensional Lie algebra g with separating filtration g D g. g.s/ ¹0º set Aut.i/ g WD ¹‰ 2 Aut g j .‰

r/

Id/.g.k/ / g.i Ck/ for all kº

and Aut.0/ g WD ¹‰ 2 Aut g j ‰.g.k/ / g.k/ for all kº: P Assume, moreover, that g D ˚ gi is graded (with induced filtration g.j / WD i j gi ). Set Aut0 g WD ¹‰ 2 Aut g j ‰.gk / gk for all kº: Regarding Aut.0/ g as a group of linear endomorphisms of the graded space g every ‰ 2 Aut.0/ g has a decomposition as a sum of homogeneous linear mappings and all P homogeneous parts of ‰ have nonnegative degree, ‰ D ‰0 .Id C ll0 ‰l /. It is straightforward that ‰0 2 Aut0 g and ‰l0 2 Der g. As a result, Aut.0/ g D Aut0 g Ë Aut.1/ g and for i 1 there is an assignment ‚i W Aut.i/ g=Aut.iC1/ g ! Deri g; ‚i .‰/ D ‰i : If ‰; ‰ 0 2 Aut.i/ g, then obviously ‚i .‰ ı ‰ 0 / D ‚i .‰/ C ‚i .‰ 0 /: If ‰ 2 Aut.i / g, ‰ 0 2 Aut.j / g, then ‰ı‰ 0 ı‰

1

ı‰ 0

1

whence ‰ ı ‰ 0 ı ‰

Id D Œ‰ Id; ‰ 0 Idı‰ 1

ı ‰0

1

1

ı‰ 0

1

D Œ‰ Id; ‰ 0 Idı.IdC /;

2 Aut.iCj / g and

‚i Cj .‰ ı ‰ 0 ı ‰

1

ı ‰0

1

/ D Œ‚i .‰/; ‚j .‰ 0 /:

By definition, ‚i is injective. We will apply these observations to the natural grading of H.2I .1; 2//. Note that every automorphism stabilizes the maximal subalgebra of codimension 2 (Theorem 4.2.6). Since the natural filtration is a standard filtration of depth 1, this filtration is uniquely determined by the maximal subalgebra. Therefore Aut H.2I .1; 2//.2/ D Aut.0/ H.2I .1; 2//.2/ and Aut H.2I .1; 2//.2/ D Aut0 H.2I .1; 2//.2/ Ë Aut.1/ H.2I .1; 2//.2/ hold.

10.2

More on H.2I .1; 2//.2/ Œp

15

In §7.3 we introduced the notion ˆ .D/ D ı D ı 1 for 2 Autc O.mI n/ and D 2 W .mI n/. Theorem 7.3.2 states the existence of a group isomorphism in the case under consideration ˆ W ¹ 2 Autc O.2I .1; 2// j .!H / 2 F !H º ! Aut H.2I .1; 2//.2/ ;

7! ˆ :

The commutative algebra O.2I .1; 2// is endowed with the natural filtration given by the lower degree of polynomials. This filtration induces a natural filtration of End O.2I .1; 2//. Let us write f D O.k/ if f 2 O.2I .1; 2//.k/ and D O.k/ if .O.2I .1; 2//.i/ / O.2I .1; 2//.i Ck/

8i

for 2 End O.2I .1; 2//. In particular, this convention implies that a derivation D 2 W .2I .1; 2// satisfies D D O.k/ if and only if D 2 W .2I .1; 2//.k/ in the natural filtration of W .2I .1; 2//. .a/ .b/ .a/ .b/ .a/ .b/ Recall that DH .x1 x2 / D @1 .x1 /x2 @2 x1 @2 .x2 /@1 is homogeneous of degree a C b 2. Lemma 10.2.1. (1) Let 1 ; 2 ; 3 2 F , 1 3 ¤ 0. The mapping .1 ;2 ;3 / given by .a/ .b/

.1 ;2 ;3 / .x1 x2 / WD .1 x1 C 2 x2 /.a/ .3 x2 /.b/ is contained in Autc O.2I .1; 2// and satisfies .1 ;2 ;3 / .!H / D 1 3 !H . (2) Let a; b 2 N, a C b 2, a < p, b < p 2 and .a; b/ ¤ .0; p/. For every ˛ 2 F and arbitrary f 2 F x2 CO.2I .1; 2//.2/ there is 2 Autc O.2I .1; 2// satisfying .x1 / D x1 C f;

.a/ .b/

.x2 / D x2 C ˛x1 x2 :

Proof. (a) Let us first clarify when an automorphism 2 Autc O..2// is contained in Autc O.2I .1; 2//. In the notation of Theorem 6.3.2 the present case gives n1 D 1 and n2 D 2. The condition (6.3.1) in that theorem is such that ˛1 .p l j / D 0 if j D 1 and l > 0 or j D 2 and l > 1, and ˛2 .p l j / D 0 if j D 1 and l 0 or j D 2 and .p l /

l > 0. This means that no summands of types x1

.p l / x1

.p l /

with l > 0 or x2

with l > 1

.p l / x2

must occur in .x1 / and no summands of types with l 0 or with l > 0 must occur in .x2 /. (b) The conditions stated in (a) are satisfied for .1 ;2 ;3 / in assertion (1). (c) We check the conditions stated in (a) for assertion (2). Since f is assumed to lie in F x2 C O.2I .1; 2//.2/ no summand of the first types does occur in f . Also, as .a/ .b/

.p l /

a C b > 1, a < p and b < p 2 , x1 x2

cannot be of the second types x1

l 0 or

.p/ x2

.p l / x2

with l 2. Finally, the case

has been excluded explicitly.

with

16

10


Theorem 10.2.2. (1) One has .a/ .b/

ˆ.1 ;2 ;3 / .DH .x1 x2 // D 1 1 3b for all 0 a p

1, 0 b p 2

1

.b/

DH ..1 x1 C 2 x2 /.a/ x2 /

1 and

Aut0 H.2I .1; 2//.2/ D ¹ˆ.1 ;2 ;3 / j 1 ; 2 ; 3 2 F; 1 3 ¤ 0º: (2) Let a; b 2 N, a C b > 2, and a < p and b < p 2 but .a; b/ ¤ .1; p/I or .a; b/ D .p; 0/I or .a; b/ D .0; p 2 /: For every ˛ 2 F there is ‰˛;a;b 2 Aut H.2I .1; 2// satisfying .a/ .b/

‰˛;a;b .D/ D D C ˛ŒDH .x1 x2 /; D C O.a C b

1 C deg D/

for every homogeneous D 2 H.2I .1; 2//Œp . (3) The assignment ‚i W Aut.i/ H.2I .1; 2//.2/ =Aut.iC1/ H.2I .1; 2//.2/ ! H.2I .1; 2//i is surjective if i ¤ p

1, and the image of ‚p

1

.i > 0/

is

.a/ .b/

span ¹DH .x1 x2 / j a C b D p C 1; .a; b/ ¤ .1; p/º: Proof. (1) Apply Theorem 7.3.6 to prove the first assertion. Write ‰ 2 Aut0 H.2I .1; 2//.2/ as ‰ D ˆ for some 2 Autc O.2I .1; 2//. Note .c/ .d / that ‰.DH .x1 x2 // 2 F DH ..x1 /.c/ .x2 /.d / / by Theorem 7.3.6. Since ‰ is assumed to be homogeneous of degree 0 so is . Therefore .x1 / D 1 x1 C 2 x2 , .x2 / D 10 x1 C 20 x2 for some 1 ; 2 ; 10 ; 20 2 F . Theorem 6.3.2 implies 10 D 0. Then D .1 ;2 ;20 / . .a/ .b/

(2) By the present assumption DH .x1 x2 / is an element of H.2I .1; 2//, homogeneous of degree a C b 2 > 0. For a definition of W .2I .1; 2// see Equa.a/ .b/ .a 1/ .b/ tion (7.3.1) of Volume 1, p. 375. One obtains that DH .x1 x2 / D x1 x 2 @2 .a/ .b 1/ x1 x2 @1 2 W .2I .1; 2// if and only if all the following conditions hold:

a

a ¤ p l .l > 0/ or b

1 ¤ 0;

1 ¤ p l .l 0/ or

b ¤ 0;

a

a ¤ 0 or b

1 ¤ p l .l > 1/;

1 ¤ 0 or

b ¤ p l .l > 0/:

10.2

More on H.2I .1; 2//.2/ Œp

17

.a/ .b/

So DH .x1 x2 / 62 W .2I .1; 2// if and only if .a; b/ is one of the following: .p l ; 1/ .l > 0/I

.0; p l C 1/ .l > 1/I

.p l C 1; 0/ .l 0/I

.1; p l / .l > 0/:

None of these pairs satisfies the assumptions of the theorem. Therefore Theorem 7.3.5 .a/ .b/ shows that there is 2 Autc O.2I .1; 2// satisfying .xj / D xj C˛DH .x1 x2 /.xj / .a/ .b/ C O.a C b/ and .!H / D !H . Obviously, 1 .xj / D xj ˛DH .x1 x2 /.xj / C O.a C b/. Then for homogeneous D one has the identity .D.xj // D D.xj / C .a/ .b/ ˛DH .x1 x2 /.D.xj // C O.a C b C deg D/ and . ı D ı 1 /.xj / .a/ .b/ D . ı D/ xj ˛DH .x1 x2 /.xj / C O.a C b/ .a/ .b/ D D.xj / ˛D DH .x1 x2 /.xj / C O.a C b C deg D/ .a/ .b/

D D.xj / C ˛ŒDH .x1 x2 /; D.xj / C O.a C b C deg D/: Set ‰˛;a;b .D/ WD ı D ı 1 . (3) The image of ‚i (for i > 0) is a subspace of Deri H.2I .1; 2//.2/ D H.2I .1; 2//i (see Theorem 7.1.2). Assertion (2) proves the claim if i ¤ p 1. Consider the case i D p 1. Again assertion (2) shows that the exposed set is contained in the image of ‚p 1 . Next take arbitrary ‰ 2 Aut.p 1/ H.2I .1; 2//.2/ n Aut.p/ H.2I .1; 2//.2/ , write ‰ D ˆ where .!H / 2 F !H . One necessarily has D Id C p 1 C O.p/. Theorem 7.3.5(1) proves p 1 .x1 /@1 C p 1 .x2 /@2 2 H.2I .1; 2// \ W .2I .1; 2//. Put p 1 .x1 /@1 C p 1 .x2 /@2 DW DH .f /. The condition DH .f / 2 W .2I .1; 2// ensures that f contains only summands of a form occurring in the assumptions of assertion (2). The statement on Im ‚p 1 follows. In the following we will investigate the structure of the Poisson Lie algebra .O.2I .1; 2//; ¹ ; º/ with multiplication ¹f; gº WD DH .f /.g/ D @1 .f /@2 .g/

@2 .f /@1 .g/:

Recall that the mapping DH W O..2r// ! W ..2r// induces a surjective homomorphism of Lie algebras ([S-F, §4.4)], Equation (4.2.4) of Volume 1, p. 188) DH W .O.2I .1; 2//; ¹ ; º/ H.2I .1; 2//.1/ with kernel F . Since ŒDH .f /; DH .g/ D DH .¹f; gº/ D DH ..DH .f /.g// r

r

holds for all f; g, one has ŒD p ; DH .g/ D DH .D p .g// for all D 2 H.2I .1; 2//.1/ p p2 and g 2 O.2I .1; 2//. As @1 .g/ D @2 .g/ D 0 for such g, we obtain in view of our introductory remarks ŒD; DH .g/ D DH .D.g//

18

10


for all D 2 H.2I .1; 2//.2/ Œp , g 2 O.2I .1; 2//. Applying both endomorphisms to f readily gives D.¹g; f º/ ¹g; D.f /º D ¹D.g/; f º, i.e., D.¹f; gº/ D ¹D.f /; gº C ¹f; D.g/º for all D 2 H.2I .1; 2//.2/ Œp , f; g 2 O.2I .1; 2//. Similarly, our introductory remarks imply that H.2I .1; 2//.2/ Œp acts restrictedly on O.2I .1; 2//. But observe that .p/

.p 2 /

H.2I .1; 2//.2/ Œp does not act restrictedly on O.2I .1; 2// ˚ F x1 ˚ F x2 Œp .p/ p .p/ for instance @1 .x1 / D 1, @1 .x1 / D 0.

, because

Lemma 10.2.3. For every 2-dimensional torus T H.2I .1; 2//.2/ Œp the intersection with H.2I .1; 2//.2/ is 1-dimensional and H.2I .1; 2//.2/ Œp D T CH.2I .1; 2//.2/ . p

Proof. Abbreviate S WD H.2I .1; 2//.2/ . Note that SŒp D S ˚ F @2 . Therefore T \ S ¤ ¹0º. Now suppose T S . If T \ S.0/ D ¹0º, then T contains elements t1 D @1 C O.0/ Œp p and t2 D @2 C O.0/. But then t2 2 @2 C S is contained in T n S, a contradiction. Therefore T \ S.0/ ¤ ¹0º. However, S.0/ =S.1/ Š sl.2/ and every toral element of this algebra acts invertibly on the natural 2-dimensional sl.2/-module S=S.0/ . Then T S.0/ . Since T \ S.1/ D ¹0º, this gives rise to a 2-dimensional torus in sl.2/, a contradiction. The final statement follows by a dimension argument. We now distinguish the three cases that T \ H.2I .1; 2//.2/ is not a torus, T \ H.2I .1; 2//.2/ is a 1-dimensional torus not contained in H.2I .1; 2//.2/ .0/ , and T \ H.2I .1; 2//.2/ H.2I .1; 2//.2/ .0/ (which necessarily is a 1-dimensional torus as H.2I .1; 2//.2/ .0/ is closed under Œp-th powers). In order to compute Œp-th powers the following remark will be very useful. Remark 10.2.4. Let g be any restricted Lie algebra and d1 ; d2 2 g. Then Œp-th powers are computed as Œp

.d1 C d2 /Œp D d1

Œp

C d2

C

p X1

si .d1 ; d2 /

i D1

Pp

1

where the correction term i D1 si .d1 ; d2 / is given by the formula mentioned in Volume 1, p. 17. In general it is rather awkward to compute this correction term. However, if Œd2 ; .ad d1 /k .d2 / D 0 for all k p 3, then the definitions imply that si .d1 ; d2 / D 0 for i p 3 and sp 2 .d1 ; d2 / D 21 Œd2 ; .ad d1 /p 2 .d2 /. In this case one has (see also Equation (1.1.1)) Œp

.d1 C d2 /Œp D d1

Œp

C d2

C .ad d1 /p

1

1 .d2 / C Œd2 ; .ad d1 /p 2

2

.d2 /:

10.2

More on H.2I .1; 2//.2/ Œp

19

Let us first compute iterated Œp-th powers of .p 1/

t WD DH .x1 /Cr0 DH .x1 x2

.p 2 1/

/Cr1 DH .x1 x2

/ 2 H.2I .1; 2//.2/ ;

r0 ; r1 2 F:

Lemma 10.2.5. The following equations hold: .i/

.j /

if 0 i; j < p k < i C j .k 1/I

ŒDH .x1 x2 /; DH .x1 x2 / D 0 .p 1/

ŒDH .x1 x2 .p 1/

ŒDH .x1 x2

.p 1/

/; DH .x1 x2 / D 2DH .x1 x2 .p 2

/; DH .x1 x2 .p 2

1/

ŒDH .x1 x2

.p 2 p/

ŒDH .x1 x2

pC1/

/I

.p 2

1/

/I

.p 2

1/

/I

/ D 2DH .x1 x2

/; DH .x1 x2 / D 2DH .x1 x2

.p 2 1/

.p/

/; DH .x1 x2 / D 2DH .x1 x2 .i/

DH .x1 x2 /p D 0;

/I

if i ¤ p l .l 0/:

Proof. In order to prove the first equation we observe that .i /

.iCj 1/

.j /

ŒDH .x1 x2 /; DH .x1 x2 / 2 H.2I .1; k//.2/ \ FDH .x1 x2

/ D ¹0º:

The third equation is obtained by Lemma 2.1.2 and .p 1/ .p 2 pC1/ ŒDH .x1 x2 /; DH .x1 x2 /

p2 D p2

1 p

!

p2 p

.p 2 1/

D 2DH .x1 x2

! 1 .p 2 DH .x1 x2 2

1/

/

/:

Equations 2, 4, 5 are obtained similarly, while the sixth follows from Lemma 7.2.1(3). Lemma 10.2.6. The element .p 1/

t WD DH .x1 / C r0 DH .x1 x2

.p 2 1/

/ C r1 DH .x1 x2

/

satisfies the equations .p 2 p/

p

t p D @2 C r0 t C r1 DH .x1 x2 tp

2

p

r0 t p

p2

.p 1/

.p 2 1/ .p 1/ ŒDH .x1 x2 /; DH .x1 x2 /

.p r0 DH .x1 x2

/;

r1 t D @2 :

Proof. Write t D @2 C r0 DH .x1 x2 that

.p 2 1/

/ C r0 r1 DH .x1 x2

.p 2 1/

/ C r1 DH .x1 x2

/. Lemma 10.2.5 shows

D 0, whence p 1/ .p 2 1/ / C r1 DH .x1 x2 / p

.p 1/ p

D r0 DH .x1 x2

p

.p 2 1/ p

/ C r1 DH .x1 x2

/ D 0:

20

10


It also shows that for 0 k p h .p r0 DH .x1 x2

1/

3 .p 2 1/

/ C r1 DH .x1 x2 .p @k2 r0 DH .x1 x2

1/

/; .p 2 1/

/ C r1 DH .x1 x2

i / D 0:

Remark 10.2.4 yields .p 2 p/

p

/ t p D @2 C r0 DH .x1 / C r1 DH .x1 x2 1 .p 1/ .p 2 / C r1 DH .x1 x2 C r0 DH .x1 x2 2

1/

/;

.p 2 pC1/

r0 DH .x1 x2 / C r1 DH .x1 x2 .p 2 p/

p

D @2 C r0 DH .x1 / C r1 DH .x1 x2 .p 2 1/

C 2r0 r1 DH .x1 x2

.p 2

p

D @2 C r0 t C r1 DH .x1 x2

/ .p 1/

/ C r02 DH .x1 x2

/

/ p/

.p 2 1/

/ C r0 r1 DH .x1 x2

/:

Similar to the preceding arguments we get .p 2 r1 DH .x1 x2 h

.p 2 1/

/ C r0 r1 DH .x1 x2

.p 2 p/

1/

/

i

D0

3, and

r0 t /p .p 2 p D @2 C r1 DH .x1 x2

p/

.p 2 1/

/ C r0 r1 DH .x1 x2

p2

.p 1/

D @2 C r1 DH .x1 / C r0 r1 DH .x1 x2 1 .p 2 C Œr1 DH .x1 x2 2 1 .p 2 C Œr1 DH .x1 x2 2 p2

D @2 C r1 t:

p /

/ .p 2 pC1/

p/

/ C r0 r1 DH .x1 x2

p/

/ C r0 r1 DH .x1 x2

.p 2 pC1/

.p 1/

D @2 C r1 DH .x1 / C r0 r1 DH .x1 x2 p2

p / D 0;

.p 2 1/

r1 DH .x1 x2 / C r0 r1 DH .x1 x2 /; .p 2 p/ .p 2 kp @2 r1 DH .x1 x2 / C r0 r1 DH .x1 x2

for 0 k p .t p

p/

.p/

/; r1 DH .x1 x2 / .2p 1/

/; r0 r1 DH .x1 x2 .p 2 1/

/ C r12 DH .x1 x2

/

/

10.2

More on H.2I .1; 2//.2/ Œp .p 1/

Theorem 10.2.7. Set tr WD DH x1 C rx1 x2 r 2 F.

21 .p 2 1/

C x1 x2

2 H.2I .1; 2//.2/ ,

(1) Every tr is a semisimple element of H.2I .1; 2/.2/ Œp and generates a 2-dimensional torus Tr . For this torus Tr \ H.2I .1; 2//.2/ D F tr is not a torus. (2) Tori Tr and Ts are conjugate under Aut H.2I .1; 2//.2/ if and only if there is 2 Fp2 such that s D 1 p r. (3) Let T H.2I .1; 2//.2/ Œp be a 2-dimensional torus and T \ H.2I .1; 2//.2/ be not a torus. Then there are ‰ 2 Aut H.2I .1; 2//.2/ and r 2 F such that ‰.T / D Tr . p2

p

Proof. (1) Due to Lemma 10.2.6 we have tr

r p tr

Œp2

Œp

p2

p2

tr D @2 . Since @2 centralizes

H.2I .1; 2//.2/ , this gives the equation tr r p tr tr D 0. Then tr is a semisimple Œp p .2/ element of H.2I .1; 2// Œp . In addition, Lemma 10.2.6 shows that tr D @2 C .p 2 p/

.p 2 1/

r0 tr C r1 DH .x1 x2 / C r0 r1 DH .x1 x2 / 62 H.2I .1; 2//.2/ . Then F tr is not a torus, but generates a torus of dimension 2. (2) Suppose there is ‰ 2 Aut H.2I .1; 2//.2/ , such that ‰.Tr / D Ts . Then ‰.F tr / D ‰.Tr \ H.2I .1; 2//.2/ / D ‰.Tr / \ H.2I .1; 2//.2/ D Ts \ H.2I .1; 2//.2/ D F ts : Let 2 F for which ‰.tr / D ts . Then ts satisfies the equations 2

tsŒp

s p tsŒp

2

ts D 0;

This gives p D ¤ 0 and s p D p p2

2

2

p tsŒp 1r p .

p r p tsŒp

ts D 0:

The first equation implies 2 Fp2 . Note

, and this gives s D 1 p r. that p 1 D p 1 Suppose s D p r for some 2 Fp2 . Define a divided power automorphism by .x1 / WD x1 , .x2 / WD 1 x2 . Theorem 7.3.6 states that there is an auto.p 1/ morphism ˆ of H.2I .1; 2//.2/ satisfying ˆ .tr / D DH x1 C r2 p x1 x2 C 2 2 .p 1/ 2 p x1 x2 D ts . Then ˆ .Tr / D Ts holds. (3)(a) Set T \H.2I .1; 2//.2/ DW F t. Write t D r@1 Cs@2 CO.0/ D DH . rx2 /C DH .sx1 / C O.0/ with r; s 2 F . If s D 0 then t Œp 2 T \ H.2I .1; 2//.2/ D F t . But then F t is a torus, which contradicts the present assumption. Define 2 Autc O.2I .1; 2// by setting .x1 / WD x1 C rx2 , .x2 / WD sx2 . Theorem 7.3.6 shows that the mapping DH .f / 7! DH .s 1 .f // is an automorphism of H.2I .1; 2//.2/ . This automorphism maps t onto DH . rx2 C .x1 C rx2 // C O.0/ D DH .x1 / C O.0/.

22

10


Thus after the application of this automorphism we may assume X .i/ .j / t D DH .x1 / C ri;j DH .x1 x2 / i Cj 2

D @2 C

X

.i/ .j /

ri;j 2 F:

ri;j DH .x1 x2 /;

i Cj 2 .a/ .bC1/

.a/ .b/

(b) Next we observe that ŒDH .x1 x2 /; t D DH .x1 x2 / C O.a C b 1/. .a/ .b/ Applying Theorem 10.2.2 several times we get rid of summands ra;b DH .x1 x2 / where a C b 2, b C 1 < p 2 and .a; b C 1/ ¤ .1; p/, or .a; b C 1/ D .0; p 2 /. Hence Q such that we find an automorphism ‰ .p Q ‰.t/ D DH .x1 / C r0 DH .x1 x2

1/

.p 2 1/

/ C DH .f x2

/

for some r0 2 F and some polynomial f D f .x1 / without constant term and deQ Then pending on x1 only. We suppress the notion of ‰. .p 1/

t D @2 C r0 x2 .p 2 1/

Since @1 .f /x2 induction that

t

j

.p 2/

@2

r0 x1 x2

.p 2 1/

@1 C @1 .f /x2

.i/

@2 .x2 / D 0 whenever 2 i p 2

.p 2 1/ .x2 /

.p 2 1 j / x2

@2

@1 :

1, one easily concludes by

2 1 pX

.mod

.p 2 2/

f x2

.i/

F x2 /

i Dp 2 Cp 2 j

for 0 j p 2 2. Since t generates a 2-dimensional torus and H.2I .1; 2//.2/ Œp acts restrictedly on O.2I .1; 2//, as an endomorphism of O.2I .1; 2// t satisfies a poly2 nomial P .X / D X p C s1 X p C s0 X. Consequently, .p 2 1 p/

s1 x2

.p 2 2/

s0 x2

2

.p 2 1/

t p .x2

/ t 2 .x2 /

.p 1/

t 1 C r0 x2

.p 2 2/

@1 .f /x2

.p 2 1/ C @1 .f /x2 X .i/ .mod F x2 /: i p 2

Then @1 .f / 2 F , which means f 2 F x1 . Set f WD r1 x1 . (c) Lemma 10.2.6 applies to t . Thus as an endomorphism of O.2I .1; 2// t satisfies 2 p the equation t p r0 t p r1 t D 0. If r1 D 0, then .t p r0 t/p D 0, but t acts semisimply and generates a 2-dimensional torus. So this is impossible. Choose 2 2 F for which r1 p 1 D 1 and set r WD r0 p 1 . Define according to Theorem 7.3.6 .i/ .j / .i/ .j / an automorphism which maps DH .x1 x2 / onto DH . 1Cj x1 x2 /. This maps t onto 1 tr .

More on H.2I .1; 2//.2/ Œp

10.2

23 2

As an endomorphism of O.2I .1; 2//, tr satisfies the polynomial Pr .X/ WD X p p r Xp X (Lemma 10.2.6). Consequently, tr has p 2 different eigenvalues on O.2I .1; 2// and Pr .X/ is its minimal polynomial. For every nonzero root ˛ of 2 2 2 r p X p X/ W .X ˛/ D X p 1 C ˛X p 2 C Pr .X / set Pr;˛ .X/ WD .X p (terms of lower degree) and

v˛ WD

.p 2 1/ Pr;˛ .tr /.x2 /

D 1 C ˛x2 C

2 1 pX

.i/

qr;˛;i 2 F:

qr;˛;i x2 ;

i D2

Theorem 10.2.8. Let .p 1/

tr D DH x1 C rx1 x2 denote Pr .X / D X p

2

rpX p

.p 2 1/

C x1 x2

2 H.2I .1; 2//.2/ ;

X its minimal polynomial and set .p 2 1/

v˛ WD Pr;˛ .tr /.x2

/

for any nonzero root ˛ of Pr .X/, .p 1/

v0 WD 1 and u WD x1 C rx1 x2

.p 2 1/

C x1 x2

:

Let ˛; ˇ be Fp -linear independent tr -eigenvalues. The following holds: (1) The set of tr -eigenvalues on O.2I .1; 2// is Fp ˛ C Fp ˇ and the tr -eigenspace for the eigenvalue j˛ C kˇ is p-dimensional and given by p 1

O.2I .1; 2//j˛Ckˇ D ˚i D0 F ui v˛ j vˇ k ;

j; k D 0; : : : ; p

1:

(2) The Poisson product of eigenvectors is given by ¹ui v˛j vˇk ; ul v˛m vˇn º D i.m˛ C nˇ/ for all 0 i; j; k; l; m; n p

l.j˛ C kˇ/ ui Cl

1 j Cm kCn v˛ vˇ

1.

Proof. (1) We first observe that tr .u/ D DH .u/.u/ D ¹u; uº D 0 and tr .v / v D .p 2 1/

Pr .tr /.x2 / D 0 for all nonzero eigenvalues . Therefore .ui v˛ j vˇ k /0i;j;kp 1 is a family of tr -eigenvectors. Eigenvectors Pp 1 of this family corresponding to a fixed eigenvalue j˛ C kˇ are of the form i D0 i ui v˛ j vˇ k . Since v˛ D 1 C O.1/, vˇ D 1 C O.1/ and ui D x1i C O.i C 1/, every vector ui v˛ j vˇ k is of the form x1i C O.i C 1/. Therefore there cannot be a linear dependency within this family. As a consequence, this family is a basis of O.2I .1; 2//. Claim 1 follows.

24

10


(2) One has ¹u; v º D DH .u/.v / D tr .v / D v for all . As every v is Pp 2 1 .i/ contained in i D0 F x2 , we get ¹v ; vı º D 0. Since every mapping ¹f; ‹º is a derivation of O.2I .1; 2//, one obtains 1 j Cm 1 kCn v˛ vˇ

¹ui v˛j vˇk ; ul v˛m vˇn º D im¹u; v˛ ºui Cl

C i n¹u; vˇ ºui Cl

1 j Cm kCn 1 v˛ vˇ

C j l¹v˛ ; uºui Cl

1 j Cm 1 kCn v˛ vˇ

C kl¹vˇ ; uºui Cl

1 j Cm kCn 1 v˛ vˇ

l.j˛ C kˇ/ ui Cl

D i.m˛ C nˇ/

1 j Cm kCn v˛ vˇ :

Recall that .g; t/ denotes the set of t-roots on g including 0 (for every torus t Der g). Corollary 10.2.9. Let T H.2I .1; 2//.2/ Œp be a 2-dimensional torus and T \ H.2I .1; 2//.2/ D F t be not a torus. Let be any nonzero T -root. The following holds: ˇ ˇ (1) ˇ.H.2I .1; 2//.2/ ; T /ˇ D p 2 , dim H.2I .1; 2//.2/ D p, dim CH.2I.1;2//.2/ .T / D p 2. (2) Every root space H.2I .1; 2//.2/ contains an element d satisfying d Œp D t Œp .t/p 1 t . (3) The 1-section H.2I .1; 2//.2/ . / is isomorphic to H.2I 1/.2/ under an isomorphism which maps T \ H.2I .1; 2//.2/ onto FDH .x1 .1 C x2 //. Proof. (1) Theorem 10.2.7 shows that T is conjugate to a torus Tr for some r 2 F . Nothing is lost if we take T D Tr and t D tr . Choose u as well as v as in Theorem 10.2.8. We mentioned that there are p 2 different eigenvalues for tr on O.2I .1; 2//. Since DH defines a tr -invariant isomorphism O.2I .1; 2// ! H.2I .1; 2//.1/ , there are p 2 roots for the torus Tr on H.2I .1; 2//.1/ . Clearly, 0 is a Tr -root on H.2I .1; 2//.2/ (as tr is a 0-root vector) and H.2I .1; 2//.1/ =H.2I .1; 2//.2/ has only 0-weight. Therefore there p 2 roots on H.2I .1; 2//.2/ . Pp are 1 .2/ One has H.2I .1; 2// D i D0 FDH .ui v /. Therefore dim H.2I .1; 2//.2/ D p. Adding dimensions we get dim CH.2I.1;2//.2/ .tr / D .p 3 2/ .p 2 1/p D p 2. (2) Set d WD DH .uv /, let , ı be Fp -independent tr -eigenvalues. It easily follows from Theorem 10.2.8(2) that in the Poisson algebra .ad uv /k .vı / D

kY1

.s C ı/v k vı

sD0

Qp

1

holds for all k D 1; : : : ; p. Note that sD0 .s C ı/ D ı p ı p 1 and v p D 1. Consequently, .ad d /p .DH .vı // D .ı p ı p 1 /DH .vı /. Similarly one sees that

More on H.2I .1; 2//.2/ Œp

10.2 j

25

j C1

j

.ad d /.DH .ui v // D .j i/ ui v , whence .ad d /p .DH .ui v // D 0. Then d Œp is semisimple and commutes with tr . Therefore it is contained in Tr . Looking at eigenvalues one gets d Œp D t Œp p 1 t. (3) We mentioned above that DH establishes a surjective Lie algebra homomorphism DH W .O.2I .1; 2//; ¹; º/ ! H.2I .1; 2//.1/ . One easily computes the identity ¹uv ; ui v p 1 º D .i C 1/ ui . Arguing by dimensions one derives from this H.2I .1; 2//

.2/

D

X

H.2I .1; 2//

.2/

C

p X2

FDH .ui /:

i D1

¤0

Therefore .2/

H.2I .1; 2//

. / D

p X1 pX1

i

j

FDH .u v // C

p X2

FDH .ui /:

i D1

i D0 j D1

Define a mapping H.2I .1; 2//.2/ . / ! H.2I 1/.2/ by DH .ui v j // 7!

1

DH .x1i .1 C x2 /i Cj /:

Theorem 10.2.8(2) readily shows ¹ui v j ; uk v l º D .il

kj /ui Ck

1

v j Cl :

Consequently, that mapping is a Lie algebra isomorphism which maps tr D DH .u/ onto 1 DH .x1 .1 C x2 //. Next we consider the case that T \ H.2I .1; 2//.2/ is a 1-dimensional torus which is not contained in H.2I .1; 2//.2/ .0/ . Lemma 10.2.10. Let rQ 2 F and gQ D g.x Q 2 / be a polynomial independent of x1 . The following holds: .p/

(1) DH .x2 C rx Q 2 /Œp D 0; .p 1/

g/ Q Œp D 0; .p/ k .p 1/ .p 1 k/ .p 1/ k (3) ad DH .x2 Crx Q 2 / DH .x1 g/ Q D . 1/k DH x1 .1Crx Q 2 / gQ for 0 k p 1; .p 1/ .p/ k .p 1/ (4) DH .x1 g/; Q ad DH .x2 Crx Q 2 / DH .x1 g/ Q D 0 for 0 k p 3; .p 1/ .p/ p 2 .p 1/ (5) DH .x1 g/; Q ad DH .x2 C rx Q 2 / DH .x1 g/ Q .p 1/ .p 1/ .p 1/ 2 .p 2/ D 2DH x1 g@ Q 2 .g/.1 Q 2rx Q 2 / rx Q 1 gQ x2 . (2) DH .x1

26

10


Proof. (1) No nonzero element in H.2I .1; 2//.2/ Œp has x1 -degree p. Therefore Œp .p 1/ .p/ /@1 vanishes. Q 2 DH .x2 C rx Q 2 /Œp D .1 C rx (2) No nonzero element in H.2I .1; 2//.2/ Œp has x1 -degree p. Therefore Œp .p 1/ Œp .p 2/ .p 1/ DH .x1 g/ Q D x1 g@ Q 2 x1 @2 .g/@ Q 1 Œp Œp .p 2/ .p 1/ D x1 g@ Q 2 x1 @2 .g/@ Q 1 D 0: (3) As @1 .g/ Q D 0, one has .p/ .p 1 k/ .p 1/ k DH .x2 C rx Q 2 /; DH x1 .1 C rx Q 2 / gQ .p/ .p 1 k/ .p 1/ k D DH DH .x2 C rx Q 2 / x1 .1 C rx Q 2 / gQ .p 1/ .p 1 k 1/ .p 1/ k D DH .1 C rx Q 2 /x1 .1 C rx Q 2 / gQ ; and the claim follows by induction. (4) If k p 3, then p 1 k 2, whence .p 1/ .p/ k .p 1/ DH .x1 g/; Q ad DH .x2 C rx Q 2 / DH .x1 g/ Q .p 1 k/ .p 1/ k .p 1/ .1 C rx Q 2 / gQ D 0: D DH .x1 g/; Q . 1/k DH x1 (5) By (3), .p DH .x1

1/

.p/ p 2 .p 1/ g/; Q ad DH .x2 C rx Q 2 / DH .x1 g/ Q .p 1/ .p 1/ p 2 D DH .x1 g/; Q DH x1 .1 C rx Q 2 / gQ ! p 1 .p 1/ .p 1/ p 2 D DH x1 g@ Q 2 .1 C rx Q 2 / gQ 1 .p 1/ .p 1/ p 2 C DH x1 @2 .g/.1 Q C rx Q 2 / gQ .p 1/ .p 1/ p 2 D 2DH x1 g@ Q 2 .g/.1 Q C rx Q 2 / .p 1/ 2 .p 1/ p 3 .p 2/ 2DH x1 gQ .1 C rx Q 2 / rx Q 2 .p 1/ 2 .p .p 1/ .p 1/ g@ Q 2 .g/.1 Q 2rx Q 2 / rx Q 1 gQ x2 D 2DH x1

2/

:

Lemma 10.2.11. Suppose T \ H.2I .1; 2//.2/ is a 1-dimensional torus which is not contained in H.2I .1; 2//.2/ .0/ . There are ‰ 2 Aut H.2I .1; 2//.2/ and toral elements t1 ; t2 such that ‰.T / D F t1 ˚ F t2 where .p/ .p 1/ .p/ .2p 1/ t1 D DH x2 x2 C x1 .x2 x2 C x2 / ; .2/

p

t2 D @2 C @2 C rDH .x2 / C O.1/; p

Moreover, t1

t1 D

p

@1 holds.

r 2 F:

10.2

More on H.2I .1; 2//.2/ Œp

27

Proof. (a) Choose a toral element t10 2 T \ H.2I .1; 2//.2/ , write t10 D DH .r1 x1 C r2 x2 / C O.0/ D r1 @2 r2 @1 C O.0/. Since t10 is assumed to be toral one has r1 D 0. As t10 62 H.2I .1; 2//.2/ .0/ then r2 ¤ 0. Define 2 Autc O.2I .1; 2// by setting .x1 / D r2 x1 , .x2 / D x2 . Theorem 7.3.6 shows that the mapping ˆ W DH .f / 7! DH .r2 1 .f // is an automorphism of H.2I .1; 2//.2/ . Substituting t10 by ˆ .t10 / we may assume t10 D DH .x2 / C O.0/: .aC1/ .b/

.a/ .b/

Note that ŒDH .x1 x2 /; t10 D DH .x1 x2 / C O.a C b 1/: Applying Theorem .a/ .b/ 10.2.2 several times we get rid of summands of the form ra;b DH .x1 x2 / in t10 where a C b 2, a C 1 < p and .a C 1; b/ ¤ .1; p/, or .a C 1; b/ D .p; 0/. Hence we find an automorphism ‰ 0 such that .p/

.p 1/

‰ 0 .t10 / D DH .x2 / C r 0 DH .x2 / C DH .x1

g/

for some r 0 2 F and a polynomial g D g.x2 / with constant term 0 and being independent of x1 . Suppressing the notion of ‰ 0 we may assume t10 D DH .x2 C g 0 /, where g 0 D O.p/ has lower degree at least p. Next choose a toral element t20 2 T n H.2I .1; 2//.2/ , write Œp

p

t20 D s@2 C s1 @1 C s2 @2 C O.0/ D s@2 C s1 @1 C s2 @2 C O.0/; Œp2

s ¤ 0:

p

Since t20 is toral and @2 D 0, one has s D s2 . Define 0 2 Autc O.2I .1; 2// by setting 0 .x1 / D x1 C s1 x2 ; 0 .x2 / D s2 x2 : Theorem 7.3.6 shows that the mapping ˆ0 W DH .f / 7! DH .s2 1 0 .f // is an automorphism of H.2I .1; 2//.2/ . Note that ˆ0 .@1 / D

ˆ0 .DH .x2 // D

DH .x2 / D @1 ;

ˆ0 .@2 / D ˆ0 .DH .x1 // D DH .s2 1 x1 C s2 1 s1 x2 / D s2 1 @2 ˆ0 .t10 / D DH .x2 C g 00 /;

s2 1 s1 @1 ;

g 00 D O.p/; p

p

ˆ0 .t20 / D s2 .ˆ0 .@2 //p C s1 ˆ0 .@1 / C s2 ˆ0 .@2 / C O.0/ D @2 C @2 C O.0/: Thus after the application of this automorphism we may assume that p

Œp

t10 D DH .x2 C g 00 /;

g 00 D O.p/;

Write g 00 D

ra;b x1 x2 . Applying automorphisms ‰ra;b ;aC1;b according

P

t20 D @2 C @2 C O.0/ D @2 C @2 C O.0/:

.a/ .b/

aCbp

.a/ .b/

to Theorem 10.2.2 we get rid of summands of the form ra;b DH .x1 x2 / in t10 , where

28

10


Q a C b p, a C 1 < p and .a C 1; b/ ¤ .1; p/. Hence we find an automorphism ‰ such that .p/ .p 1/ Q 10 / D DH .x2 / C rD t1 WD ‰.t Q H .x2 / C DH .x1 g/ Q for some rQ 2 F and a polynomial gQ D g.x Q 2 / with constant term 0 and being independent of x1 . Moreover, ‰ra;b ;aC1;b .t20 / D t20

.aC1/ .b p/ / x2

ra;b DH .x1

C O.a C b

p/:

.aC1/ .b p/

/ is regarded 0 if b < p. Hence such a term with x2 Here the term DH .x1 .a C 1/ C .b p/ D 1 can only occur if a D 0; b D p. But this is not an admissible pair. Consequently, p

Q 20 / D t20 C O.0/ D @ C @2 C O.0/: t2 WD ‰.t 2 (b) Lemma 10.2.10 in combination with Remark 10.2.4 yields Œp

t1 D t1

.p/ p 1 .p 1/ D ad DH .x2 C rx Q 2 / DH .x1 g/ Q 1 .p 1/ .p/ p 2 .p 1/ g/; Q ad DH .x2 C rx Q 2 / C DH .x1 DH .x1 g/ Q 2 .p 1/ .p 1/ .p 1/ p 1 gQ C DH x1 g@ Q 2 .g/.1 Q 2rx Q 2 / D DH .1 C rx Q 2 / .p 1/ 2 .p 2/ rD Q H x1 gQ x2 : .p/

Comparing terms independent of x1 and observing g.0/ Q D 0 one gets x2 C rx Q 2 D .p 1/ p 1 .p/ .p 1/ .p/ .1 C rx Q 2 / g. Q Consequently, gQ D .x2 C rx Q 2 /.1 C rx Q 2 / D x2 C rx Q 2 C .2p 1/ rQ 2 x2 and .p/ .p 1/ .p/ .2p 1/ Q 2 C x1 .x2 C rx Q 2 C rQ 2 x2 / : t1 D DH x2 C rx (c) Next observe .p/ .p 0 D Œt2 ; t1 D DH t2 x2 C rx Q 2 C x1 .p/

.p 1/

1/

.p/

.x2 C rx Q 2

.p/

.2p 1/

This gives t2 x2 C rx Q 2 C x1 .x2 C rx Q 2 C rQ 2 x2 toral and the action is a restricted one, we get even more .p/

t2 x2 C rx Q 2

.p 1/

C x1

.p/

.x2 C rx Q 2

P .i/ .2 p Write t2 D @2 C @2 C 2i D0 ˛i DH .x1 x2 1 in the former equation we obtain

i/

/ :

/ 2 F . But since t2 is

.2p 1/

C rQ 2 x2

.2p 1/

C rQ 2 x2

/ D 0:

/ C O.1/. Looking at terms of degree

rQ C 1 C ˛1 x2 C ˛2 x1 D 0:

10.2

Therefore rQ D .p/ x2

More on H.2I .1; 2//.2/ Œp

29

.2p 1/ x2 /

C (d) Recall

.p/

1 and ˛1 D ˛2 D 0, whence t1 D DH x2

.2/ p and t2 D @2 C @2 C ˛0 DH .x2 / C O.1/. that d p D d Œp if d 2 H.2I .1; 2//.0/ . Hence p

t1 D

.p/

Lemma 10.2.12. Put t1 WD DH x2

x2

.2/

p

Œp

p

@1 C t1

x2

.p 1/

C x1

.x2

This is the first claim.

p

D

@1 C t1 : .p 1/

C x1

.p/

.x2

x2

.2p 1/

C x2

/ and let

t2 D @2 C @2 C rDH .x2 / C O.1/ be any element of H.2I .1; 2//.2/ Œp . The following holds: .p 1/ .p 2 p/ .p 1 i/ .p 2 p/ .p 1 i/ .p 2 1/ (1) t1i .x1 x2 / D . 1/i x1 x2 C . 1/i 1 ix1 x2 C .p 1/ .p 2 p/

ıi;p 1 x1 x2 C O.p 2 C 2p 3 i/ for i D 0; : : : ; p 1; Pp 1 .p 1/ .p 2 p/ .p/ p 1 (2) . i D1 t2i / ı .t1 Id/.x1 x2 / D 1 C x2 C O.2p 2/; Pp 2 Pp 1 .p 1/ .p 2 p/ .i/ p 1 (3) .t2 Id/ ı . i D1 t1i /.x1 x2 / D i D0 . 1/i x1 C O.p

1/.

Proof. (1) The first equation is proved by induction on i . Suppose it is true for 0 i 1 p 2. Note that t1 D

.p 1/

@1 C x2

.p 1/

x1 D

.p 2/

@1 C x1

.p 1/

.1

.p 1/

@1 C x2

x2

.p/

.x2

x2

.2p 2/

C x2

.p 2/

@1 C x1

.2p 1/

C x2

/@1 .p 1/

x2 @2

/@2

x1

@1 C O.2p

3/:

Then .p 1/ .p 2 p/ x2 /

t1i .x1

D t1 . 1/i

1 .p i/ .p 2 p/ x1 x2

C O.p 2 C 2p

C . 1/i 2/

i

.p i 1/ .p 2 p/ x2

D . 1/i x1

C . 1/i

1

2

.p i/ .p 2 1/ x2

.i

1/x1

.p i 1/ .p 2 1/ x2

.i

1/x1

1 .p 1/ .p i 1/ .p 2 p/ x2 x1 x2

C O.p 2 C 2p

i

3/ C . 1/i

C O.p 2 C 3p

i

4/ C 0 C . 1/i

C O.p 2 C 3p

i

4/ C . 1/i x1

C . 1/i

1

.i

C O.p 2 C 3p

C ıi;p

.p 2/ .p i/ .p 2 1/ x1 x2

.i

1/x1

.p 1/ .p i 1/ .p 2 p/ x1 x2

.p 1/ .p i 1/ .p 2 1/ x1 x2

1/x1 i

.p 1 i/ .p 2 p/ x2

D . 1/i x1

1

4/ C O.p 2 C . 1/i

.p 1/ .p 2 p/ x2 1 x1

1

i C 2p

3/

.p 1 i/ .p 2 1/ x2

ix1

C O.p 2 C 2p

C0

3

i/:

30

10


(2), (3) Note that p

t2i D .@2 C @2

rx2 @1 C O.1//i

p.i 1/

pi

D @2 C i@2

ı .@2

rx2 @1 / C O. p.i

1/ C 1/:

In particular, we obtain the weak estimates p 1

t2

p2 p

p2 p

D @2 p 2

t2

p 2 2pC1

D @2

p 2 2p

C r@2

@2

C O. p 2 C 2p

p 2 2p

p 2 3pC1

D @2

2@2

p2 p

p 2 2pC1

t2i D @2

1/;

i p

C O. p 2 C 2p C 1/;

p 2 2p

C O. p 2 C 2p C 1/:

p 2 2p

C O. p 2 C 2p C 1/

3;

p 2 2p

C r@2

@2

p 2 2p

C O. p 2 C 3p/ D @2

t2i D O. p 2 C 2p C 1/; p X1

ı .x2 @1 / C O. p 2 C 2p C 1/

ı .x2 @1 / C @2

i D1

Applying (1) we get pX1 p t2i ı .t1

.p 1/ .p 2 p/ x2 /

1

Id/.x1

i D1

2 p D @2

p 2 2pC1

p

p 2 2p

C r@2

@2

.p 2 p/

x2 .p 1/

D 1 C x2

.p 2 1/

C x2

.p 1/

x2

.p/

C x2

ı .x2 @1 / C @2

C O.p 2 C p

2/ D 1 C x2

p 1

p 1

Id/ ı

.t2

pX1

.p t1i .x1

.p/

C O.2p

This proves (2). To prove (3) we observe that t2 and compute

2/

p2 p

Id D @2

C O.2p

2/:

C O. p 2 C 2p

1/

1/ .p 2 p/ x2 /

i D1

D

p2 p @2

2

C O. p C 2p

pX1 .p 1/ . 1/i x1

1 i/ .p 2 p/ x2

C O.p 2

1/

i D1

D

p X1 i D1

.

.p 1 i/ 1/i x1

C O.p

1/ D

p X2 i D0

.i/

. 1/i x1 C O.p

1/:

10.2

More on H.2I .1; 2//.2/ Œp .p/

.p 1/

Lemma 10.2.13. Put t1 WD DH x2 x2 Cx1

31 .p/

.2p 1/

.x2 x2 Cx2

/ and assume

.2/

p

that there is a toral element t2 D @2 C@2 CrDH .x2 /CO.1/ which commutes with t1 . Set .p/

u0 WD x2

.p 1/

x2

C x1

1

Id/ ı .

p u1 WD .t2

.p/

.x2

p X1

x2

.2p 1/

C x2

/;

.p t1i / .x1

1/ .p 2 p/ / x2

1;

.p Id/ .x1

1/ .p 2 p/ x2 /

1:

i D1

pX1 p u2 WD . t2i / ı .t1

1

i D1

The following holds: .p/

.p 1/

.p/

(1) u0 D x2 x2 C x1 x2 C O.p C 1/, u1 D x1 C O.2/, and u2 D x2 O.p C 1/. (2) ti .uj / D ıi;j .1 C uj / for i D 1; 2 and j D 0; 1; 2.

C

j

p 1

(3) O.2I .1; 2// D ˚i;j;kD0 F ui0 u1 uk2 . p 1

(4) ¹u0 ; u1 º D .1 C u1 /, ¹u0 ; u2 º D 0, ¹u1 ; u2 º D u0

.1 C u1 /.1 C u2 /.

Proof. (1) follows by definition and Lemma 10.2.12. (2) Note that t1 .u0 / D DH .u0 /.u0 / D 0. As t1 ; t2 are toral and the action on p p O.2I .1; 2// is a restricted action, t1 t1 , t2 t2 vanish on O.2I .1; 2//. Therefore p DH .t2 .u0 // D Œt2 ; DH .u0 / D Œt2 ; t1 D 0, hence t2 .u0 / 2 F and t2 .u0 / D t2 .u0 / D 0. Moreover, .p 1/ .p2 p/ p 1 p .t1 Id/.u1 / D .t2 Id/ ı .t1 t1 / .x1 x2 / .t1 Id/.1/ D 1; t1 .u2 / D

pX1

p t2i ı .t1

.p t1 / .x1

1/ .p 2 p/ x2 /

t1 .1/ D 0;

.p 1/ .p 2 p/ x2 /

t2 .1/ D 0;

i D1

p t2 .u1 / D .t2

t2 / ı

pX1

t1i

.x1

i D1

.t2

p

Id/.u2 / D .t2

p 1

t2 / ı .t1

.p Id/ .x1

1/ p 2 p/ x2 /

.t2

Id/.1/ D 1:

(3) The right hand side elements are linearly independent and therefore span a space. (4) Note that ¹u0 ; ui º D t1 .ui / D ı1;i .1 C ui / and

p 3 -dimensional

.p/

¹u1 ; u2 º D ¹ x1 C O.2/; x2

C O.p C 1/º D

.p 1/

x2

C O.p/:

Considering one gets ¹u1 ; u2 º D q.1 C u1 /.1 C u2 / where t1 .q/ D t2 .q/ D 0. Proots p 1 Then q D i Di0 si ui0 for suitable si 2 F and si0 ¤ 0, whence ¹u1 ; u2 º D si0 x2i0 C O.i0 C 1/. This gives i0 D p 1, sp 1 D .p 1 1/Š D 1.

32

10


Using the preceding lemma one derives the general multiplication formula ¹ui0 .1 C u1 /j .1 C u2 /k ; ul0 .1 C u1 /r .1 C u2 /s º D .ir

lj /uiCl 0

C .js

1

.1 C u1 /j Cr .1 C u2 /kCs

i ClCp 1

kr/u0

(10.2.1)

.1 C u1 /j Cr .1 C u2 /kCs :

If .a; b/ ¤ .0; 0/, then DH ul0 .1 C u1 /a .1 C u2 /b 2 ŒF t1 C F t2 ; FDH ul0 .1 C u1 /a .1 C u2 /b H.2I .1; 2//.2/ : Also, ŒDH u0 .1 C u1 / ; DH ul0 .1 C u1 /p p X1

X

lD0 .a;b/¤.0;0/

1

D

.l C 1/DH .ul0 /. Consequently,

pX2 FDH ul0 .1 C u1 /a .1 C u2 /b C FDH .ul0 / H.2I .1; 2//.2/ : lD1

A dimension argument then yields equality. We conclude Theorem 10.2.14. Let T H.2I .1; 2//.2/ Œp be a 2-dimensional torus and T \ H.2I .1; 2//.2/ be a torus which is not contained in H.2I .1; 2//.2/ .0/ . Let be any nonzero T -root. ˇ ˇ (1) ˇ.H.2I .1; 2//.2/ ; T /ˇ D p 2 , dim H.2I .1; 2//.2/ D p, dim CH.2I.1;2//.2/ .T / D p 2. (2) Every root space H.2I .1; 2//.2/ contains an element d for which F d Œp D T \ ker : (3) If .T \ H.2I .1; 2//.2/ / D 0, then the 1-section H.2I .1; 2//.2/ . / is abelian. (4) If .T \ H.2I .1; 2//.2/ / ¤ 0, then the 1-section H.2I .1; 2//.2/ . / is isomorphic to H.2I 1/.2/ under an isomorphism which maps T \ H.2I .1; 2//.2/ onto FDH .x1 .1 C x2 //. Proof. (1) Due to Theorem 10.1.4 all 2-dimensional tori of H.2I .1; 2//.2/ Œp have maximal dimension and Theorem 1.5.5(2) shows that they are Winter conjugate. Now Theorem 1.5.1 proves that there is a bijection of roots and root space dimensions for any two of these tori. Therefore Corollary 10.2.9(1) proves the first statement. (2) Choose according to Lemma 10.2.11 an automorphism ‰ and toral elements t1 ; t2 , and for these choices u0 ; u1 ; u2 as in Lemma 10.2.13. To simplify notation we suppress the notion of ‰. Note that T \ H.2I .1; 2//.2/ D F t1 . Define roots ˛, ˇ by setting ˛.t1 / D 1; ˛.t2 / D 0; ˇ.t1 / D 0; ˇ.t2 / D 1;

More on H.2I .1; 2//.2/ Œp

10.2

33

and put D i ˛ C jˇ ¤ 0 for some i; j 2 Fp . Lemma 10.2.13(2),(3) shows that Pp 1 H.2I .1; 2//.2/ D kD0 FDH uk0 .1 C u1 /i .1 C u2 /j . If i ¤ 0 set d WD DH .1 C u1 /i .1 C u2 /j : Due to Equation (10.2.1) one has d .u0 / D ¹.1 C u1 /i .1 C u2 /j ; u0 º D

i.1 C u1 /i .1 C u2 /j ;

d 2 .u0 / D ¹.1 C u1 /i .1 C u2 /j ; i.1 C u1 /i .1 C u2 /j º D 0I d .u1 / D ¹.1 C u1 /i .1 C u2 /j ; u1 º D

p 1

j u0

.1 C u1 /iC1 .1 C u2 /j ;

d .ul0 .1 C u1 /riC1 .1 C u2 /rj / D ¹.1 C u1 /i .1 C u2 /j ; ul0 .1 C u1 /riC1 .1 C u2 /rj º D d p .u1 / D

liul0 1 .1 C u1 /.rC1/i C1 .1 C u2 /.rC1/j j. 1/p

1

1/Ši p

.p

1

if l > 0;

.1 C u1 /piC1 .1 C u2 /pj D j.1 C u1 /I p 1

d .u2 / D ¹.1 C u1 /i .1 C u2 /j ; u2 º D iu0

.1 C u1 /i .1 C u2 /j C1 ;

d .ul0 .1 C u1 /ri .1 C u2 /rj C1 / D ¹.1 C u1 /i .1 C u2 /j ; ul0 .1 C u1 /ri .1 C u2 /rj C1 º D

liul0 1 .1 C u1 /.rC1/i .1 C u2 /.rC1/j C1

d p .u2 / D i. 1/p

1

.p

1/Ši p

1

if l > 0;

.1 C u1 /pi .1 C u2 /pj C1 D

i.1 C u2 /:

From this one concludes d p .uk / D .jt1 it2 /.uk / for k D 0; 1; 2. The elements u0 ; u1 ; u2 generate O.2I .1; 2// as an ordinary algebra. Therefore d p .jt1 it2 / vanishes on O.2I .1; 2//. As the action isa restricted one, this gives d Œp D jt1 it2 . For i D 0 put d WD DH u0 .1 C u2 /j , compute d .u0 / D d .u2 / D 0, d ..1 C u1 /.1 C u2 /rl / D .1 C u1 /.1 C u2 /.rC1/l for 0 r p 1, and argue as above to show that d Œp D t1 . The claim follows. (3) Suppose .T \ H.2I .1; 2//.2/ / D 0,P which means D jˇ for some j ¤ 0. p 1 The 1-section in direction is contained in i;kD0 FDH ui0 .1 C u2 /kj . Equation (10.2.1) shows that ŒDH ui0 .1 C u2 /kj ; DH ur0 .1 C u2 /sj D DH ¹ui0 .1 C u2 /kj ; ur0 .1 C u2 /sj º D 0: (4) Suppose .T \ H.2I .1; 2//.2/ / ¤ 0, which means D i˛ C jˇ for some i ¤ 0. Then H.2I .1; 2//

.2/

. / D

p X1 pX1 lD0 rD1

FDH ul0 .1

ri

rj

C u1 / .1 C u2 /

C

p X2 lD1

FDH .ul0 /:

34

10


Define a bijective mapping H.2I .1; 2//.2/ . / ! H.2I 1/.2/ ; DH ul0 .1 C u1 /ri .1 C u2 /rj 7! i

1

DH .x1l .1 C x2 /lCr /:

Equation (10.2.1) readily shows that this is a Lie algebra homomorphism which maps t1 D DH .u0 / onto i 1 DH .x1 .1 C x2 //. For arbitrary Lie algebras g set

1 .g/ WD g;

i C1 .g/ WD Œg; i .g/ for i > 1:

Corollary 10.2.15. Suppose H.2I .1; 2//.2/ g H.2I .1; 2// and let H.2I .1; 2//.2/ Œp gŒp H.2I .1; 2//Œp denote the respective minimal p-envelopes in Der H.2I .1; 2//.2/ . Suppose t gŒp is a 2-dimensional torus such that t \ g is a 1-dimensional torus not contained in H.2I .1; 2//.0/ . Set h WD Cg .t/. If h acts non-trigonalizably on g, then there is h 2 h such that t D t \ 3 .h/ ˚ F hŒp . In particular, no nonzero root vanishes on h. .p 2 /

.p/

Proof. (a) Since H.2I .1; 2//Œp =H.2I .1; 2//.2/ Œp Š FDH .x1 / C FDH .x2

.p 1/ .p 2 1/ FDH .x1 x2 / is malize t D F t1 ˚ F t2

10.2.13. Write u1 D

x1 C

/C

H.2I .1; 2//.2/

Œp-nilpotent, t Œp . Therefore we may noras in Lemma 10.2.11 and choose u0 ; u1 ; u2 as in Lemma

P

.a/ .b/

aCb2

˛a;b x1 x2 . Since t2 .u1 / D 0, we have ˛0;p D 0.

.a/ .b/

Then every P ˛a;b x1 x2 is a product of elements of O.2; .1; 2//.1/ . Write u1 D x1 C fi gi with fi ; gi 2 O.2; .1; 2//.1/ . Definition 2.1.1 and Equations (2.1.4), P .p/ p .p/ .p/ .p/ fi gi D x1 .mod O.2I .1; 2///. (2.1.5) therein show that u1 x1 C .p/

One similarly obtains u2

.p 2 /

x2

.mod O.2I .1; 2///. Therefore .p/

.p/

H.2I .1; 2// D DH .O.2I .1; 2/// ˚ FDH .u1 / ˚ FDH .u2 /: Since t1 D DH .u0 /, one has .p/

.p 1/

t1 .u1 / D u1

.p/

.p 1/

t1 .u2 / D 0:

t1 .u1 / D u1 t1 .u2 / D u2

.p 1/

.p 1/

.1 C u1 / D u1

Put d WD DH .1 C u1 /. In part (2) of the proof for Theorem 10.2.14 (with i D 1, j D 0) we showed that d p D t2 . The elements u0 ; u1 ; u2 generate O.2I .1; 2// as .p/ .p/ an ordinary algebra. Therefore t2 C d p vanishes on O.2I .1; 2//. However, u1 ; u2

More on H.2I .1; 2//.2/ Œp

10.2

35

are no longer contained in O.2I .1; 2//. Regarded as elements in Der O..2// t2 and d p satisfy t2 D @2 C @2 C d0 , d D DH .1 C u1 / D DH . x1 C O.2// D @2 C d00 , where 0 d0 ; d0 2 H.2I .1; 2//.0/ . As d00 p 2 H.2I .1; 2//.0/ by Lemma 7.1.1(3), t2 C d p D p p @2 @2 C dQ with dQ 2 H.2I .1; 2//. This element annihilates O.2I .1; 2//, whence dQ D 0. Consequently, t2 C d p D 0 as an element in Der O..2//. In particular, p .p/ .p/ t2 .u1 / D DH .1 C u1 / .u1 / p 1 .p 1/ D DH .1 C u1 / u1 DH .1 C u1 /.u1 / D 0 and p .p/ DH .1 C u1 / .u2 / p 1 .p DH .1 C u1 / u2 p 1 .p DH .1 C u1 / u2 p 1 .p DH .1 C u1 / u2

.p/

t2 .u2 / D D D D

1/

DH .1 C u1 /.u2 /

1/ p 1 u0 .1

C u1 /.1 C u2 / 1 .1 C u1 / :

1/ p u0

From this one proves inductively .p/

t2 .u2 / D

.p 1/

.p

1/Šu2

.p 1/

.1 C u1 /p D u2

.p 1/

.1 C u2 / D u2

t2 .u2 /:

(b) Set .p/

q1 WD u1

C

p X1

. 1/i i

1 i u1 ;

.p/

q2 WD u2

C

i D1

p X1

. 1/i i

1 i u2 :

iD1

Then t1 .q1 / D

.p 1/ u1

C

p X1

.p 1/

. 1/i ui1 1 .1 C u1 / D u1

p 1

C .u1

1/

i D1 .p 1/

D u1

.p 1/

C .p

1/Šu1

1D

1;

t1 .q2 / D 0; t2 .q1 / D 0; t2 .q2 / D

.p 1/ u2

C

p X1

. 1/i ui2 1 .1 C u2 / D

1:

i D1

This proves DH .q1 /; DH .q2 / 2 CH.2I.1;2// .t/, whence by dimension reasons CH.2I.1;2// .t/ D

p X1 i D1

FDH .ui0 / ˚ FDH .q1 / ˚ FDH .q2 /:

36

10


Pp 1 (c) Suppose h i D1 FDH .ui0 / ˚ FDH .q2 /. As ¹ui0 ; q2 º D iui0 1 ¹u0 ; q2 º D i ui0 1 t1 .q2 / D 0, in this case h is abelian. Pp 1 Otherwise h contains an element h WD DH .q1 / C i D1 i DH .ui0 / C DH .q2 /, i ; 2 F . Recall that DH .u30 / 2 CH.2I.1;2//.2/ .t/ h. Then Œh; DH .uk0 / D DH ¹q1 C

p X1

i ui0 C q2 ; uk0 º D DH

kuk0

1

¹u0 ; q1 C q2 º

i D1

D

kDH uk0

1

t1 .q1 C q2 / D kDH .uk0

1

/

is contained in 5 k .h/ for k D 3; 2. Consequently, t1 D DH .u0 / 2 3 .h/. (d) As q1 is a polynomial in u1 only, 1 D t1 .q1 / D @u1 .q1 /t1 .u1 / D @u1 .q1 /.1 C u1 /: Similarly one obtains @u2 .q2 /.1 C u2 / D

1;

and this gives p 1

¹q1 ; q2 º D @u1 .q1 /@u2 .q2 /¹u1 ; u2 º D @u1 .q1 /@u2 .q2 /u0

p 1

.1 C u1 /.1 C u2 / D u0

:

As ¹q1 ; uk0 º D

kuk0

1

kuk0

@u1 .q1 /¹u0 ; u1 º D

1

@u1 .q1 /.1 C u1 / D kuk0

1

and ¹q2 ; uk0 º D 0, one obtains k ŒDH .q2 /; ad DH .q1 / DH .q2 / D 0; p 1 ad DH .q1 / DH .q2 / D .p j

Next we observe that ¹q1 C q2 ; u0 º D

j 1

j u0

0kp

2;

1/ŠDH .u0 / D j 1

t1 .q1 C q2 / D j u0

t1 : , whence

k j ŒDH .ui0 /; ad DH .q1 C q2 / DH .u0 / D 0; p 1 j ad DH .q1 C q2 / DH .u0 / D 0 for all 1 i; j p p

1 and 0 k p p

p

2. Remark 10.2.4 gives p

h D DH .q1 / C DH .q2 /

t1 C

p X1

p

i DH .ui0 /p :

i D1

Next we compute these p-th powers. As ¹q1 ; u0 º D 1, ¹q1 ; u1 º D 0, ¹q1 ; u2 º D p 1 p 1 @u1 .q1 /u0 .1 C u1 /.1 C u2 / D u0 .1 C u2 /, one easily concludes DH .q1 /p .u0 / D 0;

DH .q1 /p .u1 / D 0;

DH .q1 /p .u2 / D

.1 C u2 / D

t2 .u2 /:

10.2

More on H.2I .1; 2//.2/ Œp

37

Similarly one shows that DH .q2 /p .ul / D 0 for l D 0; 1; 2. As ¹ui0 ; u0 º D ¹ui0 ; u2 º D 0, ¹ui0 ; u1 º D i ui0 1 .1 C u1 /, one easily shows that DH .ui0 /p .ul / D 0 for l D 0; 1; 2 if i > 1. Consequently, hp C t2 C . 1 /t1 annihilates O.2I .1; 2//. Then hŒp C t2 C .

1 /t1 .pC1/

p

2 ¹d 2 H.2I .1; 2// C F @2 j Œd; @1 D Œd; @2 D Œd; DH .x1 x2

/ D 0º D ¹0º:

In the final case let T \ H.2I .1; 2//.2/ be contained in H.2I .1; 2//.2/ .0/ . Since H.2I .1; 2//.2/ .0/ is closed under Œp-th powers, this is a 1-dimensional torus. Write t1 2 T \ H.2I .1; 2//.2/ .0/ ; t2 2 T n H.2I .1; 2//.2/

T D F t1 ˚ F t2 ;

P2 for toral elements t1 ; t2 . Put t1 D tN1 C v1 where tN1 D i;j D1 ri;j xi @j and v1 2 p H.2I .1; 2//.2/ .1/ , t2 D s@2 C v2 , v2 2 H.2I .1; 2//.2/ and s ¤ 0. Also set U WD span ¹DH .x .a/ / j a1 < pI .a2 < p/ _ .a1 D 0; a2 D p/º: Then p

¹d 2H.2I .1; 2// j Œd; @2 D 0º .p 1/ .p 1/ x2 /

U D H.2I 1/.2/ ˚ FDH .x1

.p/

˚ FDH .x2 / H.2I .1; 2//.2/

and U has dimension p 2 . Theorem 10.2.16. Let T H.2I .1; 2//.2/ Œp be a 2-dimensional torus and T \ H.2I .1; 2//.2/ .0/ ¤ ¹0º. Let be any nonzero T -root. The following holds: ˇ ˇ (1) ˇ.H.2I .1; 2//.2/ ; T /ˇ D p 2 , dim H.2I .1; 2//.2/ D p, dim CH.2I.1;2//.2/ .T / D p 2. (2) If .T \H.2I .1; 2//.2/ / D 0, then H.2I .1; 2//.2/ contains an element d which acts non-nilpotently. Moreover, the 1-section H.2I .1; 2//.2/ . / is nilpotent and acts trigonalizably on H.2I .1; 2//.2/ . (3) If .T \ H.2I .1; 2//.2/ / ¤ 0, then the 1-section H.2I .1; 2//.2/ . / is isomorphic to H.2I 1/.2/ under an isomorphism which maps T \ H.2I .1; 2//.2/ onto FDH .x1 x2 /. Proof. (a) Due to Theorem 10.1.4 all 2-dimensional tori of H.2I .1; 2//.2/ Œp have maximal dimension and Theorem 1.5.5(2) shows that they are Winter conjugate. Now Theorem 1.5.1 proves that there is a bijection of roots and root space dimensions for any two of these tori. Therefore Corollary 10.2.9(1) proves the first statement.

38

10


(b) Let be any root (we allow D 0 in this part (b)). Take any w 2 H.2I .1; 2//.2/ \ H.2I .1; 2//.2/ .k/ n H.2I .1; 2//.2/ .kC1/ for some k and write w D wk C w.kC1/ where wk ¤ 0 is homogeneous of degree k and w.kC1/ 2 H.2I .1; 2//.2/ .kC1/ . Observe p

p

.t2 /w D Œs@2 C v2 ; w D Œs@2 ; wk C O.k C 1

p/:

p

Since Œ@2 ; wk is homogeneous of degree k p, this summand vanishes. This gives wk 2 U . Consequently, there is a linear embedding gr H.2I .1; 2//.2/ ,! U . Moreover, since gr t1 D tN1 , every t1 -eigenvector for the eigenvalue .t1 / is mapped onto a tN1 -eigenvector for the same eigenvalue .t1 /. Next observe that F tN1 is a 1-dimensional torus in U . Due to Theorem 7.5.8 it is conjugate under Aut H.2I 1/.2/ to one of the tori FDH .x1 x2 / or FDH ..1 C x1 /x2 /. This normalization shows that every tN1 -eigenspace of U is p-dimensional. (c) Suppose ¤ 0 and .t1 / D 0. We proved in (a) that H.2I .1; 2//.2/ is pdimensional. A dimension argument shows that the linear mapping mentioned in (b) is an isomorphism onto CU .tN1 /. In particular, there exists a root vector d WD tN1 Cw.1/ 2 H.2I .1; 2//.2/ , w.1/ 2 H.2I .2; 1//.2/ .1/ . This element acts non-nilpotently. In addition, these deliberations show that every root space and therefore the full 1-section H.2I .1; 2//.2/ . / has the property that H.2I .1; 2//.2/ . / F t1 C CH.2I.1;2//.2/ .1/ .t1 /: Then H.2I .1; 2//.2/ . / is nilpotent and acts trigonalizably on H.2I .1; 2//.2/ . (d) Suppose .t1 / ¤ 0. We mentioned in (a) that H.2I .1; 2//.2/ i is p-dimensional for every i 2 Fp . A dimension argument shows that the linear mapping mentioned in (b) is an isomorphism onto Ui .t1 / . Then we have H.2I 1/.2/ gr H.2I .1; 2//.2/ . / H.2I 1/: Definition 4.2.4 shows that H.2I .1; 2//.2/ . / is a Lie algebra of Hamiltonian type, .1/ Theorem 4.2.7 yields that H.2I .1; 2//.2/ . / is simple and Theorem 6.2.2 im.1/ .2/ plies that H.2I .1; 2// . / is of type H.2I 1I '/.2/ . In addition, as a 1-section with respect to a torus of maximal dimension this algebra has absolute toral rank 1 (see Theorem 1.3.11(3)). Theorem 10.1.4 shows that there is an isomorphism .1/ H.2I .1; 2//.2/ . / Š H.2I 1/.2/ . Moreover we have dim H.2I .1; 2//.2/ . / D .2/ 2 p 2 D dim H.2I 1/ by assertion (1), and this gives rise to an isomorphism ˆ W H.2I .1; 2//.2/ . / Š H.2I 1/.2/ : The uniquely determined maximal subalgebras of codimension 2 (see Theorem 4.2.6) H.2I .1; 2//.2/ . / \ H.2I .1; 2//.2/ .0/ and H.2I 1/.2/ .0/

10.3

2-dimensional tori in H.2I 1I ˆ.//.1/

39

are mapped onto each other under this isomorphism. In particular, t1 is mapped into H.2I 1/.2/ .0/ . Due to Theorem 7.5.8 there is an automorphism of H.2I 1/.2/ which maps F ˆ.t1 / onto FDH .x1 x2 /. We mention that each type of tori described in Corollary 10.2.9, Theorem 10.2.14 and Theorem 10.2.16, respectively, does occur. For tori as in Corollary 10.2.9 we just quote Theorem 10.2.7. Next we turn to Theorem 10.2.14. Let g WD .O.2I .1; 2//; ¹; º/ denote the Poisson Lie algebra given by the vector space O.2I .1; 2// and the multiplication as in Equation (10.2.1). Theorem 6.3.2 proves that the mapping x1 7! y1 WD u1 D x1 C O.2/ .p 1/ and x2 7! y2 WD u0 Cu2 D x2 Cx1 x2 CO.pC1/ induces an automorphism of the flag algebra O.2I .1; 2//. Put q WD ¹y ; 1 y2 º 2 O.2I .1; 2// and observe that ¹f; gº D q @y1 .f /@y2 .g/ @y2 .f /@y1 .g/ . The mapping f 7! q@y1 .f /@y2 q@y2 .f /@y1 im.1/ poses a Hamiltonian structure on g=C.g/. Hence g=C.g/ Š H.2I nI ˆ/.1/ for some choices of n and ˆ. Next we determine dimensions. It is not hard to show that p 1 u0 is not contained in g.1/ . Since u00 D ¹u0 .1 C u1 /p 1 ; u1 º 2 g.1/ \ C.g/, one has .1/ dim g=C.g/ p 3 2. So it can only be that gQ WD g.1/ =C.g.1/ / Š H.2I .1; 2//.2/ . The proof of Theorem 10.2.14(2) shows that the p-envelope of gQ in Der gQ contains the torus F t1 ˚ F t2 as described in Lemma 10.2.13(2). That lemma also shows that the multiplication in gQ and the action of T on gQ are uniquely determined up to the above mentioned automorphism of H.2I .1; 2//.2/ . Consequently, there is just one orbit of such 2-dimensional tori. To deal with Theorem 10.2.16 we take a 2-dimensional torus t as in Theorem 10.2.14. There is a t-1-section H.2I .1; 2//.2/ . / Š H.2I 1/.2/ of Hamiltonian type. Choose a toral element t1 in H.2I .1; 2//.2/ .0/ which is mapped onto DH .x1 x2 /. Set T WD F t1 ˚ t \ ker . This is a 2-dimensional torus having the T -1-section H.2I .1; 2//.2/ . /. By construction this torus is not as in Corollary 10.2.9 and Theorem 10.2.14. So it is as in Theorem 10.2.16.

10.3


In §6.3 we have dealt with the Special algebras S.mI nI '/.1/ , however we have mentioned on p. 299 of Volume 1 that the case m D 2 covers the respective Hamiltonian algebras. Namely, contrary to the general assumption that we speak of Special algebras only when m 3, we may read there S.2I nI '/ D H.2I nI '/. One has to keep in mind though that S.2I nI '/.1/ is not always simple. In the course of the deliberations on p. 308/309 of Volume 1 we showed that H.2I 1I ˆ.// is the set of elements of W .2I 1/ annihilating the Hamiltonian form ! WD .1 C x ..1// /dx1 ^ dx2 . This means H.2I 1I ˆ.// D H.2I 1I !/; ! WD .1 C x . .1// /dx1 ^ dx2 :

40

10


Note that .1 C x ..1// / 1 D 1 x ..1// and .1 a .1/ and jaj 1. Theorem 6.3.7(1) gives x . .1// /@1 C F .1

H.2I 1I ˆ. // D F .1

x ..1// /x .a/ @1 D x .a/ @i whenever

x . .1// /@2 C

X

H.2I 1/k

k0

D F .1

x

..1//

/@1 C F .1

x

..1//

X

/@2 C

FDH .x .a/ /

a.1/; jaj2

C

.p/ FDH .x1 /

C

.p/ FDH .x2 /:

P2

!i;j dxi ^ dxj with !i;j D

!1;2 D

1 !2;1 D .1 C x ..1// /; 2

Next put ! D

i;j D1

!j;i , where

!i;i D 0 for i D 1; 2;

and compute .gi;j / D .!i;j / 1 , g1;2 D

g2;1 D

2.1

x ..1// /;

gi;i D 0 for i D 1; 2:

Equation (6.5.4) in Volume 1, p. 338 gives DH;! .f / D 2.1

x ..1// / @1 .f /@2

@2 .f /@1 :

(10.3.1)

The above deliberations show .p/

.p/

H.2I 1I ˆ.// D DH;! .O.2I 1// ˚ FDH;! .x1 / ˚ FDH;! .x2 /; while Theorem 6.3.10(1) gives H.2I 1I ˆ. //.1/ D F .1

x ..1// /@1 C F .1

X

x ..1// /@2 C

FDH .x .a/ /

a.1/;jaj2

D DH;! .O.2I 1//: Recall that H.2I 1I ˆ.//.1/ Œp denotes the p-envelope of H.2I 1I ˆ.//.1/ in Der H.2I 1I ˆ.//.1/ . Theorem 10.3.1. We have H.2I 1I ˆ.//.1/ Œp D Der H.2I 1I ˆ.//.1/ . Proof. The derivation algebra in question is described in Theorem 7.1.3(3). In that theorem one has !.˛/ WD ! D

1 dx1 ^ dx2 2

dx2 ^ dx1 C x ..1// dx1 ^ dx2

x ..1// dx2 ^ dx1 ;

10.3


41

i.e., ˛1;1 D ˛2;2 D 0, ˛1;2 D ˛2;1 D 1=2. Since the algebra in question is simple, one has H.2I 1I ˆ.//.1/ D H.2I 1I !.˛//.2/ . The theorem states Der H.2I 1I ˆ.//.1/ Š H.2I 1I !/.0/ C FDH;! .x2 / C FDH;! .x1 / D H.2I 1I ˆ.// .p/

.p/

D DH;! .O.2I 1// C FDH;! .x1 / C FDH;! .x2 / .p/

.p/

D H.2I 1I ˆ.//.1/ ˚ FDH;! .x1 / ˚ FDH;! .x2 /: In particular, we get dim Der H.2I 1I ˆ.//.1/ =H.2I 1I ˆ.//.1/ D 2: On the other hand, Theorem 7.2.2(6) shows that dim H.2I 1I ˆ.//.1/ Œp =H.2I 1I ˆ.//.1/ rank.˛/ D 2: As H.2I 1I ˆ. //.1/ Œp Der H.2I 1I ˆ.//.1/ , the result follows.

Recall that H.2I 1I ˆ.//.1/ W .2I 1/. This latter algebra is a restricted Lie algebra. Let H.2I 1I ˆ.//.1/p denote the p-envelope of H.2I 1I ˆ.//.1/ in W .2I 1/. Then H.2I 1I ˆ. //.1/p Š H.2I 1I ˆ.//.1/ Œp ˚ C , where H.2I 1I ˆ.//.1/ is centralized by C (Theorem 1.1.6(2)). The image of C in W .2I 1/ therefore is annihilated by FDH;! .x1 / C FDH;! .x2 / C FDH;! .x1 x2 /. But then C D ¹0º. We may regard H.2I 1I ˆ. //.1/ Œp as the p-envelope of H.2I 1I ˆ.//.1/ in W .2I 1/. Introduce a Poisson multiplication on O.2I 1/ by setting ¹f; gº! WD DH;! .f /.g/ D 2.1

x ..1// / @1 .f /@2 .g/

@2 .g/@1 .f / :

Equation (6.5.5) in Volume 1, p. 339 shows that DH;! W O.2I 1/ ! H.2I 1I ˆ.//.1/ defines a (surjective) Lie algebra homomorphism with kernel F 1 when O.2I 1/ is equipped with the multiplication ¹ ; º! . It is clear that DH;! .f / 2 W .2I 1/ is a derivation of O.2I 1/, whence ¹f; ghº! D ¹f; gº! h C g¹f; hº!

8f; g; h 2 O.2I 1/:

Let denote any automorphism of O.2I 1/ and put yi WD .xi / for i D 1; 2. Then y1 , y2 are generators for O.2I 1/ as an algebra and are polynomials in x1 , x2 , yi D ri;1 x1 C ri;2 x2 C fi ;

fi 2 O.2I 1/.2/ ;

ri;j 2 F;

det.ri;j / ¤ 0:

42

10


Since ¹f; ‹º! and ¹‹; gº! are derivations, one can express the Poisson bracket in terms of y1 , y2 , ¹f; gº! D @y1 .f /@y1 .g/¹y1 ; y1 º! C @y1 .f /@y2 .g/¹y1 ; y2 º! C @y2 .f /@y1 .g/¹y2 ; y1 º! C @y2 .f /@y2 .g/¹y2 ; y2 º! D @y1 .f /@y2 .g/ @y2 .f /@y1 .g/ ¹y1 ; y2 º! :

Observe that ¹y1 ; y2 º! D 2.1 x ..1// / @1 .y1 /@2 .y2 / @2 .y1 /@2 .y2 / 2 det.rij / .mod O.2I 1/.1/ /. In particular, ¹y1 ; y2 º! is an invertible element in O.2I 1/. Theorem 10.3.2. Let T be a torus of H.2I 1I ˆ.//.1/ Œp of maximal toral rank. (1) There is 2 Aut O.2I 1/, such that for the generators yi WD the Poisson bracket has the form ¹y1 ; y2 º! D ˇ.1 C y1 /.1 C y2 /;

.xi /, i D 1; 2,

ˇ 2 F ;

and T D F .1 C y1 /@y1 ˚ F .1 C y2 /@y2 : (2) Every T -root space of H.2I 1I ˆ.//.1/ for a nonzero root is 1-dimensional while CH.2I1Iˆ.//.1/ .T / D ¹0º. In particular, H.2I 1I ˆ.//.1/ Œp D T ˚ H.2I 1I ˆ.//.1/ and .H.2I 1I ˆ.//.1/ ; T / [ ¹0º is a 2-dimensional Fp -vector space. (3) Every T -1-section is abelian. (4) Let T D F t1 ˚ F t2 , where t1 ; t2 are toral. There exists a T -eigenvector basis .d / such that Œd ; d D .t1 /.t2 / .t2 /.t1 / dC if C ¤ 0: (5) Every nonzero T -root vector v 2 H.2I 1I ˆ.//.1/ satisfies F v Œp D T \ ker : Proof. (1) Regard H.2I 1I ˆ.//.1/ Œp as a subalgebra of W .2I 1/. Note that T is 21 ıT ı dimensional (Theorem 10.1.4). Choose 2 Aut O.2I 1/ such that T 0 WD is one of the tori F .ı1 C x1 /@1 ˚ F .ı2 C x2 /@2 where either ı1 D ı2 D 0, or ı1 D 1, ı2 D 0, or ı1 D ı2 D 1 (Corollary 7.5.2). Put tQi WD

ı .ıi C xi /@i ı

1

2T

and

yi WD

.xi / 2 O.2I 1/:

10.3


43

Then tQj .yi / D

ı .ıj C xj /@j ı

1

.yi / D ıi;j .ıj C xj / D ıi;j .ıj C yj /:

This means that T D F .ı1 Cy1 /@y1 ˚F .ı2 Cy2 /@y2 . We have to compute the Poisson bracket ¹y1 ; y2 º! DW d . Since T is contained in the p-envelope of H.2I 1I ˆ.//.1/ , Equation (6.5.5) on p. 339 of Volume 1 yields Œt; DH;! .f / D DH;! .t.f // Then t .¹f; gº! / Consequently,

8t 2 T:

¹f; t.g/º! D ¹t.f /; gº! for all f; g 2 O.2I 1/ and all t 2 T .

.ı1 C y1 /@y1 .d / D .ı1 C y1 /@y1 .¹y1 ; y2 º! / D ¹.ı1 C y1 /@y1 .y1 /; y2 /º! C ¹y1 ; .ı1 C y1 /@y1 .y2 /º! D ¹y1 ; y2 º! D d: ˇij .ı1 C y1 /i .ı2 C y2 /j we obtain X d D .ı1 C y1 /@y1 .d / D iˇij .ı1 C y1 /i .ı2 C y2 /j :

Thus writing d D

P

0i;j p 1

0i;j p 1

This shows ˇij D 0 for i ¤ 1. By symmetry we also have ˇij D 0 for j ¤ 1 and hence d D ˇ1;1 .ı1 C y1 /.ı2 C y2 /. The constant term of this polynomial is ˇ1;1 ı1 ı2 . It has been mentioned earlier that this constant term is nonzero. As a result ˇ WD ˇ1;1 ¤ 0 and ı1 D ı2 D 1. This proves (1). (2) Every monomial .1 C y1 /a .1 C y2 /b is an eigenvector with respect to T for a root , which is given by ..1 C y1 /@y1 / a, ..1 C y2 /@y2 / b mod .p/. Different exponents yield different roots. We mentioned before that Œt; DH;! .f / D DH;! .t .f // for all t 2 T . Hence every root space in H.2I 1I ˆ.//.1/ for a nonzero a b root is given as FDH;! .1 C y1 / .1 C y2 / with .a; b/ ¤ .0; 0/, whence is 1dimensional. Since F 1 is the zero root space and DH;! .1/ D 0, this yields the assertion about .H.2I 1I ˆ.//.1/ ; T / and the dimensions of root spaces. Recall that dim H.2I 1I ˆ.//.1/ Œp =H.2I 1I ˆ.//.1/ D 2. Then H.2I 1I ˆ.//.1/ Œp D T ˚ H.2I 1I ˆ. //.1/ . (3) By (2) every 1-section of H.2I 1I ˆ.//.1/ is of the form X FDH;! .1 C y1 /ia .1 C y2 /ib 1i p 1

for fixed a; b. We compute ŒDH;! .1 C y1 /ia .1 C y2 /ib ; DH;! .1 C y1 /ra .1 C y2 /rb D DH;! ˇ.iarb

raib/.1 C y1 /.i Cr/a .1 C y2 /.Cr/ib D 0:

44

10


(4) Write ti D ai;1 .1 C y1 /@y1 C ai;2 .1 C y2 /@y2 (i D 1; 2). Since these span the 2-dimensional torus T , a WD a1;1 a2;2 a1;2 a2;1 ¤ 0. As t1 ; t2 are toral, ai;j 2 Fp and a 2 Fp . Set u1 WD .1 C y1 /a2;2 =a .1 C y2 /

a2;1 =a

;

u2 WD .1 C y1 /

a1;2 =a

.1 C y2 /a1;1 =a :

Then t1 .u1 / D

a1;1 a2;2

t1 .u2 / D

a1;2 a2;1 a

u1 D u1 ;

t2 .u1 / D

a1;1 a1;2 C a1;2 a1;1 u1 D 0; t2 .u2 / D a

and ¹u1 ; u2 º! D

ˇ .a2;2 a1;1 a2

a2;2 a2;1

a2;1 a2;2 a

u1 D 0;

a2;1 a1;2 C a2;2 a1;1 u2 D u2 a

a2;1 a1;2 /u1 u2 D

ˇ u1 u2 : a

Put .t1 / .t2 / u2 ;

u WD .ˇ=a/.t1 /C.t2 / 1 u1

2 .H.2I 1I ˆ.//.1/ ; T /:

Direct computation shows that ¹u ; u º! D .t1 /.t2 / .t2 /.t1 / uC . Set d WD DH;! .u /. (5) Since root spaces are 1-dimensional, it suffices to prove the claim for the element d mentioned in (4). As Œd ; dCi D .t1 /.t2 / .t2 /.t1 / dC.iC1/ for p i 2 Fp , we get .ad d /p .d / D .t1 /.t2 / .t2 /.t1 / d . Note that .ti /; .ti / 2 Fp . Therefore F -dependency means Fp -dependency. Hence this expression does not vanish if and are Fp -independent. Consequently, d Œp is nonzero. Due to (2) it is contained in T . It is contained in ker . Corollary 10.3.3. All 2-dimensional tori in H.2I 1I ˆ.//.1/ Œp are conjugate under Aut H.2I 1I ˆ.//.1/ . Proof. Let T and TQ be 2-dimensional tori in H.2I 1I ˆ.//.1/ Œp . Choose toral elements t1 ; t2 ; tQ1 ; tQ2 such that T D F t1 ˚ F t2 and TQ D F tQ1 ˚ F tQ2 , define a bijective linear mapping T ! TQ by rt1 C st2 7! r tQ1 C s tQ2 , define the induced bijective mapQ tQ/ WD .t/ and choose ping .H.2I 1I ˆ.//.1/ ; T / ! .H.2I 1I ˆ.//.1/ ; TQ / via . elements d , dQ as in Theorem 10.3.2(4). The mapping ' W H.2I 1I ˆ.//.1/ Œp ! H.2I 1I ˆ. //.1/ Œp , ' rt1 C st2 C

X 2.H.2I1Iˆ.//.1/ ;T /

a d WD r tQ1 C s tQ2 C

X

a dQ

.1/ ;T Q/ 2.H.2I1Iˆ.// Q

is an automorphism of the restricted algebra which maps T onto TQ .

10.4

10.4

45

Semisimple elements in H.2I 1I ˆ.1//Œp


P .p/ .p/ .i/ .p/ Recall that exp.x1 / D 1 and !H;1 WD .exp.x1 //dx1 ^ dx2 . The i D0 .x1 / Hamiltonian algebra H.2I 1I ˆ.1// is defined as ¹D 2 W .2I 1/ j D.!H;1 / D 0º. It has been described in Theorem 6.5.8 using Equation (6.5.8). Namely, 1 .p DH;1 .f / D DH .f / C x1 2

1/

D DH .f / C

.p 1/ x1 f

D @1 .f /@2

@2 .f /@1

2f @2 C @2 .f /.x1 @1 C x2 @2 / .x1 @1 C x2 @2 /.f /@2 p 1 x1 f

@2 D DH .f / p 1

x1

(10.4.1)

@2

f @2

for all f 2 O.2I 1/, and H.2I 1I ˆ.1// D ¹DH;1 .f / j f 2 O.2I 1/º. This is a simple Lie subalgebra of W .2I 1/ of dimension p 2 (Theorem 6.5.8). We mention explicitly p 1

DH;1 .1/ D

x1

.p 1/

@2 D x1

.p/

@2 D DH .x1 /:

The minimal p-envelope H.2I 1I ˆ.1//Œp is described in Theorem 7.2.2(7) and accordingly is isomorphic via the ad-representation to H.2I 1I ˆ.1//Œp Š H.2I 1I ˆ.1// ˚ F .x1 @1 C x2 @2 /: As in the case of H.2I .1; 2//.2/ one has to be careful when dealing with p-th powers. Arguing as in that section one has D Œp D D p if D 2 H.2I 1I ˆ.1//Œp \ W .2I 1/.0/ . However, Œp p DH;1 .x1 /Œp D @2 D 0; DH;1 .x1 /p D @2 ¤ 0; and (cf. Remark 10.2.4) p 1

x2 @2

p

p 1

x2 @2

Œp

DH;1 .x2 /p D

@1

x1

DH;1 .x2 /Œp D

@1

x1

.p/

.p/

.p/

.p 1/

Since @1 .exp.x1 // D exp.x1 /@1 .x1 / D x1 .i /

.p/

p

@1 C x2 @2 ;

D x2 @2 : .p/

exp.x1 / and

.i 1/

exp.x1 / C x1 exp.x1 /@1 .x1 /

.i 1/

exp.x1 /

p

.p/

@1 .x1 exp.x1 // D x1 D x1

.p/

D

.i/

.p/

.p/

.p/

for 0 < i < p, one gets inductively @1 .exp.x1 // D exp.x1 / and .p/ .p/ p @1 exp.x1 /dx1 ^ dx2 D exp.x1 /dx1 ^ dx2 :

.p/

46

10

Tori in Hamiltonian and Melikian algebras .p/

This explains why DH;1 .x2 /p does annihilate exp.x1 /dx1 ^dx2 , while DH;1 .x2 /Œp does not. Let us compute products, ŒDH;1 .f /;DH;1 .g/ p 1

D ŒDH .f /

x1

D DH .DH .f /.g// p 1

C x1

p 1

g@2 .@1 .f //@2 p 1

p 1

x1

@1 .f /@2 .x1

C @1 .g/@2 .x1 x1

p 1

f @2 ; DH .g/

f /@2

g@2 p 1

g/@2 C @2 .f /@1 .x1

p 1

x1

g@2 .@2 .f //@1 p 1

@2 .g/@1 .x1 p 1

f @2 .@1 .g//@2 C x1 p 1

p 1

@2 .f /g

f /@2

f @2 .@2 .g//@1

D DH .DH .f /.g// C DH .x1 @2 .f /g/ p 1 x1 DH .f /.g/ @2 D DH;1 DH .f /.g/ C x1

g/@2

p 1

x1

p 1

DH .x1

f @2 .g//

f @2 .g/ :

Consequently, p ŒDH;1 .f /; DH;1 .g/ D DH;1 DH .f /.g/ x1

1

p 1

f @2 .g/Cx1

g@2 .f / : (10.4.2)

In a first step we will describe Aut H.2I 1I ˆ.1//. Theorem 7.3.2 states the existence of a group isomorphism ˆ W ¹ 2 Autc O.2I 1/ j .!H;1 / 2 F !H;1 º ! Aut H.2I 1I ˆ.1//;

7! ˆ :

Q P ai Proposition 10.4.1. Let D D ˛a;j m i D1 xi @j 2 W .mI 1/.1/ where ˛a;j 2 Fp . Suppose k is an index such that ak ıj;k 1 whenever ˛a;j ¤ 0. Then exp.tD/ D P tn n n0 nŠ D is a well-defined element of Autc O.mI 1/ for every t 2 F . Proof. P (a) Choose ˛Q a;j 2 Z of ˛a;j and define a continuous derivation Q preimages ai X @ of the power series ring QŒŒX1 ; : : : ; Xm . Let t denote a DQ WD ˛Q a;j m i D1 i j further indeterminate and put Q 2 Aut Q.t/ŒŒX1 ; : : : ; Xm : WD exp.t D/ In particular, .f /p D .f p / holds for every f 2 QŒŒX1 ; ˇ: : : ; Xm . For every polynomial q.t/ with coefficients in QŒŒX1 ; : : : ; Xm let q.t/ˇ t n denote the coefficient of t n . The above equation yields 1 X p ˇ pX1 t j p ˇ 1 Qn p tj Q j ˇ ˇ D .f / D D .f / ˇ n D DQ j .f / ˇ n ; t t nŠ jŠ jŠ j D0

j D0

n D 0; : : : ; p

1:

10.4

47


(b) Consider the Z-lattice WD ˚Z.p/

am X1a1 Xm ; a1 Š am Š

Z.p/ WD

Note that is a ring as products are given by isomorphism of algebras W F ˝Z.p/ ! O.m/;

°r

Xia Xib aŠ bŠ

.1 ˝

s

± j r; s 2 Z; p6 j s :

D

aCb aCb Xi a .aCb/Š ,

and there is an

am X1a1 Xm / D x .a/ : a1 Š am Š i

(c) We will apply Lemma 2.1.2 to discuss the mappings f 7! f .i/ WD fiŠ on P QŒŒX1 ; : : : ; Xm . The definition immediately shows .f C g/.i/ D ji D0 f .j / g .i j / .r/

and .fg/.i / D i Šf .i/ g .i/ . Note that Xi 2 for all r 2 N by definition. Therefore is closed under all these divided mappings. P power j For arbitrary r 2 N let r D rj p (0 rj < p) be the p-adic expansion. Due to Lemma 2.1.2(3) and induction on the number of summands there is ˛ 2 Z with Q ˛ 1.p/ and rŠ D ˛ .rj p j /Š. This means Y j f .r/ D ˛ 1 f .rj p / : Due to Lemma 2.1.2(2) there are ˇj ; j 2 Z with ˇj j 1.p/ and .rj p j /Š D ˇj .rj Š/.p j Š/rj , p j C1 Š D j .pŠ/.p j Š/p . This means f .rj p

j/

j

j

D ˇj 1 .f .p / /.rj / D ˇj 1 .rj Š/ 1 .f .p / /rj and f .p

j C1 /

j

D j 1 .f .p / /.p/ :

Due to the above relations and Equations (2.1.10), (2.1.9) the isomorphism respects this divided power structure. (d) The assumption on k implies that for every f 2 O..m// and i 1 there is p ip fQi 2 O..m//.1/ such that D i .f / D xki fQi . As xk D 0 and .xki fQ/.p/ D xk fQ.p/ D 0, one has D p D 0; .D i .f //.p/ D 0 if i 1; and D i1 .f1 / D il .fl / D 0 8f1 ; : : : ; fl 2 O..m// if l 1; i1 C C il p: P Pp 1 n n Then exp.tD/ D n0 tnŠ D n D nD0 tnŠ D n is a well-defined automorphism of O..m//. It is continuous and maps O.mI 1/ into itself, because D has these properties. (e) The final set of equations in (a) gives in particular a set of equations in for Q Xiai f D ai Š and therefore gives the following set of equations in F ˝Z.p/ Š O.m/ with f D x .a/ , a > 0, pX1 t j .p/ ˇ 1 n .p/ ˇ D .f / D D j .f / ˇ n; t nŠ jŠ j D0

n D 0; : : : ; p

1:

48

10


.p/ These equations show that WD exp.tD/.f .p/ / exp.tD/.f / is of t -degree p. More exactly, it is contained in the divided power algebra generated by elements of the form .D i .f //.p/ and D i1 .f / D il .f /, where i 1 and i1 C C il p. Due to (d) every such element vanishes. Then D 0 and exp.tD/.f .p/ / D .p/ exp.tD/.f / holds if f is of the form f D x .a/ . Using Equations (2.1.4), (2.1.10) and (2.1.9) one obtains that exp.tD/ is a continuous divided power automorphism of O..m//. Lemma 10.4.2. Let f 2 O.2I 1/.1/ , g 2 O.2I 1/.s/ and s 1. (1) .f C g/.p/ D f .p/ C g .p/ C h where h 2 O.2I 1/.p (2) If f 2 O.2I 1/.2/ then f

.p/

1Cs/ .

D 0, and exp.f / D 1 C f if f 2 O.2I 1/.p/ .

(3) Suppose s 2. Then exp..f C g/.p/ / D .exp.f .p/ //.1 C h/ where h 2 O.2I 1/.p 1Cs/ . (4) Let fQ 2 O.2I 1/.r/ and 2 r < s. Then exp..f CfQCg/.p/ / D .exp.f .p/ //.1C f .p 1/ fQ C h/ where h 2 O.2I 1/.pCr/ . Pp 1 Proof. (1) Equation (2.1.4) shows .f C g/.p/ D f .p/ C g .p/ C i D1 f .i/ g .p i/ . Recall that q .i/ 2 O.2I 1/.il/ if q 2 O.2I 1/.l/ (see Volume 1, p. 59). Let h WD Pp 1 .i / .p i/ g 2 O.2I 1/.p 1Cs/ . iD1 f (2) Claim 2 follows from the fact that O.2I 1/.2p/ D ¹0º. (3) By (2), g .p/ D 0. Then (1) shows the existence of h 2 O.2I 1/.p 1Cs/ such that exp..f C g/.p/ / D exp.f .p/ C h/ D .exp.f .p/ //.exp.h//. Apply (2) to the second factor. (4) By Equation (2.1.4), .f C fQ C g/.p/ D f .p/ C .fQ C g/.p/ C f .p

1/

.fQ C g/ C

p X2

f .i/ .fQ C g/.p

i/

:

i D1

Note that .fQ C g/.p/ D 0 by (2), f .p

1/ g

2 O.2I 1/.p

1Cs/

O.2I 1/.pCr/ . As

i C r.p i / D p C .r 1/.p i/ p C 2.r 1/ p C r if i p 2; Pp 2 we obtain i D1 f .i/ .fQ C g/.p i/ 2 O.2I 1/.pCr/ . Consider h WD f .p 1/ g C Pp 2 .i / Q .f Cg/.p i/ . Then exp..f C fQ Cg/.p/ / D .exp.f .p/ //.exp.f .p 1/ fQ C i D1 f h// D .exp.f .p/ //.1 C f .p 1/ fQ C h/ by (2). Theorem 10.4.3. (1) Let 1 ; 2 2 F , 2 ¤ 0. The mapping .1 ;2 / 2 Autc O.2I 1/ given by .1 ;2 / .x1 / WD x1 ; satisfies .1 ;2 / .!H;1 / D 2 !H;1 .

.1 ;2 / .x2 / WD 1 x1 C 2 x2

10.4

49


(2) For ˛ 2 F the mappings ´ .p exp.DH;1 .˛// D Id C ˛x1 .˛ja;b/ WD exp.DH;1 .˛x1a x2b //;

1/

@2 ;

a D p; b D 0; a; b p 1; a C b 3;

are contained in Autc O.2I 1/. They satisfy .˛ja;b/ .!H;1 / D !H;1 . (3) Let ' denote any assignment which to every fk 2 O.2I 1/k C ık;p F (3 k 2p 2) attaches 'fk 2 Autc O.2I 1/ satisfying 'fk .!H;1 / 2 F !H;1 and 'fk .xi / xi C DH;1 .fk /.xi /

i D 1; 2:

.mod O.2I 1/.k/ /;

Every 2 Autc O.2I 1/ satisfying .!H;1 / 2 F !H;1 has a presentation D

2p Y2

'fk ı .1 ;2 /

kD3

for suitable 1 ; 2 2 F and fk 2 O.2I 1/k C ık;p F . Proof. (1) Recall that .!H;1 / D exp..x1 /.p/ /d..x1 // ^ d..x2 // holds for any 2 Autc O.2I 1/. In the present case this gives .p/

.1 ;2 / .!H;1 / D exp.x1 /dx1 ^ d.1 x1 C 2 x2 / D 2 !H;1 : p 1

(2) Note that DH;1 .1/ D x1 @2 is an element of W .2I 1/ of x1 -degree p 1. Proposition 10.4.1 shows that exp.DH;1 .˛// 2 Autc O.2I 1/. p 1Ca b x2 @2 Suppose b 2. Then DH;1 .x1a x2b / D ax1a 1 x2b @2 bx1a x2b 1 @1 x1 has x2 -degree b 1 1. Apply Proposition 10.4.1. One argues similarly if a 2. Since DH;1 .x1a x2b /.!H;1 / D 0, one has .˛ja;b/ .!H;1 / D !H;1 . (3) (a) Write .xi / D ri;1 x1 C ri;2 x2 C fi ;

i D 1; 2:

fi 2 O.2I 1/.2/ ;

Using Lemma 10.4.2 one gets .p/

p

.p/

.x1 / D .x1 /.p/ D .r1;1 x1 C r1;2 x2 C f1 /.p/ D r1;1 x1

p

.p/

C r1;2 x2

C g;

where g 2 O.2I 1/.p/ . Therefore there is 2 F such that .p/

.p/

exp.x1 /dx1 ^ dx2 D exp.x1 /dx1 ^ dx2

.p/

D exp..x1 //d.x1 / ^ d.x2 / p

.p/

p

.p/

D .exp.r1;1 x1 //.exp.r1;2 x2 //.1 C g/ @1 ..x1 //@2 ..x2 //

@2 ..x1 //@1 ..x2 // dx1 ^ dx2 :

50

10


Consequently, p

D exp..r1;1

.p/ p .p/ 1/x1 / exp.r1;2 x2 / .r1;1 r2;2

r1;2 r2;1 /.1 C q/

for some q 2 O.2I 1/.1/ , which gives p

.1 C .r1;1 p

.1 C .r1;1

.p/

.p/

p

1/x1 /.1 C r1;2 x2 /.r1;1 r2;2 .p/

1/x1

p

.p/

C r1;2 x2 /.r1;1 r2;2

r1;2 r2;1 /.1 C q/

r1;2 r2;1 /

.mod .O..2//.pC1/ C O.2I 1/.1/ //: We arrive at p

r1;1 D 1;

r1;2 D 0;

r1;1 r2;2

r1;2 r2;1 D ;

whence r1;1 D 1, r2;2 D ¤ 0. Then .r2;1 ;r2;2 / 1 ı .xi / xi 2 O.2I 1/.2/ for i D 1; 2. (b) We now may assume that .xi / xi 2 O.2I 1/.2/ for i D 1; 2. Then .x1a x2b / D x1a x2b C O.a C b C 1/ for all a; b. Looking at as an endomorphism of the graded algebra O.2I 1/ in its natural grading we decompose into the sum of its homogeneous parts, X D Id C l C k ; l 1: k>l

Comparing the term dx1 ^dx2 of lowest degree in !H;1 it is easy to see (as .!H;1 / 2 F !H;1 ) that .!H;1 / D !H;1 . Then (see Lemma 10.4.2, let O.s/ stand for elements in O.2I 1/.s/ ) !H;1 D .!H;1 / D exp..x1 C O.l C 1//.p/ / d.x1 C l .x1 / C O.l C 2// ^ d.x2 C l .x2 / C O.l C 2//

.p/

D exp.x1 /.1 C O.p C l// dx1 ^ dx2 C dl .x1 / ^ dx2 C dx1 ^ dl .x2 / C O.l C 1/dx1 ^ dx2 D 1 C @1 .l .x1 // C @2 .l .x2 // C O.l C 1/ !H;1 ;

and this gives 1 D 1 C @1 .l .x1 // C @2 .l .x2 // C O.l C 1/. Comparing terms of degree l we get @1 .l .x1 // C @2 .l .x2 // D 0. Set D WD l .x1 /@1 C l .x2 /@2 . Then D.dx1 ^ dx2 / D 0, whence .p/ .p/ D 2 H.2I 1/l D DH .O.2I 1/lC2 / C ılC2;p FDH .x1 / C FDH .x2 / :

51


10.4

Observe that .xi / D xi C D.xi / C O.l C 2/;

i D 1; 2:

If l ¤ p 2, then there is f 2 O.2I 1/lC2 such that D D DH .f /. Suppose l D p 2. Then there are ˛; ˇ 2 F and f 2 O.2I 1/p such that .p/ .p/ D D ˛DH .x1 / C ˇDH .x2 / C DH .f /. By Lemma 10.4.2(4) (with r D p 1), !H;1 D .!H;1 / D exp..x1 C D.x1 / C O.p//.p/ / d..x1 // ^ d..x2 // .p/

.p 1/

D exp.x1 /.1 C x1

D.x1 / C O.2p

@1 ..x1 //@2 ..x2 // .p 1/ .p 1/ x2

D .1

ˇx1

@1 ..x1 //@2 ..x2 // Note that O.2I 1/.2p

1/

@2 ..x1 //@1 ..x2 // dx1 ^ dx2

.p 1/

x1

1//

@2 .f / C O.2p

1// @2 ..x1 //@1 ..x2 // !H;1 :

D ¹0º and

q WD @1 ..x1 //@2 ..x2 //

@2 ..x1 //@1 ..x2 // D 1 C O.1/: .p 1/

Moreover, as f 2 O.2I 1/p , it must have a factor x1 , whence x1 above equation then gives 1 D .1

.p 1/ .p 1/ x2 /q

ˇx1

Dq

@2 .f / D 0. The

.p 1/ .p 1/ x2 :

ˇx1

P Set V WD a 3 define 'fk by the equation exp.ad DH;1 .fk // D ˆ'fk . (c) Let ‰ 2 Aut H.2I 1I ˆ.1// be arbitrary, find 2 Autc O.2I 1/ satisfying .!H;1 / 2 F !H;1 and ‰ D ˆ . By Theorem 10.4.3(3), has a presentation of the form 2p Y2 D 'fk ı .1 ;2 / : kD3

54

10


Choose ˛; ˇ; ; ı 2 F such that f3 D ˛x13 C ˇx12 x2 C x1 x22 C ıx33 . One gets ‰ D ˆ D

2p Y2

ˆ'fk ı ˆ.1 ;2 / D

2p Y2

Q ˛;ˇ; ;ı ı ˆ. ; / : exp.ad DH;1 .fk // ı ‰ 1 2

kD4

kD3

Q ˛;ˇ; ;ı 2 G . Set ‰f3 WD ‰

For further applications we need the following Corollary 10.4.5. p 1

(1) If f 2 F x23 C x1 x22 O.2I 1/, then exp.ad DH;1 .f // is a well-defined automorphism of H.2I 1I ˆ.1//. (2) For every f 2 O.2I 1/k C ık;p F (k 3) there is an automorphism ‰f of H.2I 1I ˆ.1// satisfying ‰f .DH;1 .x1a x2b // D DH;1 .x1a x2b / C ŒDH;1 .f /; DH;1 .x1a x2b / C DH;1 .ga;b /; where ga;b 2 O.2I 1/.aCbCk ‰f D exp.ad DH;1 .f //.

1Cıa;0 ıb;0 p/ .

p 1 2 x2 g

Proof. (1) Let f D ˛x23 C x1

E1 WD DH;1 .˛x23 / D

and put

3˛x22 @1

p 1 2 x2 g/

E2 WD DH;1 .x1

Moreover, if k 4, one may take

p 1 3 x2 @2

˛x1

2 H.2I 1I ˆ.1//.p

2 H.2I 1I ˆ.1//.1/ ;

1/ :

Let d1 ; d2 2 H.2I 1I ˆ.1// be arbitrary. Note that .ad DH;1 .f //p .d1 / is a linear combination of terms .ad E1 /r1 .ad E2 /s1 .ad E1 /rq .ad E2 /sq .d1 /; P P where r WD rj , s WD sj satisfy r C s D p. Any such term is contained in H.2I 1I ˆ.1//.rC.p 1/s 1/ . As the highest term in the filtration of H.2I 1I ˆ.1// has degree 2p 4, one obtains for any nonzero such term 2p

4 r C .p

1/s

1 D p C .p

2/s

1:

This forces s D 0. Hence any such term is of the form .ad E1 /p .d1 /. However, ad E1 raises the x2 -degree by at least 2, and therefore such term vanishes as well. Next observe that (for k p) Œ.ad DH;1 .f //i .d1 /; .ad DH;1 .f //k i .d2 / is a linear combination of terms r;s D ŒS1 ; S2 , where S1 WD .ad E1 /r1 .ad E2 /s1 .ad E1 /rq .ad E2 /sq .d1 /; S2 WD .ad E1 /rqC1 .ad E2 /sqC1 .ad E1 /rl .ad E2 /sl .d2 /

10.4

55


P P and r WD rj , s WD sj satisfy r C s D k p. For any such nonzero term r;s 2 H.2I 1I ˆ.1//.rC.p 1/s 2/ one has 2p

4 r C .p

1/s

1 p C .p

2/s

2:

This forces s 1 and r D k s p 1. In addition, ad E1 raises the x2 -degree by at least 2 and ad E2 raises the x2 -degree by at least 1. Therefore r;s has x2 degree at least 2r C s 2 2p 4 > p. Again, no such nonzero terms exist. Consequently, DH;1 .f / satisfies the requirements which are sufficient to ensure that exp.ad DH;1 .f // is an automorphism. (2) is obvious for k 4 by Corollary 10.4.4(2), and for k D 3 one may take an Q ˛;ˇ; ;ı mentioned in the proof of Corollary 10.4.4(3a). automorphism of the form ‰ We now turn to the determination of 2-dimensional tori and semisimple elements. Theorem 10.4.6. Let T H.2I 1I ˆ.1//Œp be a 2-dimensional torus. The following holds: (1) dim T \ H.2I 1I ˆ.1// D 1 and H.2I 1I ˆ.1//Œp D T C H.2I 1I ˆ.1//. ˇ ˇ (2) ˇ.H.2I 1I ˆ.1//; T /ˇ D p 2 and dim H.2I 1I ˆ.1// D 1 for all roots . (3) CH.2I1Iˆ.1//Œp .T / D T . Proof. (1) Abbreviate S WD H.2I 1I ˆ.1//. Since dim SŒp =S D 1, one has T \ S ¤ ¹0º. To derive a contradiction we assume T S. If T \ S.0/ D ¹0º, then S D T C S.0/ and this would imply that S is restricted, which is not true. Therefore T \ S.0/ ¤ ¹0º. However, S.0/ =S.1/ Š sl.2/ and every toral element of this algebra acts invertibly on the natural 2-dimensional sl.2/-module S=S.0/ . Then T S.0/ . Since T \ S.1/ D ¹0º, this gives rise to a 2-dimensional torus in sl.2/, the final contradiction. The second statement in (1) follows from a dimension argument. (2) Due to Theorem 10.1.4 all 2-dimensional tori of H.2I 1I ˆ.1//Œp have maximal dimension and Theorem 1.5.5(2) shows that they are Winter conjugate. Now Theorem 1.5.1 proves that there is a bijection of roots and root space dimensions for any two of these tori. Therefore it is sufficient to prove the statement for any particular 2-dimensional torus T 0 . In order to do so we take T 0 D F x1 @1 ˚ F x2 @2 . The space FDH;1 .x1a x2b / is the full root space for the root given by .x1 @1 / D a 1,

.x2 @2 / D b 1. (3) For dimension reasons it can only be that dim CH.2I1Iˆ.1//Œp .T / D 2. This gives CH.2I1Iˆ.1//Œp .T / D T . Proposition 10.4.7. Each element t D DH;1 .r0 C r1 x1 C r2 x2 C h/;

r0 ; r1 ; r2 2 F; h 2 O.2I 1/.2/ ;

56

10


is conjugate under Aut H.2I 1I ˆ.1// to p 1

(a) r1 DH;1 .r0 r1 1 C x1 C f x2 p 1

(b) r2 DH;1 .x2 C x1

/, f D f .x1 /,

if r1 ¤ 0, r2 D 0,

g/, g D g1 x2 C g2 x22 , g1 ; g2 2 F , p 1

(c) r2 DH;1 .1 C x2 C x1 r2 ¤ 0.

g/, g D g1 x2 C gp

p 1 , 1 x2

if r0 D 0, r2 ¤ 0,

g1 ; gp

1

2 F,

if r0 ¤ 0,

Proof. (1) Suppose r1 ¤ 0, r2 D 0. (a) Set t 0 WD r1 1 t , r00 WD r1 1 r0 , h0 WD r1 1 h. Then t 0 D DH;1 .r00 C x1 C h0 /. P j l j Choose l maximal with h0 2 O.2I 1/.l/ . If l < p 1 then write h0 D jl D0 ˛j x1 x2 P ˛j j l j C1 C hQ with hQ 2 O.2I 1/.lC1/ . Set f WD l x x , which is of degree j D0 l j C1 1

2

l C 1 3. The automorphism ‰f (cf. Corollary 10.4.5(2)) maps t 0 onto DH;1 .r00 C x1 C hQ 0 / with hQ 0 2 O.2I 1/.lC1/ . Thus we may assume h0 2 O.2I 1/.p 1/ . (b) Write h0 D

X

XX ˛a x1a x2b0 C ˛a;b x1a x2b ;

˛k ¤ 0;

a0 b>b0

ak

for suitable k; b0 0. Observe that k C b0 p 1. If b0 < p 1 then apply the k x1k x2b0 C1 (cf. Corollary 10.4.5(2)). automorphism exp.ad DH;1 .f // with f D b0˛C1 There is g 2 O.2I 1/ such that exp.ad DH;1 .f //.t 0 / D t 0 C

˛k ŒDH;1 .x1k x2b0 C1 /; DH;1 .r00 Cx1 /CDH;1 .gx22b0 /; b0 C 1

where deg.g/ C 2b0 2p 3. Therefore b0 ¤ 0 or g has a factor x2 or g D 0, whence gx22b0 D gx Q 2b0 C1 for suitable g. Q Consequently, exp.ad DH;1 .f //.t 0 / DDH;1 r00 C x1 C

X

p ˛a x1a x2b0 C r00 ık;0 ˛0 x1

1 b0 x2

a>k

C

XX

˛Q a;b DH;1 .x1a x2b /:

a0 b>b0

Induction on k and on b0 brings us to the case b0 D p 1. Then t is conjugate to an element of the form (a). (2) Consider the case r0 D 0, r2 ¤ 0. (a) By Corollary 10.4.4(1) the automorphism ˆ.r1 ;r2 / maps r2 DH;1 .x2 / onto 1 DH;1 .r1 x1 C r2 x2 /. Therefore ˆ.r maps t onto r2 DH;1 .x2 C h0 / with h0 2 1 ;r2 / O.2I 1/.2/ . So it is enough to prove the claim for t 0 WD DH;1 .x2 C h0 /. Choose l P j l j maximal with h0 2 O.2I 1/.l/ . If l < p 1 then write h0 D jl D0 ˛j x1 x2 C hQ

57


10.4

P with hQ 2 O.2I 1/.lC1/ . Set f WD jl D0

˛j j C1 l j x2 , j C1 x1

which is of degree l C 1 3. The automorphism ‰f (cf. Corollary 10.4.5(2)) maps t 0 onto DH;1 .x2 C hQ 0 / with hQ 0 2 O.2I 1/.lC1/ . Thus we may assume h0 2 O.2I 1/.p 1/ . (b) Write X XX h0 D x1a0 ˛b x2b C ˛a;b x1a x2b a>a0 b0

bk

for suitable a0 ; k 0. Observe that a0 C k p 1. If a0 < p 1 then apply the k automorphism exp.ad DH;1 .f // with f D a˛0 C1 x1a0 C1 x2k (cf. Corollary 10.4.5(2)), exp.ad DH;1 .f //.t 0 / D t 0

˛k ŒDH;1 .x1a0 C1 x2k /; DH;1 .x2 / C DH;1 .x12a0 g/; a0 C 1

where 2a0 C deg.g/ 2p 3. Therefore a0 ¤ 0 or g has a factor x1 or g D 0, whence x12a0 g D x1a0 C1 gQ for suitable g. Q Consequently, exp.ad DH;1 .f //.t 0 / D DH;1 x2 C x1a0

X

Induction on k and on a0 brings us to the case a0 D p (c) By the previous step t 0 is conjugate to p 1

gD

g/;

˛Q a;b x1a x2b :

a>a0 b0

b>k

t 00 WD DH;1 .x2 C x1

XX

˛b x2b C

p X1

1.

gb x2b ; gb 2 F:

bD0

Applying exp.ad DH;1 P.g0 // brings us to the case g0 D 0. Choose b0 3 such that g D g1 x2 C g2 x22 C bb0 gb x2b . Equation (10.4.2) yields DH;1

gb0 b0

D DH;1 D DH;1

1

X b0 gb p 0 gb x1 b0 1

x2b0

1 bCb0 1 00 x2 ;t

b0

gb0 b0

1

x2b0


1 bCb0 1 x2 ; DH;1 .x2

p 1

C x1

g/

b0

p 1 b0 x2

gb0 x1

gb0 b0

1

b0 x2b0

1 p 2 x1 g

C


2 bCb0 1 x2

b0

p 1 b0 x2 /:

D DH;1 . gb0 x1 Take f D

gb0 b0 x b0 1 2

P

b0 gb0 p 1 bCb0 1 x2 . b0 b0 1 gb x1

Due to Corollary 10.4.5(1)

58

10


exp.ad DH;1 .f // is an automorphism of H.2I 1I ˆ.1//. One computes p 1

exp.ad DH;1 .f // x2 C g1 x1

p 1 2 x2

x2 C g2 x1

p 1

X

C x1

gb x2b

bb0

D DH;1 x2 C

p 1 g1 x1 x2

C

p 1 g2 x1 x22

p 1

X

C x1

gb x2b

gb x2b

b>b0

C

p X1

p 1 b0 x2 //

s .ad DH;1 .x2b0 //s .DH;1 .x1

sD1 p 1

D DH;1 x2 C g1 x1

p 1 2 x2

x2 C g2 x1

p 1

X

C x1

b>b0

C

p X1

p 1 s b0 Cs.b0 1/ x2 /:

s0 DH;1 .x1

sD1

Repeating the argument in (b) we now obtain that t 0 is conjugate to DH;1 .x2 C P p 1 x1 .g1 x2 C g2 x22 C b>b0 gQ b x2b //. In the present case the proof is completed by induction on b0 . (3) Consider the case r0 ¤ 0, r2 ¤ 0. The automorphism ˆ.r 1 r ;r 1 r / maps DH;1 .r0 C r0 x2 / onto r0 r2 1 DH;1 .r0 C r1 x1 C r2 x2 /. Therefore ˆ1 h0 /

1

0

1

2

0

.r0 1 r1 ;r0 1 r2 /

h0 /

maps

h0

t onto r0 r2 DH;1 .r0 C r0 x2 C r0 D r2 DH;1 .1 C x2 C with 2 O.2I 1/.2/ . 0 0 So it is enough to prove the claim for t WD DH;1 .1 C x2 C h /. Next proceed almost word-by-word as in (2) to show that t 0 is conjugate to p 1

t 00 D DH;1 1 C x2 C g1 x1

p 1

x2 C x1

X

gb x2b

bb0

with b0 2. If b0 D p DH;1

gb0 b0 C1 x b0 C 1 2

D DH;1

1, the claim follows. Otherwise observe that X

p 1 bCb0 00 x2 ;t

gb0 gb x1

b0

gb0 b0 C1 x b0 C 1 2

X

p 1 bCb0 x2 ; DH;1 .1

gb0 gb x1

p 1

C x2 C x1

b0

gb0 p p 1 b0 x1 D DH;1 gb0 x1 x2b0 b0 C 1 X p 2 C DH;1 gb0 gb x1 x2bCb0

1 b0 C1 x2

p 2

gb0 x2b0 x1

g

b0

D DH;1

p 1 b0 x2

gb0 x1

gb0 p b0 x1 b0 C 1

1 b0 C1 x2

p 1

DW DH;1 .x1

q/:

g/

10.4


59

P gb0 b0 C1 p 1 bCb0 Take f D b0 C1 x2 x2 . Due to Corollary 10.4.5(1) b0 gb0 gb x1 exp.ad DH;1 .f // is an automorphism of H.2I 1I ˆ.1//. One computes exp.adDH;1 .f //.t 00 / p 1

D DH;1 1 C x2 C g1 x1 C

p X1

p 1

x2 C x1

X

gb x2b

b>b0 p 1

s .ad DH;1 .x2b0 C1 //s .DH;1 .x1

gb0 p b0 x1 b0 C 1

1 b0 C1 x2

gb0 p b0 x1 b0 C 1

1 b0 C1 x2

q//

sD1 p 1

D DH;1 1 C x2 C g1 x1 C

p X1

p 1

x2 C x1

X b>b0

gb x2b

p 1 s

s0 DH;1 .x1 qx2sb0 /:

sD1

Repeating the argument in (2b) we now obtain that t 0 is conjugate to DH;1 .1 C x2 C P p 1 x1 .g1 x2 C b>b0 gQ b x2b //. The proof is completed by induction on b0 . Next we determine Œp-th powers of the elements normalized according to Proposition 10.4.7. Lemma 10.4.8. Let r; s1 ; s2 2 F , f D f .x1 / be a polynomial in x1 and g D g.x2 / a polynomial in x2 only. The following holds: k p 1 p k p 1 k (1) ad DH;1 .x2 / DH;1 .r C x1 g/ D DH;1 ..k 1/Šrx1 C kŠx1 g/ for k D 1; : : : ; p 1; k p 1 p 1 (2) ŒDH;1 .r C x1 g/; ad DH;1 .x2 / DH;1 .r C x1 g/ D 0 for k p 3; p 2 p 1 p 1 (3) ŒDH;1 .r C x1 g/; ad DH;1 .x2 / DH;1 .r C x1 g/ D p 1 2DH;1 .x1 @2 .g/g/; p 1 Q D DH;1 .1 C s1 x2 C s2 x p 1 /@2 .h/ Q (4) Œ.1 C s1 x2 C s2 x1 /@2 ; DH;1 .h/ s1 hQ 1 for all hQ 2 O.2I 1/; k p 1 p 1 p 1 k p 1 (5) ad.1Cs2 x1 /@2 DH;1 .f x2 / D . 1/k kŠDH;1 f .1Cks2 x1 /x2 for k D 0; : : : ; p 1; p 1 p 1 p 1 (6) ad.1 C s1 x2 C s2 x1 /@2 DH;1 .f x2 / D p 1 p 1 p 1 p 1 2s1 s2 x1 x2 / if s1 ¤ 0; s1 DH;1 .f x2 / DH;1 f .1 C s1 x2 s2 x1 k p 1 p 1 p 1 (7) ŒDH;1 .f x2 /; ad.1Cs1 x2 Cs2 x1 /@2 DH;1 .f x2 / D 0 for k p 3; p 2 p 1 p 1 p 1 (8) ŒDH;1 .f x2 /; ad.1 C s2 x1 /@2 DH;1 .f x2 / D p 1 p 1 p 2 p 1 p 1 2f .0/2 DH;1 .x1 x2 / 2f .0/2 s2 DH;1 .x1 x2 / 2DH;1 .@1 .f /f x2 /. Proof. (1) is proved by induction on k using Equation (10.4.2). p 1 (2) Equation (10.4.2) shows that ŒDH;1 .r C x1 g/; DH;1 .x1i x2b / D 0 for i 2 and arbitrary b. Apply (1) for i D p k 3 and i D p 1 k 2.

60

10


(3) Applying (1) we obtain from Equation (10.4.2) (as .p 1/Š 1.p/) p 2 p 1 p 1 ŒDH;1 .r C x1 g/; ad DH;1 .x2 / DH;1 .r C x1 g/ p 1 D ŒDH;1 .r C x1 g/; DH;1 .p 3/Šrx12 C .p 2/Šx1 g r p 1 p 1 D ŒDH;1 .x1 g/; DH;1 .x12 / C DH;1 .x1 g/ D 2DH;1 .x1 @2 .g/g/: 2 (4) Equations (10.4.1) and (10.4.2) easily show that Q D DH;1 .@2 .h//; Q Œ@2 ; DH;1 .h/ Q D DH;1 .x2 @2 .h/ Q Œx2 @2 ; DH;1 .h/ p 1

Œx1

Q h/;

Q D ŒDH;1 . 1/; DH;1 .h/ Q D DH;1 .x p @2 ; DH;1 .h/ 1

1

Q @2 .h//:

(5) Due to (4), p 1

ad.1 C s2 x1

/@2

k

p 1

DH;1 .f x2

p / D DH;1 ..1 C s2 x1

1

p 1

/@2 /k .f x2

/

p 1 k p 1 k / x2

D . 1/k kŠDH;1 f .1 C s2 x1

p 1

D . 1/k kŠDH;1 f .1 C ks2 x1

p 1 k

/x2

:

(6) Note that p 1 p 1 x2

s1

p 1

D .1 C s1 x2 C s2 x1 / ! p X1 p 1 D . 1/p 1 j

p 1

.1 C s2 x1 j

p /

1

p 1 j

.1 C s1 x2 C s2 x1

p 1 p 1 j

/ .1 C s2 x1

/

j D0

D

p X1

p 1

.j C 1/s2 x1

.1

p 1 j

/.1 C s1 x2 C s2 x1

/ :

j D0

Therefore (4) yields p 1

ad.1 C s1 x2 C s2 x1

/@2

p

1

p 1

D DH;1 ..1 C s1 x2 C s2 x1 D

1 p s1

D

X

p X1

DH;1 f 1

p 1

DH;1 .f x2 /@2

s1 Id/p

/

1

p 1

.f x2

p 1

.j C 1/s2 x1

.js1

/

s1 /p

1

p 1 j

.1 C s1 x2 C s2 x1

j D0

DH;1 f 1

p 1

.j C 1/s2 x1

p 1 j

.1 C s1 x2 C s2 x1

/

j ¤1 p 1

DH;1 .f x2

p 1

DH;1 .f x2

D s1 D s1

p 1

/

DH;1 f .1

p 1

/

DH;1 .f .1 C s1 x2

p 1

2s2 x1

p 1

/.1 C s1 x2 C s2 x1 p 1

s2 x1

p 1

2s1 s2 x1

/

x2 //:

/

61


10.4

k p 1 p 1 (7) Since k p 3, (4) implies ad.1 C s1 x2 C s2 x1 /@2 DH;1 .f x2 / D Q 2 / for some hQ 2 O.2I 1/. As ŒDH;1 .f x p 1 /; DH;1 .hx Q 2 / D 0, (7) follows. DH;1 .hx 2 2 2 p 1 2 Q Q (8) Since ŒDH;1 .f x2 /; DH;1 .hx2 / D 0 for every h 2 O.2I 1/, statement (5) in combination with Equation (10.4.2) gives p DH;1 .f x2

1

p 1

/; ad.1 C s2 x1

/@2

p

2

p 1

DH;1 .f x2

p 1

/

p 1

D ŒDH;1 .f x2 /; DH;1 .f .1 2s2 x1 /x2 / p 1 p 1 D DH;1 @1 .f /f .1 2s2 x1 /x2 C f @1 .f .1 p 1 p 1 2f .0/2 x1 x2 p 1

D

2DH;1 .@1 .f /f x2

p 1 p 1 x2 /

p 2 p 1 x2 /

p 2

p 1

//x2

/ C 4s2 DH;1 .@1 .f /f x1

2s2 DH;1 .f 2 x1 Finally observe that f 2 x1

p 1

2s2 x1

p 2

D f .0/2 x1

p 1 p 1 x2 /:

C 2f .0/2 DH;1 .x1 p 1

C 2@1 .f /f x1

.

Proposition 10.4.9. The element p 1

t WD DH;1 .r C x1 C f x2 is semisimple if and only if f D t Œp D st holds.

s.r C x1

/;

r 2 F; f D f .x1 /; p 1

r 2 x1

/ for some s 2 F . In this case

Œp p 1 p 1 Proof. (a) Note that DH;1 .r Cx1 /Œp D .1 rx1 /@2 D 0 and DH;1 .f x2 /Œp 2 W .2I 1/.p 3/ Œp W .2I 1/.p2 3p/ D ¹0º. Lemma 10.4.8(5)–(8) in combination with Remark 10.2.4 shows that p 1 Œp /@2 C DH;1 .f x2 / Œp p 1 p 1 D .1 rx1 /@2 C DH;1 .f x2 /Œp p 1 p p 1 DH;1 .f x2 C ad.1 rx1 /@2

t Œp D .1

p 1

rx1

1 p C ŒDH;1 .f x2 2

D

1

/

p 2 p 1 p /; ad.1 rx1 /@2 DH;1 .f x2 p 1 p 1 p 1 DH;1 f .1 C rx1 / C f .0/2 DH;1 .x1 x2 / p 1 p 2 p 1 C rf .0/2 DH;1 .x1 x2 / DH;1 @1 .f /f x2 : 1

1

/

(b) Clearly t is semisimple if and only if it fits into a torus T H.2I 1I ˆ.1//Œp . If T is 1-dimensional, then t Œp 2 F t. If T is 2-dimensional, then Theorem 10.4.6

62

10


shows that T \ H.2I 1I ˆ.1// is 1-dimensional. Since t Œp 2 H.2I 1I ˆ.1// by (a), this proves that t is semisimple if and only if t Œp 2 F t in any case. Suppose t is semisimple, whence t Œp D st for some s ¤ 0. Since the mapping DH;1 is injective, the computations in (a) yield p 1

s.r C x1 C f x2

p 1

/D

f .1 C rx1

p 1 p 1 x2

/ C f .0/2 x1

p 1

@1 .f /f x2

p 2 p 1 x2

C rf .0/2 x1

:

Comparing terms without x2 gives s.r C x1 / D

p 1

f .1 C rx1

/;

p 1

p 1

whence f D s.r C x1 /.1 rx1 / D s.r C x1 r 2 x1 /. p 1 Suppose f D s.r C x1 r 2 x1 /. The computation in (a) yields t Œp D

p 1

DH;1 f .1 C rx1

p / C f .0/2 DH;1 .x1

1 p 1 x2 /

p 2 p 1 x2 /

C rf .0/2 DH;1 .x1

p

r 2 x1

D sDH;1 .r C x1

p 1 DH;1 @1 .f /f x2 1 p 1 p /.1 C rx1 / C s 2 r 2 DH;1 .x1

p 2 p 1 x2 /

p 2

C s 2 r 3 DH;1 .x1

s 2 DH;1 .1 C r 2 x1

p 1 p 1 x2

D sDH;1 r C x1 C sr 2 x1

1 p 1 x2 / p 1

r 2 x1

/.r C x1

p 1

s.r C x1 /x2

D st:

p 1

/x2

Proposition 10.4.10. The element p 1

t D DH;1 x2 C x1

.g1 x2 C g2 x22 / ;

g1 ; g2 2 F;

satisfies the equations t Œp D x2 @2

p 1

g2 DH;1 .x22 /

2g2 DH;1 x1

.g1 x22 C g2 x23 /

g1 t

and 2

p

t Œp C .g1

1/t Œp

g1 t D 0:

It is semisimple if and only if g1 ¤ 0. Proof. (a) We mentioned at the beginning of this section that DH;1 .x2 /Œp D x2 @2 : p 1

Since DH;1 x1

.g1 x2 C g2 x22 / raises the x1 -degree by p

2, one obtains

63


10.4

Œp p 1 DH;1 x1 .g1 x2 C g2 x22 / D 0. Lemma 10.4.8(1)–(3) in combination with Remark 10.2.4 gives p 1

t Œp D DH;1 .x2 /Œp C DH;1 .x1 .g1 x2 C g2 x22 //Œp p 1 p 1 C ad DH;1 .x2 / DH;1 .x1 .g1 x2 C g2 x22 // 1 p C ŒDH;1 .x1 2

1

.g1 x2 C g2 x22 //;

p ad DH;1 .x2 /

D x2 @2

DH;1 .g1 x2 C g2 x22 / DH;1 g1 x2 C g2 x22

D x2 @2

DH;1 .g2 x22 /

D x2 @2

DH;1

2

p 1

DH;1 .x1

.g1 x2 C g2 x22 //

p 1

.g1 C 2g2 x2 /.g1 x2 C g2 x22 / p 1 DH;1 x1 .g12 x2 C 3g1 g2 x22 C 2g22 x23 / p 1 x1 .2g1 g2 x22 C 2g22 x23 / g1 t: DH;1 x1

Set p 1

t1 WD t Œp C g1 t D x2 @2

DH;1 g2 x22 C x1 p

g2 DH;1 .x22 /

D x2 @2

2g2 DH;1 x1

.2g1 g2 x22 C 2g22 x23 / 1 .g1 x22 C g2 x23 / :

(b) We intend to show that t1 is toral. Write according to Volume 1, p. 17, .x2 @2

p 1 2 x2 @2 /Œp

g2 DH;1 .x22 //Œp D x2 @2 C .2g2 x2 @1 C g2 x1 C

p X1

p 1 2 x2 @2 /:

si .x2 @2 ; 2g2 x2 @1 C g2 x1

i D1 p 1

Note that 2g2 x2 @1 C g2 x1 x22 @2 raises the x2 -degree by at least 1. This shows that p 1 .2g2 x2 @1 C g2 x1 x22 @2 /Œp D 0. Since p 1 2 x 2 @2

Œx2 @2 ; 2g2 x2 @1 C g2 x1 the correction term

Pp

.ad x2 @2 /p

1 i D1

1

p 1 2 x2 @2 ;

D 2g2 x2 @1 C g2 x1 p 1 2 x2 @2 /

si .x2 @2 ; 2g2 x2 @1 C g2 x1 p 1 2 x2 @2 /

.2g2 x2 @1 C g2 x1

equals p 1 2 x2 @2

D 2g2 x2 @1 C g2 x1

(see Remark 10.2.4). Hence .x2 @2

g2 DH;1 .x22 //Œp D .x2 @2

g2 DH;1 .x22 //:

In particular, ad.x2 @2 g2 DH;1 .x22 // is semisimple. Put X j p V WD FDH;1 .x1i x2 / 3 DH;1 .x1 pC1i Cj 2p 2

1

.g1 x22 C g2 x23 //:

./

64

10


Note that in V only summands occur for which i; j 2. Therefore p 1

ŒDH;1 .x1

.g1 x22 C g2 x23 //; V D ¹0º:

j

./

j

j

Since Œx2 @2 ; DH;1 .x1i x2 / D .j 1/DH;1 .x1i x2 / and ŒDH;1 .x22 /; DH;1 .x1i x2 / D j C1 2iDH;1 .x1i 1 x2 /, we conclude that V is invariant under both x2 @2 and DH;1 .x22 /, that adV x2 @2 has only eigenvalues in Fp (recall that j 2) and that g2 DH;1 .x22 / acts nilpotently on V . Consequently, the eigenvalues on V of ad x2 @2 g2 DH;1 .x22 / are nonzero elements of the prime field. But then the eigenvalue of adV .x2 @2 p 1 is just 1. Since this is a semisimple endomorphism, we obtain g2 DH;1 .x22 // p g2 DH;1 .x22 /

adV x2 @2

1

D IdV :

./

./–./ yield Œp

t1

D x2 @2

g2 DH;1 .x22 / C ad.x2 @2

D x2 @2

g2 DH;1 .x22 /

g2 DH;1 .x22 //

p p

2g2 DH;1 x1

1

p 1 . 2g2 DH;1 x1 .g1 x22 C g2 x23 / 1 .g1 x22 C g2 x23 / D t1 :

(c) We have now proven the equation 0 D .t Œp C g1 t/Œp

p

2

.t Œp C g1 t/ D t Œp C .g1

1/t Œp

g1 t:

2

If g1 ¤ 0, then t 2 F t Œp C F t Œp , whence t is semisimple. If g1 D 0, then .t Œp t /Œp D 0 holds. Our previous computations readily give t Œp ¤ t . Then t Œp t is nonzero and nilpotent. In this case t is not semisimple. Proposition 10.4.11. The element p 1

t D DH;1 1 C x2 C x1

.g1 x2 C gp

p 1 / ; 1 x2

g1 ; gp

1

2 F;

satisfies the equations t Œp D x2 @2 C DH;1 x1

g1 x2

gp

p 1 1 x2

p 1

g12 x1

x2

and 2

p

t Œp C .g1

1/t Œp

It is semisimple if and only if g1 C gp

1

.g1 C gp

1 /t

D 0:

¤ 0.

Proof. (a) We mentioned at the beginning of this section that DH;1 .x2 /Œp D x2 @2 . p 1 p 1 As ad DH;1 1 C x1 .g1 x2 C gp 1 x2 / raises the x1 -degree by at least p 2,

65


10.4 p 1

p 1

obviously DH;1 1 C x1 .g1 x2 C gp 1 x2 in combination with Remark 10.2.4 gives t Œp D x2 @2 C DH;1 .p

Œp / D 0 holds. Lemma 10.4.8(1)–(3)

2/Šx1 C .p

p 1

DH;1 x1

D x2 @2 C DH;1 x1

.g1

p 2 /.g1 x2 1 x2

gp

g1 x2

p 1 1 x2

gp

p 1 / 1 x2

1/Š.g1 x2 C gp

p 1 / 1 x2

C gp p 1

g12 x1

x2 :

Set t1 WD t Œp C g1 t D x2 @2 C DH;1 g1 C x1 p 1

D .1 C x2

g1 x1

p 1 1 x2

gp

p 1 1 DH;1 ..g1 x1

/@2 C gp

p 1 p 1 x2 1 x1

C g1 gp p 1

1/x2

/:

Lemma 10.4.8(6)–(8) in combination with Remark 10.2.4 gives .1 C x2

p 1

g1 x1

/@2

Œp

p 1

D x2 @2 C .1

g1 x1

Œp

p 1

D .x2 @2 /Œp C .1 D .1 C x2

/@2

g1 x1

p 1

/@2

p

1

g1 x1

Œp

p 1

/@2 C @2

.g1 x1

@2 /Œp

and Œp

t1

p 1

D .1 C x2

g1 x1

C gp

1

gp2

1

C

2

/@2

Œp p 1

ad.1 C x2

g1 x1

h p DH;1 .g1 x1 ad.1 C x2 p 1

D .1 C x2

g1 x1

gp

1 DH;1

C gp2 gp2 D t1

p 1

1

1/x2

;

p 1

p

g1 x1

/@2 C gp p 1

.g1 x1

/@2

/@2

p 1

2

p 1

p 1

1/.1 C x2 C g1 x1 gp2

1/x2

DH;1 .g1 x1 p 1 p 1 1/x2 .g1 x1

1 DH;1

p 1 p 1 x2 / 1 DH;1 .x1

p 1

DH;1 ..g1 x1

1.

p 1

C 2g1 x1

p 1

1/x2

x2 /

p 2 p 1 x2 /

g1 /DH;1 .x1

p 2 p 1 x2 / 1 DH;1 .g1 x1

g1 gp

1 DH;1

p 1

x1

1 DH;1

C gp2

p 1 p 1 x2 / 1 DH;1 .x1 1 DH;1

D t1 C gp

1 t:

x2

p 1

C gp

D t1 C gp

p 1

C x1

1 C x2 C g1 x1

p 1

1 C x2 C g1 x1

p 1

C 2g1 x1

x2 C gp2

x2

/

p 1 p 1 x2 / 1 DH;1 .x1

i

66

10


(c) We have now proven the equation 0 D .t Œp C g1 t/Œp

.t Œp C g1 t/

gp

1t

2

p

D t Œp C .g1

1/t Œp

.g1 C gp

As in the proof of the preceding proposition t is semisimple if and only if g1 Cgp 0 holds.

1 /t: 1

¤

The conclusion of all these deliberations is the following Theorem 10.4.12. Every semisimple element t 2 H.2I 1I ˆ.1// is conjugate under Aut H.2I 1I ˆ.1// to an element of one and only one of the following sets: (1) F DH;1 .x1 x2 /; (2) F DH;1 .x1 x2 C x22 /; (3) F DH;1 x1 .1 C x2 / ; p 1 p 1 .r C x1 r 2 x1 /x2 , r 2 F ; p 1 x2 C rx1 x2 , r 2 F ; p 1 x2 C x1 .rx2 C x22 / , r 2 F ; p 1 p 1 1 C x2 C x1 .rx2 C sx2 / , r; s 2 F , r C s ¤ 0.

(4;r) F DH;1 r C x1 (5;r) F DH;1 (6;r) F DH;1 (7;r,s) F DH;1

Proof. (1) In order to prove uniqueness let t and t 0 be contained in one of the exposed sets and suppose there is an automorphism ‰ such that t 0 D ‰.t/. Recall that ‰ H.2I 1I ˆ.1//.0/ D H.2I 1I ˆ.1//.0/ : Then also ‰ H.2I 1I ˆ.1//.i/ D H.2I 1I ˆ.1//.i/ holds for all i > 0. It is not hard to show that X ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1// FDH;1 .x1a x2b /: a;bp 1 aCb2

In particular, the elements DH;1 .1/; DH;1 .x1 /; DH;1 .x2 / are linear independent modulo ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1//. Write according to Corollary 10.4.4 ‰D

2p Y2

exp .ad DH;1 .fk // ı ‰f3 ı ˆ.ˇ; / ;

kD4

where fk 2 O.2I 1/k C ık;p F for k 3. According to that Corollary one also has a more subtle description of ‰f3 , namely ‰f3 D

Y

.

i

ı

0 i /;

i

D ˆ.i ;i0 / ;

0 i

D exp.ad DH;1 .fi //; fi 2 F x13 [ F x23 :


10.4

67

Moreover, ‰f3 D Id C O.1/, and this gives (as all i are homogeneous of degree 0) Q i D Id. Observe that for every d 2 H.2I 1I ˆ.1// the following congruences hold modulo ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1// exp .ad DH;1 .fk //.d / d;

fk 2 O.2I 1/k C ık;p F; k 4;

and ‰f3 .d / D

Y

.

i

ı

0 i /.d /

Y

i

.d / D d:

Then ‰.t / ˆ.ˇ; / .t/

.mod ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1///:

(a) Suppose t D DH;1 .x1 x2 /. The introductory remark shows that t 0 D ‰.t/ can only be of type (1) or type (2). On the other hand, t 0 D ‰.t/

1

DH;1 x1 .ˇx1 C x2 / .mod H.2I 1I ˆ.1//.1/ /:

Since t 0 is contained in one of the sets of our list, it can only be that ˇ D 0 and t 0 is contained in F DH;1 .x1 x2 /. So t 0 is of type (1) as well. One argues likewise if t D DH;1 .x1 x2 C x22 /, namely t0

1

DH;1 .ˇ C ˇ 2 /x12 C . C 2ˇ /x1 x2 C 2 x22 / D DH;1 2 .ˇ C ˇ 2 /x12 C 1 .1 C 2ˇ/x1 x2 C x22 :

Then it must be that ˇ C ˇ 2 D 0 and t0 is contained in F DH;1 .x1 x2 C x22 /. (b) Suppose t D DH;1 x1 .1 C x2 / . Then t 0 D ‰.t/

1

DH;1 x1 .1 C ˇx1 C x2 /

modulo ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1//. Since t 0 is contained in one of the exposed sets, the introductory remark shows that t 0 is of type (3). One argues likewise if t is of type .4I r/. p 1 (c) Suppose t D DH;1 .x2 C rx1 x2 / is of type .5I r/. Then t 0 D ‰.t/

1

p 1

DH;1 .ˇx1 C x2 C r x1

x2 /

modulo ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1//. As a consequence, t 0 is of type .5I r 0 / or p 1 .6I r 0 /. Then ˇ D 0 and the above gives t 0 D DH;1 .x2 Cr 0 x1 x2 / or t 0 D DH;1 .x2 C p 1 x1 .r 0 x2 C x22 //. Proposition 10.4.10 shows that t Œp D x2 @2

rt

68

10


and t 0Œp D x2 @2

r 0t 0

or t 0Œp D x2 @2

DH;1 .x22 /

p 1

2DH;1 .x1

.r 0 x22 C x23 //

r 0t 0:

Since H.2I 1I ˆ.1//Œp is centerless, ‰ is an automorphism of restricted algebras. Moreover, as an automorphism of H.2I 1I ˆ.1//, ‰ D .Id C O.1// ı ˆ.0; / (recall that ˇ D 0). Then ‰.DH;1 .x1 // D

1

DH;1 .x1 / C O.0/; ‰.DH;1 .x2 // D DH;1 .x2 / C O.0/

and this gives modulo H.2I 1I ˆ.1//.0/ (where ı1 WD and ı2 WD 1) Œ‰.x2 @2 /; DH;1 .xi / Œ‰.x2 @2 /; ıi ‰.DH;1 .xi // D ‰.Œx2 @2 ; ıi DH;1 .xi // ´ ‰. ı1 @2 / D DH;1 .x1 /; D 0;

i D 1; i D 2:

Then ‰.x2 @2 / D x2 @2 C O.1/. We obtain modulo H.2I 1I ˆ.1//.1/ ‰.t Œp / x2 @2

r‰.t/ D x2 @2

rt 0 x2 @2

rDH;1 .x2 /;

while t 0Œp D x2 @2

r 0 t 0 x2 @2

r 0 DH;1 .x2 /

DH;1 .x22 /

r 0 t 0 x2 @2

DH;1 .x22 /

or t 0Œp x2 @2

r 0 DH;1 .x2 /:

As t 0Œp D ‰.t Œp /, one obtains that r D r 0 and that the last case is impossible. Then t 0 is of type .5I r/. p 1 (d) Suppose t D DH;1 x2 C x1 .rx2 C x22 / is of type .6I r/. With the obvious changes in (c) we get that t 0 is of the same type .6I r/. p 1 p 1 (e) Suppose t D DH;1 1 C x2 C x1 .rx2 C sx2 / . By the results (a)–(d) it can p 1 only be that t 0 is of type .7I r 0 ; s 0 /. More exactly, t 0 D qDH;1 1 C x2 C x1 .r 0 x2 C p 1 s 0 x2 / for some q 2 F . But then qDH;1 .1 C x2 / t 0 D ‰.t/

1

p 1

DH;1 1 C ˇx1 C x2 C x1

.r x2 C s p

1 p 1 x2 /

10.4


69

modulo ŒH.2I 1I ˆ.1//.1/ ; H.2I 1I ˆ.1//. This gives ˇ D 0, q D D 1 and t 0 D p 1 p 1 DH;1 1 C x2 C x1 .r 0 x2 C s 0 x2 / . Next conclude from Proposition 10.4.11 that 2

t Œp C .r p t

0Œp2

C .r 0p

1/t Œp 1/t 0Œp

.r C s/t D 0; .r 0 C s 0 /t 0 D 0:

Since ‰ is a Œp-homomorphism, these equations imply r D r 0 and s D s 0 . (2) Next we show that every conjugacy class of semisimple elements meets one of the exposed sets nontrivially. Let t be semisimple. Suppose t 2 H.2I 1I ˆ.1//.0/ and write t D DH;1 .˛1 x12 C ˛2 x1 x2 C ˛3 x22 / C DH;1 .r/ C DH;1 .h/ where ˛1 ; ˛2 ; ˛3 ; r 2 F and h 2 O.2I 1/.3/ . If both ˛2 ; ˛3 were 0, then t were Œpnilpotent, which is not true. Hence at least one of these scalars is nonzero. Therefore the equation ˛1 C ˛2 T C ˛3 T 2 D 0 has a solution 1 2 F . Then ˆ.1 ;1/ .t/ is of the form DH;1 .˛20 x1 x2 C ˛30 x22 / C DH;1 .r 0 / C DH;1 .h0 /. Here ˛20 ¤ 0 because ˆ.1 ;1/ .t/ is not Œp-nilpotent. If ˛30 ¤ 0, set 2 WD ˛20 ˛30 1 . Then ˆ.0;2 / maps ˆ.1 ;1/ .t/ onto 2 1 DH;1 .2 ˛20 x1 x2 C 22 ˛30 x22 / C DH;1 .r 00 / C DH;1 .h00 / D ˛20 DH;1 .x1 x2 C x22 / C DH;1 .r 00 / C DH;1 .h00 /. We set t0 WD DH;1 .x1 x2 / and t0 WD DH;1 .x1 x2 C x22 / in the respective cases and now may assume that t D t0 C DH;1 .r/ C DH;1 .h/. Next we observe that p 1 t0 D x2 @2 x1 @1 and t0 D x2 @2 x1 @1 2x2 @1 x1 x22 @2 , respectively. From this Œp it is easy to derive that t0 D t0 holds. Hence DH;1 .r/ C DH;1 .h/ decomposes into a sum of eigenvectors with respect to t0 , DH;1 .r/ C DH;1 .h/ D

X

DH;1 .hi /;

Œt0 ; DH;1 .hi / D ihi :

Choose inductively an eigenvector hi for nonzero eigenvalue i ¤ 0 for which hi P has lowest degree, hi D j j0 hi;j with hi;j homogeneous of degree j . Applying Corollary 10.4.5(2) with f D 1i hi;j0 brings t to the form t hi;j0 C terms of higher degree, and thereby eventually brings t to the form t 0 D t0 C DH;1 .r 0 C h0 /;

Œt0 ; Dh;1 .r 0 C h0 / D 0:

But then t 0Œp t 0 D DH;1 .r 0 C h0 /Œp DH;1 .r 0 C h0 / 2 W .2I 1/.1/ is Œp-nilpotent. Hence it vanishes as t 0 is semisimple. As DH;1 .r 0 C h0 / 2 W .2I 1/.1/ itself is Œpnilpotent, the equation DH;1 .r 0 C h0 / D DH;1 .r 0 C h0 /Œp yields DH;1 .r 0 C h0 / D 0. Consequently, t is conjugate to t0 , and this shows that it is of type (1) or (2). (3) Suppose t 62 H.2I 1I ˆ.1//.0/ . Proposition 10.4.7 shows that a nonzero multiple of t is ruled by Propositions 10.4.9–10.4.11. We assume that t is such.

70

10


p 1 p 1 (a) If t is as in Proposition 10.4.9, then t D DH;1 r Cx1 s.r Cx1 r 2 x1 /x2 and s ¤ 0. Choose 2 F such that p 1 s D 1. Applying ˆ.0;/ one may assume s D 1. If r ¤ 0, then t is of type .4I r/. p 1 So assume r D 0, whence t is conjugate to DH;1 .x1 x1 x2 /. Proposition 10.4.9 shows that t is toral. We now compare this element with DH;1 x1 .1 C x2 / . Note that DH;1 x1 .1 C x2 / D .1 C x2 /@2 x1 @1 is toral. Choose an automorphism ‰ such that ‰ DH;1 .x1 .1 C x2 // is as in Proposition 10.4.7(a) with r0 D 0, r1 D 1, r2 D 0. Therefore ‰ DH;1 .x1 .1 C x2 // is as in Proposition 10.4.9 with r D 0. The beginning of this part (a) then shows that DH;1 x1 .1 C x2 / is conjugate to a multiple p 1 of DH;1 .x1 x1 x2 /, and this in turn yields that t is of type (3). p 1

(b) If t is as in Proposition 10.4.10, then it is conjugate to DH;1 .x2 Cx1 .g1 x2 C g2 x22 // with g1 ¤ 0. Set r WD g1 . If g2 D 0, then t is of type .5I r/. So assume g2 ¤ 0. The automorphism ˆ.0;g 1 / maps t onto g2 DH;1 .g2 1 x2 C p 1

p 1

2

.rg2 1 x2 C g2 1 x22 // D DH;1 .x2 C x1 .rx2 C x22 //. Then t is of type .6I r/. (c) If t is as in Proposition 10.4.11, it is of type .7I r; s/ with r WD g1 and s WD gp 1 .

x1

10.5

Melikian algebras

In this section we present parts of Skryabin’s paper [Skr 01], extending his result on the automorphism group of M.1; 1/ to arbitrary Melikian algebras. Namely, we describe the automorphism groups of Melikian algebras M.n1 ; n2 /, and the orbits of 1-dimensional tori and its centralizers in the restricted Melikian algebra M.1; 1/. Throughout this section the characteristic of the ground field is 5. We normalize n1 n2 . For properties of Melikian algebras see §4.3. We summarize: There is defined the natural grading degM on M.n1 ; n2 /. Its homogeneous component of degree k is denoted by M.n1 ; n2 /Œk . This grading defines the natural filtration .M.n1 ; n2 /.i/ /i 3 , where X M.n1 ; n2 /.i/ D M.n1 ; n2 /Œk : ki

This filtration has depth 3 and height s D 3.5n1 C5n2 / 7. It is natural in the sense that it is the only standard filtration of depth 3 and the 0-component having codimension 5 (Theorem 4.3.3). There is another grading mentioned on p. 208 of Volume 1, the Melikian grading. Note that M.n1 ; n2 / is a free O.2I .n1 ; n2 //-module with basis .@1 ; @2 ; 1; @Q1 ; @Q2 /. Define a degree function on O.2I .n1 ; n2 // by setting deg x1 D 1, deg x2 D 2, and give a degree . 1; 2; 1; 0; 1/ on the respective basis elements. The Melikian grading is

10.5

71

Melikian algebras

induced by this degree function. The homogeneous components of degree k are ® M.n1 ; n2 /hki WD span x .a/ @j ; x .b/ @Qj ; x .c/ j a1 C 2a2

1

ıj;2 D b1 C 2b2

ıj;2 D c1 C 2c2

¯ 1Dk :

In particular, M.n1 ; n2 /hi i D ¹0º

if i
0 ‰Œl /. One has ‰Œ0 2 Aut M.n1 ; n2 / and ‰Œl0 2 Der M.n1 ; n2 /. The homogeneous derivations of positive degree are contained in ad M.n1 ; n2 / (Theorem 7.1.4). Hence ‰Œl0 D ad y with y 2 M.n1 ; n2 /Œl0 . As a result, Aut M.n1 ; n2 / D AutŒ0 M.n1 ; n2 / Ë Aut.1/ M.n1 ; n2 / and for i 1 there is an assignment ‚i;Œ; W Aut.i / M.n1 ; n2 /=Aut.i C1/ M.n1 ; n2 / ! M.n1 ; n2 /Œi ; ‰Œi D ad ‚i;Œ; .‰/: In contrast, Aut M.n1 ; n2 / might not map the subalgebra M.n1 ; n2 /¹0º into itself. To apply the preceding deliberations one has to consider Aut¹0º M.n1 ; n2 /. As a result, Aut¹0º M.n1 ; n2 / D Auth0i M.n1 ; n2 / Ë Aut¹1º M.n1 ; n2 /: Note that the Melikian filtration is the standard filtration determined by M.n1 ; n2 /¹0º and M.n1 ; n2 /¹ 1º (see Definition 3.5.1). As M.n1 ; n2 /¹ 1º D ŒM.n1 ; n2 /¹0º .1/ ; M.n1 ; n2 / C M.n1 ; n2 /¹0º , the Melikian filtration therefore is uniquely determined by M.n1 ; n2 /¹0º only. Consequently, Aut¹0º M.n1 ; n2 / D ¹‰ 2 Aut M.n1 ; n2 / j ‰.M.n1 ; n2 /¹0º / M.n1 ; n2 /¹0º º: We recall that, if ‰; ‰ 0 2 Aut.i/ M.n1 ; n2 / then ‚i;Œ; .‰ ı ‰ 0 / D ‚i;Œ; .‰/ C ‚i;Œ; .‰ 0 /

10.5

73

Melikian algebras

and, if ‰ 2 Aut.i / M.n1 ; n2 /, ‰ 0 2 Aut.j / M.n1 ; n2 / then ‰ ı ‰ 0 ı ‰ Aut.i Cj / M.n1 ; n2 / and ‚i Cj;Œ; .‰ ı ‰ 0 ı ‰

1

ı ‰0

1

1

ı ‰0

1

2

/ D Œ‚i;Œ; .‰/; ‚j;Œ; .‰ 0 /:

By definition, ‚i;Œ; is injective. These deliberations also hold for the Melikian grading. In order to determine the automorphism group of M.n1 ; n2 / we have to describe AutŒ0 M.n1 ; n2 / and the image of all ‚i;Œ; , i > 0. Recall that gO U.g/ denotes the universal p-envelope of any Lie algebra g. By definition of a universal p-envelope every homomorphism W g ! g0 extends uniquely to a restricted homomorphism O W gO ! gb0 . The following theorem, which is of independent interest, describes the automorphisms of W .mI n/. It is clear that the assumptions of the theorem are satisfied by any automorphism. Theorem 10.5.1. Let a; a0 W .mI n/ be subalgebras such that W .mI n/ D W .mI n/.0/ ˚ a D W .mI n/.0/ ˚ a0 : Set K WD NoraO W .mI n/.0/ , K 0 WD Norb0 W .mI n/.0/ : For every algebra isomorphism a W a ! a0 satisfying O.K/ D K 0 there is exactly one automorphism of W .mI n/ which extends .

3 4 3 u.W .mI n// D U.W .mI n// D U.W .mI n/

Proof. (a) Observe that W .mI n/ D W .mI n/.0/ ˚ aO and

.0/ /

and similarly,

3 4

˝F U.a/;

3

W .mI n/ D W .mI n/.0/ ˚ ab0 ; u.W .mI n// D U.W .mI n/.0/ / ˝F U.a0 /: Set H WD Nor

1 W .mI n/

W .mIn/

.0/

4

4

D W .mI n/.0/ ˚ K D W .mI n/.0/ ˚ K 0 :

Theorem 2.3.2 and the subsequent example yields the existence of a divided power isomorphism ' W Homu.H / .u.W .mI n//; F / Š O.mI n/;

3

such that the action ..D/f /.u/ WD f .uD/

3

3

for D 2 W .mI n/, f 2 Homu.H / .u.W .mI n//; F /, u 2 u.W .mI n// induces the identity on W .mI n/, i.e., ' ı .D/ ı ' 1 D D for all D 2 W .mI n/.

74

10


Extend to an algebra isomorphism U.a/ ! U.a0 / and then to a linear automorphism W U.W .mI n// by setting .w ˝ a/ WD w ˝ .a/;

w 2 U.W .mI n/.0/ /; a 2 U.a/:

For u 2 U.W .mI n// and h 2 H put X uD u0 ˝ u1 ; u0 2 U.W .mI n/.0/ /; u1 2 U.a/

4

and

h D h0 C h1 ; h0 2 W .mI n/.0/ ; h1 2 K: Then hu D

X

.h0 u0 / ˝ u1 C

X

h1 .u0 ˝ u1 /

X

.h0 u0 / ˝ u1 C

X

u0 ˝ .h1 u1 /

modulo W .mI n/.0/ U.W .mI n//. Consequently, X X .hu/ .h0 u0 / ˝ .u1 / C u0 ˝ .h1 /.u1 / .h/.u/

.mod W .mI n/.0/ U.W .mI n///:

Our assumptions ensure that .H / D H . As W .mI n/.0/ acts trivially on the module F , induces a linear automorphism of Homu.H / .u.W .mI n//; F / satisfying .f /.u/ D f ..u//. Next observe that for a 2 a one has X X .ua/ D u0 ˝ .u1 a/ D u0 ˝ .u1 /.a/ D .u/.a/:

3

3

Recall the multiplication P on Homu.H / .u.W .mI n//; F / from Volume 1, p. 64. The comultiplication .u/ D u.1/ ˝ u.2/ obviously commutes with , X ..u// D .u.1/ / ˝ .u.2/ /:

3

For u 2 u.W .mI n// one obtains X X .f /.u.1/ /.g/.u.2/ / D f ..u.1/ //g..u.2/ // .f /.g/ .u/ D D .fg/..u// D .fg/.u/; .a/.f / .u/ D f ..ua// D f .u/.a/ D ..a//f .u/

8a 2 a:

This shows that is an algebra automorphism, which (as .a/ D .a/ for a 2 a) satisfies .a/.f / D ..a//f 8a 2 a: The assignment D 7! ' ı If D 2 a then 'ı

1

1 ı .D/ ı

ı .D/ ı ı '

1

ı'

1

defines an automorphism of W .mI n/.

D ' ı ..D// ı '

1

D .D/:

10.5

75

Melikian algebras

(b) In order to show uniqueness it suffices to assume a D a0 and D Id. Let be an extension. Since it stabilizes the unique maximal subalgebra of codimension 2, it also stabilizes all subspaces W .mI n/.i/ . Therefore induces an automorphism N of gr W .mI n/ Š W P.mI n/ (Corollary 6.1.7). It is not hard to see that, since N is the identity on gr a D F @i , one has N D Id. This means D Id C O.1/. In particular, acts identically on the highest term W .mI n/.P .pni 1/ 1/ . It also acts identically on a. However, the subalgebra generated by W .mI n/.P .pni 1/ 1/ and a is W .mI n/ itself. Then D Id. Next we extend isomorphisms of suitable subalgebras of M.n1 ; n2 / to automorphisms of M.n1 ; n2 / itself. Theorem 10.5.2. Let W and W 0 be subalgebras of M.n1 ; n2 / satisfying gr W D gr W 0 D M.n1 ; n2 /0 : Every isomorphism W W ! W 0 of algebras can be extended to an automorphism Q of M.n1 ; n2 /. Let 3 D 1 ¤ . If Q is an extension of , then there is an extension Q which satisfies Q .v/ D .v/ Q for v 2 M.n1 ; n2 /Œs . Moreover, Q , Q , Q 2 are the only extensions of . Proof. (a) Set M WD M.n1 ; n2 / and n WD .n1 ; n2 /. Since M0 Š W .2I n/ as graded algebras (where the grading of W .2I n/ D ˚i 1 W .2I n/ \ MŒ3i is equivalent to the natural one), it is easy to conclude that W has no subalgebra of codimension 1 and exactly one of codimension 2 (which is W \ M.0/ ), and W =W \ M.0/ is W \ M.0/ irreducible. Then this subalgebra defines a separating standard filtration. Looking at the graded algebras one observes that this filtration is given by the original one, namely W.i / D W \ M.3i/ . Definition 4.2.4 shows that W is of Witt type, and then Corollary 6.1.7 gives an isomorphism of filtered algebras

' W W ! gr W D M0 Š W .2I n/: This isomorphism imposes a grading on W D ˚i W3i D '

1

.W .2I n/i /

and

1 W3i ,

where

M.3i/ D W3i ˚ M.3i C1/ ;

i

1:

P Set W.3i / WD j i W3j . We now consider the W -module M. Observe that W.3/ annihilates M.s 2/ , and x1 @1 C x2 @2 2 gr0 W acts on gri M as 2iId. Therefore the W.0/ -module M.s 2/ decomposes M.s

2/

D Vs ˚ Vs

1

˚ Vs

2;

76

10


where

B

Vs WD MŒs D W .2I n/.3.5n1 C5n2 / Vs

1

WD ¹v 2 M.s Š MŒs

Vs

2

1

1

j ad '

2/

5 .x1 @1 C x2 @2 / .v/ D

D O.2I n/.3.5n1 C5n2 /

D ¹v 2 M.s

D W3.5n1 C5n2

1

j ad '

2/

3/

Š MŒs

D F x ..n// @Q1 ˚ F x ..n// @Q2 ;

7/

8/

vº

D F x ..n// ;

5 .x1 @1 C x2 @2 / .v/ D 2vº 2

D F x ..n// @1 ˚ F x ..n// @2 :

These are pairwise non-isomorphic and irreducible W.0/ -modules. Arguing in the graded algebra one obtains CM .W.3/ / D M.s 2/ . Let D1 WD ' 1 .@1 /; D2 WD n ' 1 .@2 / 2 W 3 denote the images of @1 , @2 2 M0 . Then ŒDi5 i ; W D ¹0º and ni hence ŒDi5 ; M.s 2/ CM .W.3/ / D M.s 2/ (i D 1; 2). As a consequence, there are ˛i ; ˇi 2 F such that n1

n2

D15 jVi D ˛i IdVi ;

D25 jVi D ˇi IdVi ;

i D s; s

1; s

2:

Moreover, ˛s 2 D ˇs 2 D 0. Since the W -module generated by Vs C Vs 1 C Vs 2 nj nj is M and Dj5 centralizes W (j D 1; 2), it follows that Dj5 respects the natural filtration of M. Now argue in gr M Š M. As Œgri M; gr 3 M D gri 3 M for all i , n it is clear that D15 1 acts on W .2I n/ as ˛s Id and on O.2I n/ as ˛s 1 Id. Considering the multiplication of M one obtains 3˛s D ˛s 2 D 0, hence ˛s D 0. Similarly n n we conclude ˛s 1 D ˇs D ˇs 1 D 0. This means that D15 1 and D25 2 annihilate Vs C Vs 1 C Vs 2 . Again, since the W -module generated by Vs C Vs 1 C Vs 2 is M nj nj and Dj5 centralizes W , it follows that Dj5 D 0 for j D 1; 2. (b) Set X X j j b K WD FD 5 C FD 5 C W.0/ M:

B

1

b

2

j n1

j n2

We proved above that Vs , Vs 1 , Vs 2 are K-modules. By definition of the p-mapping they are restricted modules. Note that s 2 .mod 3/. Therefore there are Kmodule isomorphisms Vs Š W .2/ (for the definition of W .2/

1;1 ; 1;˛

Vs

1

Š F 3;

Vs

2

Š W .2/

1; 1

and F˛ see §5.3). Theorem 5.3.7 shows

b / ˝u.K/ Vs Š W .2I n/.2 div/ ; u.W b / ˝u.K/ Vs u.W

1

Š O.2I n/.

c/ ˝u.K/ Vs u.W

2

Š W .2I n/:

2 div/ ;

10.5

77

Melikian algebras

These are pairwise nonisomorphic and irreducible W -modules (Proposition 4.3.2). Hence the universal property of induced modules shows that there is an injective mapping of W -modules ' 0 W W .2I n/.2 div/ ˚ W .2I n/ ˚ O.2I n/.

2 div/

! M:

By dimension reasons this mapping is an isomorphism. (c) We now set inductively for i s Vi

WD ŒD1 ; Vi C ŒD2 ; Vi :

3

Then M.i / D Vi ˚ M.i C1/ for all i 3 and X V2 WD V3i C2 D ' 0 .W .2I n/.2 div/ /; i 1

V

2

WD

X

V3i C1 D ' 0 .O.2I n/.

2 div/ /

i 1

P and W D i 1 V3i D ' 0 .W .2I n//. One should mention here that the decomposition is compatible with the grading of W . Namely, since Vs 2 D W3.5n1 C5n2 3/ , V3i D W3i holds for all i and, in particular, V 3 D W 3 D FD1 C FD2 . In addition, nj Vi 3 D ŒVi ; V 3 holds for all i 0. As Dj5 D 0 for j D 1; 2 one concludes for k D 1; 2; 3 n1

Vk D D15

1

n2

D25

1

.VsC1Ck / and ŒVk ; V 3 D ¹0º:

Recall that the image of x1 @1 C x2 @2 in W0 acts on Vi as 2iId. Therefore X ŒVi ; Vj Vi Cj C ViCj C5k ; k>0

whence ŒVi ; Vj Vi Cj if i 2 and j s 2. By construction this inclusion is also true for i D 3. It is now easy to show by induction that ŒVi ; Vj Vi Cj

for all i; j D

3; : : : ; s:

(d) Proceed similarly for W 0 . Choose the grading given by the grading of W and the isomorphism . Construct likewise subspaces V 0 i , where V 0 i Š MŒi for i D s; s 1; s 2 and V 0 i 3 D ŒF .D1 / C F .D2 /; V 0 i . Define analogously W 0 -modules V 0 2 and V 0 2 . The isomorphism W W ! W 0 gives rise to linear isomorphisms 2 W V2 ! V 0 2 ; 2 W V 2 ! V 0 2 which map Vi onto V 0 i for all i and satisfy i Œx; y D Œ .x/; i .y/ 8 x 2 W ; y 2 Vi .i D ˙2/:

78

10


Note that Vs and V 1 are 2-dimensional irreducible W0 -modules and Vs 1 is a 1dimensional W0 -module. As W0 Š gl.2/ (hence W0 .1/ Š sl.2/), these modules are indeed irreducible (respectively, trivial) sl.2/-modules. The pairings Vs V

1

! V 0s

1

Š F;

.v; w/ 7! 2 .Œv; w/ and .v; w/ 7! Œ2 .v/; 2 .w/

give rise to sl.2/-module homomorphisms Vs ! V there is ˛ 2 F such that

1

.

Schur’s lemma shows that

Œ2 .v/; 2 .w/ D ˛ 2 .Œv; w/ 8 v 2 Vs ; w 2 V 1 : Fix nonzero u 2 Vs V

1

1.

There are sl.2/-module isomorphisms

! V 0s

2;

w 7! .Œu; w/ and w 7! Œ 2 .u/; 2 .w/:

As before Schur’s lemma shows that there is ˇ 2 F such that Œ 2 .u/; 2 .w/ D ˇ.Œu; w/ 8 w 2 V 1 : Choose 2 ; 2 2 F satisfying 32 D .˛ˇ/ .v/ 8 v 2 W and Q .v/ WD 2 2 .v/ 8 v 2 V2 ;

1

and

2

D .ˇ2 / 1 . Set .v/ Q D

.v/ Q WD 2 2 .v/ 8 v 2 V 2 :

w/ D Œ.v/; Q .w/ Q for all v 2 Vs C Vs 1 , w 2 V 1 . This By construction, .Œv; Q equation is, by definition, also true if v 2 M and w 2 W . Applying W 3 D V 3 several times and observing that ŒV 3 ; V 1 D ¹0º shows that the above equation holds for all v 2 M. Then applying W several times gives the equation for all w 2 M, whence Q is an automorphism of M. (e) The choice of 2 ; 2 is not unique, namely 02 WD 2 , 0 2 WD 2 2 is another possible choice and yields an extension Q . Finally, let Q and Q 0 be extensions of . Then W Q 1 ı Q 0 is an automorphism of M, which extends the identity of W . Since VŒs is an irreducible W0 -module, there is 2 F such that .v/ D v for v 2 VŒs . As VŒs generates the W -module V2 , one has even more .v/ D v for v 2 V2 . The multiplication in M now yields 3 D 1. Then 2 D Id on V2 . From this one concludes 2 D Id, whence Q 1 ı Q 0 2 D 2 D Id and Q D Q 0 2 . Combining the preceding two theorems we deduce Corollary 10.5.3. Let a; a0 M.n1 ; n2 / be subalgebras such that M.n1 ; n2 / D M.n1 ; n2 /.

2/

˚ a D M.n1 ; n2 /.

and dim aO = NoraO M.n1 ; n2 /.0/ n1 C n2 :

2/

˚ a0

10.5

79

Melikian algebras

Set K WD NoraO M.n1 ; n2 /.0/ , K 0 WD Norb0 M.n1 ; n2 /.0/ . The following holds: a

(1) Every algebra isomorphism W a ! a0 satisfying O.K/ D K 0 can be extended to an automorphism ‰ of M.n1 ; n2 /. (2) Let 3 D 1 ¤ . If ‰ is an extension of , then there is an extension ‰ given by ‰ .v/ WD ‰.v/ for v 2 M.n1 ; n2 /Œs . Moreover, ‰, ‰ , ‰2 are the only extensions of . (3) If .a/ a 2 M.n1 ; n2 /. 2/ for all a 2 a, then there exists an extension ‰ 2 Aut.1/ M.n1 ; n2 /. If there is y 2 M.n1 ; n2 /Œk (k > 0) such that .a/ a Œy; a 2 M.n1 ; n2 /.k 2/ for all a 2 a, then there exists an extension ‰y satisfying ‰y .v/

v

Œy; v 2 M.n1 ; n2 /.i CkC1/

8 v 2 M.n1 ; n2 /Œi ; 8 i

3:

Proof. (a) Set M WD M.n1 ; n2 /, M.i/ WD M.n1 ; n2 /.i/ and n D .n1 ; n2 /. Let ı W u.Oa/ ! End M denote the unitary algebra homomorphism extending the adrepresentation. Define a linear mapping W M ! Homu.K/ .u.Oa/; M=M.0/ /;

.y/.u/ WD ı.u/.y/ C M.0/ :

For a 2 a, y 2 M and u 2 u.Oa/ one has .Œa; y/.u/ D ı.u/ı.a/.y/ C M.0/ D ı.ua/.y/ C M.0/ D .y/.ua/: Note that .y/.1/ D y C M.0/ for all y 2 M. Therefore ker is an a-invariant subspace of M.0/ . The assumption on a yields gr a D MŒ 3 , whence gr ker is an MŒ 3 -invariant subspace of M.0/ . Equation (4.3.1) then gives gr ker D ¹0º, whence ker D ¹0º. Counting dimensions dim M dim Homu.K/ .u.Oa/; M=M.0/ / O D 5dim a=K .dim M=M.0/ / 5n1 Cn2 5 D dim M

shows that is a linear isomorphism. Set W WD

1

Homu.K/ .u.Oa/; a C M.0/ =M.0/ / M:

Clearly, a W as a is a subalgebra. Also, W C M.0/ D ı.1/.W / D .W /.1/ a C M.0/ . Consequently, W D a ˚ .W \ M.0/ /. As above .Œa; y/.u/ D .y/.ua/ 2 a C M.0/ for all a 2 a, y 2 W and u 2 u.Oa/, whence Œa; W W . From this one easily concludes ŒW ; W W C M.0/ D a C M.0/ . Since ŒW ; W is invariant under a, the definition of now gives ŒW ; W W . Note that gr W is .gr a/-invariant. Hence it is an MŒ 3 -invariant subspace of MŒ 3 C M.0/ . This can only be if gr W M0 . As is a linear isomorphism and dim a D 2, one has dim W D 5n1 Cn2 2 D dim M0 . We conclude gr W D M0 .

80

10


(b) According to (a) there are subalgebras W and W 0 of M satisfying W D .W \ M.0/ / ˚ a;

W 0 D .W 0 \ M.0/ / ˚ a0

and gr W D gr W 0 D M0 Š W .2I n/. Definition 4.2.4 shows that both W and W 0 are of Witt type, and then Corollary 6.1.7 yields isomorphisms of filtered algebras W W ! W .2I n/;

0 W W 0 ! W .2I n/:

Note that W \ M.0/ has codimension 2 in W . The uniqueness of such algebra gives .W \ M.0/ / D W .2I n/.0/ ;

W .2I n/ D W .2I n/.0/ ˚ .a/;

0

W .2I n/ D W .2I n/.0/ ˚ 0 .a0 /:

0

.W \ M.0/ / D W .2I n/.0/ ; Consequently,

2 1 W .2I n/ 4 D W \ M ˚ aO , whence Nor

O = Nor O .W \ M.0/ / D dim W .2I n/= Nor dim W W

W .2In/

O On the other hand, one has W .0/ Q where KQ WD NoraO .W \ M.0/ /. This gives W \ M.0/ ˚ K,

4

.0/

O W

D n1 C n2 :

.W \ M.0/ / D

O = Nor O .W \ M.0/ / D n1 C n2 : dim aO =KQ D dim W W Q The assumption dim aO =K n1 C n2 now gives KQ D K. By definition, K K. Since is an isomorphism, O W aO ! ab0 is an isomorphism as well. By assumption, O.K/ D K 0 . Arguing as before gives K 0 D Norb0 .W 0 \ M.0/ /. We are now in the a position to apply Theorem 10.5.1. Hence there is an isomorphism W W ! W 0 which extends . (c) Theorem 10.5.2 shows that there is an extension Q 2 Aut M of . This proves claim (1). In order to determine all extensions of , we may assume that D Id. Choose any extension ‰ 2 Aut M of . The constructions in (a) and (b) yield a D a0 , W D W 0 Š W .2I n/ and ‰.W / D W . Moreover, the uniqueness of the extension (Theorem 10.5.1) gives ‰jW D IdjW . Theorem 10.5.2 now shows that ‰ is one of Id, Id , or Id2 , where 3 D 1 ¤ . This proves (2). (d) Suppose .a/ a 2 M. 2/ for all a 2 a. Choose any extension ‰ 2 Aut M N of of . Since ‰ stabilizes the natural filtration of M, it induces an automorphism ‰ P N defines an automorphism of gr M Š M. Note that, by restriction, ‰ i 1 MŒ3i D N M0 Š W .2I n/. The present assumption shows that ‰jMŒ 3 D Id. The only automorphism of W .2I n/ which acts as the identity on F @1 CP F @2 is the identity itself N D Id on M . As M D (Theorem 10.5.1). Hence ‰ i 1 MŒ3i C2 , one has 0 2 N ‰.M2 / D M2 . As this is an irreducible M0 -module, there is 2 F such that N D Id on M . The multiplication of M gives 3 D 1. Then ‰2 is an extension of ‰ 2

10.5

81

Melikian algebras

which satisfies ‰2 .v/ D v for all v 2 MŒs . Arguing as before we conclude ‰2 D Id on M2 . As M2 generates the algebra, one gets ‰2 D Id. Hence ‰2 2 Aut.1/ M is the required extension of . (e) Suppose there is y 2 MŒk (k > 0) such that .a/ a Œy; a 2 M.k 2/ for all a 2 a. Let ‰ 2 Aut.1/ M be an extension. If ‰ D Id, then Œy; a M.k 2/ . Looking at the graded algebra we obtain X Œy; MŒ 3 MŒk 3 \ MŒj D ¹0º; j k 2

whence y D 0. The claim follows. P Otherwise decompose ‰ D Id C j l ‰Œj , where ‰Œj is a homogeneous linear mapping of degree j > 0 and ‰Œl ¤ 0. Recall that ‰Œl is a derivation of M. If l > k, then ‰.v/ v 2 M.l 3/ M.k 2/ for all v 2 M. In particular, for v 2 a this gives Œy; v 2 M.k 2/ . As before, y D 0 and the claim follows. So assume l k. Set d WD ‰Œl ad y 2 Der M. Suppose d D 0. As ‰Œl ¤ 0, this gives k D l and ‰Œl D ad y. This is the claim. So assume d ¤ 0. Note that d acts on gr M in the natural way as a nonzero homogeneous degree l derivation. Since X d.a/ D ‰.a/ a ‰Œj .a/ Œy; a j >l

D .a/

a

X

Œy; a

‰Œj .a/ 2 M.l

2/

8a2a

j >l

one obtains d jMŒ 3 D 0. As l > 0, one has d.MŒs / D ¹0º. Then M2 D P j j 0 .ad MŒ 3 / .MŒs / is annihilated by d . As this set generates M as an algebra, we get d D 0, a contradiction In the following we will apply this corollary. Before doing this let us pursue a somewhat different approach. It is of some interest how one can extend divided power automorphisms of O.2I .n1 ; n2 // to mod.3/-homogeneous automorphisms of M.n1 ; n2 /. In order to get hand on that problem let 2 Autc O.2I .n1 ; n2 // and set ki;j WD .@i . 1 .xj ///; Then ı @i ı

1

i; j D 1; 2:

D ki;1 @1 C ki;2 @2 . Let us assume

det.ki;j / D 1;

div.ki;1 @1 C ki;2 @2 / D 0 for i D 1; 2:

(10.5.1)

Define ˆ W M.n1 ; n2 / ! M.n1 ; n2 / by setting ˆ .D/ WD ı D ı

1

Q WD ı D ı ˆ .D/

1

2 M.n1 ; n2 /0 ;

8D 2 M.n1 ; n2 /0 ;

1

8DQ 2 M.n1 ; n2 /2 ;

E 2 M.n ; n / ;

ˆ .f / WD .f / 2 M.n1 ; n2 / 2 ;

2 2

8f 2 M.n1 ; n2 / 2 :

(10.5.2)

82

10


Then ˆ is an algebra automorphism of M.n1 ; n2 /0 . It is easy to see that div.fD/ D f div.D/ C D.f / and ˆ .fD/ D .f /ˆ .D/;

B

Q D .f /ˆ .D/; ˆ .f D/ .D.f // D ˆ .D/..f // hold for f 2 O.2I .n1 ; n2 // and D 2 M.n1 ; n2 /0 . Equation (10.5.1) gives div.ˆ .f @i // D div .f /.ki;1 @1 C ki;2 @2 / D .f / div.ki;1 @1 C ki;2 @2 / C .ki;1 @1 C ki;2 @2 /..f // D .ki;1 @1 C ki;2 @2 /..f // D . ı @i ı 1 /..f // D .@i .f //: As a consequence, div.ˆ .D// D .div.D// for all D 2 M.n1 ; n2 /0 . This shows that ˆ is an M.n1 ; n2 /0 -module automorphism of M.n1 ; n2 /2 and M.n1 ; n2 / 2 , respectively. Moreover, putting .1; 1/ WD .2; 2/ WD 0;

.1; 2/ WD

.2; 1/ WD 1

one computes Œˆ f @Qr ; ˆ g @Qs D Œ.f /.kr;1 @Q1 C kr;2 @Q2 /; .g/.ks;1 @Q1 C ks;2 @Q2 / D .f /.g/.kr;1 ks;2 D .fg/.r; s/ D ˆ ˆ Œf; g D ˆ 2g@2 .f /@Q1

kr;2 ks;1 / D .fg/ det .ki;j /.r; s/ Œf @Qr ; g @Qs I

2f @2 .g/@Q1 C 2f @1 .g/@Q2

2g@1 .f /@Q2

D 2.g/.k2;1 @1 C k2;2 @2 /..f //.k1;1 @Q1 C k1;2 @Q2 / 2.f /.k2;1 @1 C k2;2 @2 /..g//.k1;1 @Q1 C k1;2 @Q2 / C 2.f /.k1;1 @1 C k1;2 @2 /..g//.k2;1 @Q1 C k2;2 @Q2 / 2.g/.k1;1 @1 C k1;2 @2 /..f //.k2;1 @Q1 C k2;2 @Q2 / D 2.k2;1 k1;2

k1;1 k2;2 / .g/@1 ..f //

C 2.k2;2 k1;1

.f /@1 ..g// @Q2

k1;2 k2;1 / .g/@2 ..f //

.f /@2 ..g// @Q1

D Œ.f /; .g/ D Œˆ .f /; ˆ .g/I Q D ˆ .fD/ D .f /ˆ .D/ D Œˆ .f /; ˆ .D/: Q ˆ .Œf; D/

10.5

83

Melikian algebras

As a result, every 2 Autc O.n1 ; n2 / satisfying (10.5.1) and (10.5.2) defines in the indicated way an automorphism of M.n1 ; n2 / which is homogeneous with respect to the mod.3/-grading. Theorem 10.5.4. Let ˛ 2 F and 2 SL.F x1 ˚ F x2 /. Extend to an element of Autc O.2/. The assignment .a/ .b/

ˆ˛; .x1 x2 @i / D ˛ 3.aCb/ 3 .x1 /.a/ .x2 /.b/ . ı @i ı

1

.a/ .b/ ˆ˛; .x1 x2 @Qi / D ˛ 3.aCb/ 1 .x1 /.a/ .x2 /.b/ . ı @i ı

1

/;

F/;

.a/ .b/

ˆ˛; .x1 x2 / D ˛ 3.aCb/ 2 .x1 /.a/ .x2 /.b/ defines a surjective homomorphism of groups F SL.F x1 ˚ F x2 / ! AutŒ0 M.n1 ; n2 / if n1 D n2 ; F ¹ 2 SL.F x1 ˚ F x2 / j .x2 / 2 F x2 º ! AutŒ0 M.n1 ; n2 / if n1 < n2 with kernel ¹.1; Id/; . 1; Id/º. Proof. (a) Observe that ˆ˛; D ˆ˛;Id ı ˆ1; , and ˆ˛;Id maps v onto ˛ i v for all v 2 M.n1 ; n2 /Œi . The latter obviously is an automorphism. Let 2 SL.F x1 ˚F x2 /. If n1 D n2 then defines a divided power automorphism of O.2I .n1 ; n2 //, and if n1 < n2 then every ¹ 2 SL.F x1 ˚ F x2 / j .x2 / 2 F x2 º does so because of Theorem 6.3.2. Put in the respective cases .xi / D si;1 x1 C si;2 x2 and k1;1 WD s2;2 ; k2;2 WD s1;1 ; k1;2 WD s2;1 ; k2;1 WD s1;2 : Since s1;1 s2;2 s1;2 s2;1 D 1, then 1 .xj / D k1;j x1 C k2;j x2 and .@i . 1 .xj /// D ki;j 2 F . Therefore Equation (10.5.1) holds. In these cases ˆ1; is the homogeneous automorphism ˆ defined by Equation (10.5.2). As a result, all ˆ˛; are automorphisms of M.n1 ; n2 / and therefore ˆ is a group homomorphism. (b) Suppose ˆ˛; D Id. Then ˛.xi / D ˆ˛; .xi / D xi ;

i D 1; 2:

Computing determinants on the space F x1 ˚ F x2 gives ˛ 2 D ˛ 2 det D 1, whence ˛ D ˙1 and D ˙Id. .a/ .b/ .a/ .b/ Suppose D Id. Then .x1 x2 / D . 1/aCb x1 x2 , whence ı @i ı 1 D @i . Consequently, ˆ 1; Id D ˆ1;Id D Id. (c) Let 2 AutŒ0 M.n1 ; n2 /. The homogeneity of gives .M.n1 ; n2 /Œ 1 / D M.n1 ; n2 /Œ 1 . Write .@Qi / D ki;1 @Q1 C ki;2 @Q2 for i D 1; 2 where ki;j 2 F . Set ˛ WD

q

det .ki;j /; s1;1 WD ˛

1

k2;2 ; s2;2 WD ˛

1

k1;1 ; si;j WD

˛

1

kj;i if i ¤ j:

84

10


Then .1/ D Œ .@Q1 /; .@Q2 / D k1;1 k2;2

k2;1 k1;2 D ˛ 2 ;

.@i / D Œ .1/; .@Qi / D ˛ 2 .ki;1 @1 C ki;2 @2 /: This gives .@1 / D ˛ 3 .s2;2 @1

.@2 / D ˛ 3 . s1;2 @1 C s1;1 @2 /;

s2;1 @2 /;

and from this we conclude by induction .a/ .b/

.x1 x2 / D ˛

3.aCb/C2

.s1;1 x1 C s1;2 x2 /.a/ .s2;1 x1 C s2;2 x2 /.b/ ;

for a 5n1 1; b 5n2 1. Since s2;1 D 0 if n1 < n2 . Define 2 SL.F x1 ˚ F x2 /;

maps M.n1 ; n2 / into itself, we conclude

.xi / WD si;1 x1 C si;2 x2 :

Note that .x2 / 2 F x2 if n1 < n2 . Extend to an element of O.2/. Then

.@i .

1

.x1 / D s2;2 x1

1

.x2 / D

1

.xj /// D ˛

s1;2 x2 D ˛

1

s2;1 x1 C s1;1 x2 D ˛ 1

ki;j

.k1;1 x1 C k2;1 x2 /; 1

.k1;2 x1 C k2;2 x2 /;

8i; j D 1; 2;

.@Qi / D ˛.˛ 1 ki;1 @Q1 C ki;2 @Q2 / D .@Qi /. Put Q WD ˆ˛ 1 1 ; ı . We have shown that Q acts on M.n1 ; n2 /Œ 1 as the identity. Assume inductively that Q .w/ D w for all w 2 M.n1 ; n2 /Œi (i 1). Then and ˆ˛

1 ;

Œu; v D Q .Œu; v/ D Œu; Q .v/ 8u 2 M.n1 ; n2 /Œ

1 ;

v 2 M.n1 ; n2 /Œi C1 :

Q Q acts Using P (g1) of Notation 3.5.2 we immediately conclude .v/ D v. Hence on i 1 M.n1 ; n2 /Œi as the identity. Using (g2) of Notation 3.5.2 this implies D ˆ˛ 1 ; . Next we aim to determine Aut.1/ M.n1 ; n2 /. Lemma 10.5.5. (1) If u 2 M.n1 ; n2 / and .ad u/5 D 0.

2

[ M.n1 ; n2 /2 , then one has .ad u/3 .M.n1 ; n2 / 2 / D ¹0º

.r/

(2) .ad xi @j /5 D 0 if r 1 and i ¤ j . (3) adM.n1 ;n2 / M.n1 ; n2 /.0/ is a restricted subalgebra of Der M.n1 ; n2 /.

10.5

85

Melikian algebras

Proof. (1) Take u; g 2 M.n1 ; n2 / 2 D O.2I .n1 ; n2 //. The multiplication (4.3.1) yields .ad u/3 .g/ D 2.ad u/2 g@2 .u/ u@2 .g/ @Q1 C u@1 .g/ g@1 .u/ @Q2 D 2Œu; ug@2 .u/ u2 @2 .g/ @1 C u2 @1 .g/ ug@1 .u/ @2 D 2ug@2 .u/@1 .u/ C 2u2 @2 .g/@1 .u/ C 4@1 ug@2 .u/ u2 @2 .g/ u 2u2 @1 .g/@2 .u/ C 2ug@1 .u/@2 .u/ C 4@2 u2 @1 .g/ ug@1 .u/ u D 0: It follows that .ad u/4 .M.n1 ; n2 /0 / D ¹0º and .ad u/5 .M.n1 ; n2 /2 / D ¹0º. Take u D EQ 2 M.n1 ; n2 /2 D W .2I .n1 ; n2 //, g 2 M.n1 ; n2 / 2 D O.2I .n1 ; n2 //. Then

F

Q 2 . gE/ D ŒE; Q E.g/EQ C 2 div.gE/E Q .ad u/3 .g/ D .ad E/ Q E Q D 0: D . E.g/ C 2 div.gE//ŒE; As before, the result follows. .r/ .r/ (2) For f 2 O.2I .n1 ; n2 // one has .xi @j /5 .f / 2 .xi /5 O.2I .n1 ; n2 // D .r/ .r/ ¹0º. Then .ad xi @j /5 .f @k / D f .ad xi @j /5 .@k / D 0. Note that M.n1 ; n2 /0N Š .r/ W .2I .n1 ; n2 //. Set d WD .ad xi @j /5 . We proved that d vanishes on M.n1 ; n2 /0N . Then d induces an M.n1 ; n2 /0N -module endomorphism of M.n1 ; n2 /2N . The irreducibility of this module forces that d acts as ˛Id on this module for some ˛ 2 F . As d is a derivation and d vanishes on M.n1 ; n2 /0N , one has ˛ 3 D 0, i.e., ˛ D 0. Since M.n1 ; n2 /2N generates the algebra, this gives d D 0. (3) Recall that M.n1 ; n2 /.0/ is the unique subalgebra of codimension 5 defining the depth 3 filtration. Observe that X M.n1 ; n2 /.0/ D M.n1 ; n2 /.0/ \ M.n1 ; n2 /iN : i D0;˙2

By the first assertion, .ad u/5 2 ad M.n1 ; n2 /.0/ if u 2 M.n1 ; n2 /.0/ \ M.n1 ; n2 /iN , i D ˙2. Note that M.n1 ; n2 /.0/ \M.n1 ; n2 /0N Š W .2I .n1 ; n2 //.0/ . Theorem 7.1.1(3) shows that for every D 2 M.n1 ; n2 /.0/ \M.n1 ; n2 /0N there is E 2 M.n1 ; n2 /.0/ \M.n1 ; n2 /0N such that d WD .ad D/5 .ad E/ vanishes on M.n1 ; n2 /0N . Now argue as in (2) to show that d D 0. Lemma 10.5.6. Let g be a restricted LiePalgebra over a field of characteristic 5. If a; b 2 g then .a C b/Œ5 D aŒ5 C b Œ5 C 4i D1 si .a; b/, where s1 D .ad b/4 .a/;

s2 D Œb; Œb; Œa; Œb; a

2Œa; Œb; Œb; Œb; a;

s4 D .ad a/4 .b/;

s3 D Œa; Œa; Œb; Œa; b

2Œb; Œa; Œa; Œa; b:

86

10


Proof. Equation (1.1.1) describes s4 .a; b/ D .ad a/4 .b/ and 1 Œa; Œa; Œb; Œa; b C Œa; Œb; Œa; Œa; b C Œb; Œa; Œa; Œa; b 2 1 D .2Œa; Œa; Œb; Œa; b C Œb; Œa; Œa; Œa; b/: 2

s3 .a; b/ D

The terms s1 and s2 are obtained symmetrically.

Lemma 10.5.7. Let k 1, ˛ 2 F and l D 1; 2. .i /

.j /

(1) Œxl ; xl D 0 if i; j < 5k < i C j ;

4 4 (3) .@ ˛x / @ .mod C.M .n ; n ///; 4 (4) if i < 5 then .@ ˛x @Q / @ .mod C.M .n ; n /// and Q 4 .@ ˛x @ / @ .mod C.M.n ; n ///; 4 (5) .@ ˛x @ / 2 C.M .n ; n // if n C k n . (2) .1

.5k 1/

˛x1

.5k 1/

@Q2 /5 ; .1

˛x2

.5k 1/ 5kC1 l

l

k

2

2

@Q1 /5 2 C.M .n1 ; n2 //;

5kC1 l .i/ 5k l 1

1 .i/ 5k 5k l 2 2 n .5k 1/ 2 5 1 2

1

2

5k

1

1

1

1

2

2

2

1

2

Proof. (1) We treat the case l D 1. .i /

.j /

.i/ .j 1/

Œx1 ; x1 D 2.x1 x1

i Cj D2 i since i; j < 5k i C j 1. .5k (2) Put a WD 1 and b D x1 .5k 1/

Œa; Œa; b D Œ1; x1

1/

.j / .i 1/

x1 x1 /@Q2 ! ! i C j 1 .iCj 1 x1 j

1/

@Q2 D 0

@Q2 . Then .5k 1/

@2 D 0; Œb; Œa; b D Œx1

k

.5 @Q2 ; x1

1/

@2 D 0:

Therefore one has in Lemma 10.5.6 s1 D D s4 D 0. Hence .1

.5k 1/

˛x1

.5k 1/

@Q2 /5 D 15

˛ 5 .x1

4

@Q2 /5 2 C.M.n1 ; n2 //

by Lemma 10.5.5. The second inclusion is treated similarly. i .5k 5i / (3) Let i < k and a WD @51 , b WD x1 . Take 0 r 3 if i < k 0 r 2 if i D k 1. Part (1) of this lemma yields .5k 5i /

Œb; .ad a/r .b/ D Œx1

.5k .rC1/5i /

; x1

1 and

D 0: .5k 5iC1 /

Apply Lemma 10.5.6. If i < k 1, then s1 D s2 D s3 D 0 and s4 D x1 induction and Lemma 10.5.5 one obtains .@1

.5k 1/ 5k

˛x1

/

1

k 1

@51

.5k 5k

˛x1

1/

4

.mod C.M.n1 ; n2 ///:

. By

10.5

If i D k

87

Melikian algebras

1, Lemma 10.5.6 gives

s1 D s2 D 0;

.5k 5k

s3 D

2Œx1

1/

.5k 45k

; x1

1/

.5k 1/

2 F xl

@Q2 ;

s4 D 1:

As a result, .5k 1/ 5kC1

.@1

˛x1

/

k

@51

.5k 1/

˛.1 C ˛ 0 x1

˛ 5 .1 C ˛ 0 x1 .5k 1/

We mentioned in (2) that .1 C ˛ 0 x1

.5k 1/ 5kC1

˛x1

4 4 0 .mod C.M .n ; n ///. Then 4 .mod C.M .n ; n ///: .5k 1/

kC1

@51

.@1

5 @Q2 /

/

@Q2 /5

@Q2 /5

.mod C.M.n1 ; n2 ///:

kC1

1

@51

1

2

2

The case l D 2 is treated similarly. .i/ .j / (4) Note that Œx1 @Ql ; x1 @Ql D 0 for all i; j . Using Lemmas 10.5.6 and 10.5.5 j j .i/ .i 5j / Q one proves inductively .@1 ˛x1 @Ql /5 @51 ˛x1 @l modulo C.M.n1 ; n2 // for all j . The first equation follows. The second equation is treated similarly.

4

.5k 5i /

i

(5) Put a WD @52 , b WD x2

.r/

.s/

@1 . As Œx2 @1 ; x2 @1 D 0 for all r; s, one has

4

.r/

s1 D s2 D s3 D 0 in Lemma 10.5.6. Lemma 10.5.5 gives .x2 @1 /5 2 C.M.n1 ; n2 // if r > 0. Then .5k 1/

.@2

˛x2

4

n2

@1 /5

.@52

k

n2

˛@1 /5

k

n2

@52

n2

˛5

k

n2

@51

k

0

modulo C.M.n1 ; n2 //.

Theorem 10.5.8. The mapping ‚i;Œ; W Aut.i/ M.n1 ; n2 /=Aut.iC1/ M.n1 ; n2 / ! M.n1 ; n2 /Œi ; is surjective except if i D 3.5k the image of ‚i;Œ; is

1/ for some k > 0. If i D 3.5k

.a/ .b/

span ¹x1 x2 @l ; a; b ¤ 0; a C b D 5k ; l D 1; 2º

1/ and k > 0 then

if n1 C k > n2 ; .5k /

.a/ .b/

i 1;

span ¹x1 x2 @l ; a; b ¤ 0; a C b D 5k ; l D 1; 2º ˚ F x2

@1 if n1 C k n2 :

Proof. Set M WD M.n1 ; n2 /. (1) We prove first that the exposed sets are contained in the image of ‚i;Œ; . (a) For l D 1; 2, and 0 k nl 1, and ˛ 2 F define W MŒ

3

! M;

.5k /

.@s / WD @s C Œ˛xl

; @s D @s

.5k 1/

˛ıs;l xl

;

s D 1; 2:

88

10

Then .MŒ

3 /

Tori in Hamiltonian and Melikian algebras ns

is abelian, and .@s /5

O by Lemma 10.5.7(3). Corollary 2 C.M/

.5k / F xl

10.5.3 applies. Hence Im ‚35k 2;Œ; for l D 1; 2 and 0 k nl (b) For l; j D 1; 2, and 1 i 5nj 1, and ˛ 2 F define W MŒ

3

! M;

.i/ .@s / WD @s C Œ˛xj @Ql ; @s D @s

.i 1/

˛ıs;j xj

1.

@Ql :

ns O by Lemma 10.5.7(4). Corollary 10.5.3 Then .MŒ 3 / is abelian and .@s /5 2 C.M/ .i/ Q applies. Hence F xj @l Im ‚3i 1;Œ; for l; j D 1; 2 and 1 i 5nj 1. (c) Suppose n1 C k n2 , k > 0. Define

W MŒ

3

.5k /

! M;

.@s / WD @s C Œ˛x2

.5k 1/

@1 ; @s D @s

˛ıs;2 x2

5ns 2 C.M/ O by Lemma 10.5.7(5). 3 / is abelian and .@s / k .5 / Hence F x2 @1 Im ‚35k 3;Œ; .

Then .MŒ

@1 ;

s D 1; 2:

Corollary 10.5.3

applies. (d) Define T to be the Lie set generated by .5k /

F xl

r

.i/ .5 / [ F xj @Ql [ F x2 @1 ;

where j; l D 1; 2;

0 k nl

1;

1 i 5nj

1;

0 < r n2

n1 :

We have shown that T ˚i 1 Im ‚i;Œ; . As .i/ .j / .i/ .j / Œx1 @Q1 ; x2 @Q2 D x1 x2 ; ! j .j / .s/ Q .j s/ Q Œxl @1 ; xl @2 D x ; s l

l D 1; 2;

and the latter is nonzero for some 1 s j 1 if j is not a power of 5, we get .i / .j / F x1 x2 2 T (i C j > 0, 0 i < 5n1 , 0 j < 5n2 ) except if i D 0; j D 5k or i D 5k ; j D 0. The exceptions are contained in T due to (a). Hence M 2 \ M.1/ T . Next we compute .i C1/

Œx1

.i/

.j /

; x2 D .j /

Œx1 ; x1 x2 D D

.2/

.5n1

Œx1 @1 ; x1

.i C1/ .j 1/ Q x2 @1

2x1

.i/

.j 1/

2x1 x1 x2

.j / .i/ 2x2 x1 @Q2 ;

.i/ .j / @Q1 C 2.x1 x2

.i C1/ .j 1/ Q x2 @1

2.i C 1/x1

C 2.1

.2/ Œx1 ; x1 @Q1 D x12 @1 D 2x1 @1 ; ! ! 5n1 1 5n1 1 .5n1 2/ .j / Q x2 @2 D C2 x1 2 1

D

.5n1 1/ .j / Q x2 @2 :

x1

.j / .i 1/

x1 x2 x1

/@Q2

.i/ .j / i/x1 x2 @Q2 ;

1/ .j / Q x2 @2

10.5

89

Melikian algebras

.i/ .j / Combining the first and the second equation we get F x1 x2 @Q2 T if i < 5n1 1. Combine this result with the third equation and apply the fourth equation. One obtains .5n1 1/ .j / Q that F x1 x2 @2 T . By symmetry we get M2 \ M.1/ T . Next we compute .i/ .j / .i/ .j / Œx1 ; x2 @Ql D x1 x2 @l ; ! j .j / .s/ .j s/ Q Œxr ; xr @l D x @l ; s r

l D 1; 2; r; l D 1; 2:

The latter is nonzero for some 1 s j 1 if j is not a power of 5. We get .i / .j / F x1 x2 @l T except if i D j D 0 or i D 0, j D 5k or i D 5k , j D 0. The .5k /

missing set F x2 @1 has been treated in (c). (2) In order to prove the reverse inclusion we have to show that Im ‚r;Œ; cannot be bigger than claimed. Choose ‰ 2 Aut.r/ M arbitrary. There is nothing to prove except r D 3.5k 1/ for some k > 0. Due to what we proved in (1) one may adjust ‰ so that X .5k / ˛i;j Œxi @j ; u 2 M.35k Cl 2/ 8u 2 MŒl ; l 3 ‰.u/ u i;j D1;2

and not all ˛i;j are 0. Taking u D @1 ; @2 we find q1 ; q2 2 M.35k ‰.@i / D @i

.5k 1/

xi

5/

such that

.˛i;1 @1 C ˛i;2 @2 / C qi :

Note that k < ni if ˛i;1 ¤ 0 or ˛i;2 ¤ 0 because otherwise this is not an element n b On the other hand, in M. As ‰ is an automorphism, one has ‰.@i /5 i 2 C.M/. ni ni k 5ni 5 5 b is ‰.@i / D @i .˛i;1 @1 C ˛i;2 @2 / C terms of higher degree. Since C.M/ n ni 5 b graded, all homogeneous components of ‰.@i / are contained in C.M/. As @5i i 2 b we get .˛i;1 @1 C ˛i;2 @2 /5ni k 2 C.M/. b If ˛i;1 ¤ 0 this gives ni k n1 C.M/, and if ˛i;2 ¤ 0 this gives ni k n2 . The normalization n1 n2 made at the beginning of this section and the assumption k > 0 imply ˛1;1 D 0 and ˛i;2 D 0 for all i D 1; 2. Therefore only ˛2;1 can be nonzero, and n2 k n1 holds. Then .5k /

‚r;Œ; .‰/ 2 F x2

@1 and n1 C k n2 . This is the claim.

Let us determine the analogous group Aut¹0º M.n1 ; n2 / with respect to the Melikian grading. Theorem 10.5.9. Let ˆ denote the homomorphism mentioned in Theorem 10.5.4 and B the image of F ¹ 2 SL.F x1 ˚ F x2 / j .x2 / 2 F x2 º under ˆ. Then Aut¹0º M.n1 ; n2 / D B Ë Aut.1/ M.n1 ; n2 /:

90

10


Proof. (a) We intend to show that B Aut¹0º M.n1 ; n2 /. Clearly, ˆ˛;Id is contained in Aut¹0º M.n1 ; n2 /. Take 2 SL.F x1 ˚F x2 / satisfying .x2 / 2 F x2 . Put .x1 / D s1;1 x1 C s1;2 x2 , .x2 / D s2;2 x2 . Then (we did a similar short calculation already in part (a) of the proof of Theorem 10.5.4) ˆ1; .@Q1 / D ı @Q1 ı

1

D s2;2 @Q1 ;

ˆ1; .@Q2 / D

s1;2 @Q1 C s1;1 @Q2 ;

ˆ1; .@1 / D ı @1 ı

1

D s2;2 @1 ;

ˆ1; .@2 / D

s1;2 @1 C s1;1 @2 ;

.a/ .b/

.b/

ˆ1; .x1 x2 / D .s1;1 x1 C s1;2 x2 /.a/ s2;2 b x2 : These equations immediately imply that ˆ1; does not lower the Melikian degree. (b) We intend to show that Aut.1/ M.n1 ; n2 / Aut¹0º M.n1 ; n2 /. Take ‰ 2 Aut.1/ M.n1 ; n2 /. Then there is y 2 M.n1 ; n2 /Œ1 D F x1 C F x2 such that ‰.v/ D v C Œy; v C O.i C 2/ for all v 2 M.n1 ; n2 /Œi . For v 2 M.n1 ; n2 /¹0º one has v 2 F @Q1 C M.n1 ; n2 /.0/ , whence ‰.v/ 2 v C F Œy; @Q1 C M.n1 ; n2 /.1/ v C F x1 @1 C F x2 @1 C M.n1 ; n2 /.1/ M.n1 ; n2 /¹0º : This gives ‰.M.n1 ; n2 /¹0º / M.n1 ; n2 /¹0º . We mentioned at the beginning of this section that this result implies the claim. (c) In order to prove the reverse inclusion consider ‰ 2 Aut¹0º M.n1 ; n2 /. Due to (b) we may adjust ‰ by an element of Aut.1/ M.n1 ; n2 /, i.e., assume that ‰ 2 AutŒ0 M.n1 ; n2 /. Theorem 10.5.4 shows that there are ˛ 2 F and 2 SL.F x1 ˚ F x2 / such that ‰ D ˆ˛; . As ‰ stabilizes M.n1 ; n2 /¹0º , it induces a linear transformation on M.n1 ; n2 /Œ 1 \ M.n1 ; n2 /¹0º D F @Q1 . Hence ˆ˛; .@Q1 / 2 F @Q1 . By definition this implies ˆ˛; .@1 / 2 F @1 which in turn means .x2 / 2 F x2 . We now restrict ourselves to the Melikian algebra M.1; 1/. In this case Theorem 10.5.8 says the following: ‚i;Œ; is surjective except if i D 3.5k 1/ for some k > 0, and as n1 D n2 , only the first of the exceptional cases is possible. Since n1 D n2 D 1, .5/ .5/ the elements x1 @l , x2 @l are not contained in M.1; 1/, which means that the image .a/ .b/ span ¹x1 x2 @l ; a C b ¤ 0; a C b D 5k ; l D 1; 2º of ‚i;Œ; is the full space M.1; 1/Œi . Consequently, Im ‚i;Œ; D M.1; 1/Œi

8 i 1:

This result gives us sufficient information on Aut M.1; 1/. Next we are going to derive a similar result for the Melikian grading. We mentioned earlier that the natural and the Melikian grading are compatible. Therefore every endomorphism has a decomposition X X X X X D Œi ;hj i D Œi D hj i ; Œi D Œi;hj i ; hj i D Œi;hj i ; i;j

i

j

j

i

10.5

91

Melikian algebras

where Œi ; hj i ; Œi;hj i are the homogeneous components of with respect to the standard, P the Melikian and both gradings. If 2 Aut¹0º M.1; 1/, then we have that D i;j 0 Œi ;hj i has only non-negative components with respect to both gradings and h0i is an automorphism of M.1; 1/. Define distinguished automorphisms h;i;ˇ .v/ WD ˇ j v;

v 2 M.1; 1/hj i ; ˇ 2 F :

(10.5.3)

These are contained in Auth0i M.1; 1/. Let y 2 F x1 [ F x1 @Q2 [ F x12 @2 . Then y 2 M.1; 1/Œi0 \ M.1; 1/h0i for some i0 > 0. Due to Theorem 10.5.8 and an earlier remark there is y 2 ‚i0 ;Œ;

1

.y/ Aut.i0 / M.1; 1/:

Let us fix an assignment y 7! y . ObservePthat y 2 Aut.1/ M.1; 1/ Aut¹0º M.1; 1/ by Theorem 10.5.9 and y D IdCad y C i>i0 y;Œi by definition. The above shows that its homogeneous degree 0 component y WD y;h0i ;

y 2 F x1 [ F x1 @Q2 [ F x12 @2 M.1; 1/h0i ;

(10.5.4)

is an element of Auth0i M.1; 1/ \ Aut.i0 / M.1; 1/ and that X y D Id C ad y C y;Œi;h0i i>i0

holds. Theorem 10.5.10. (1) Every element of Auth0i M.1; 1/ acts identically on C.M.1; 1/h0i /; the restriction Auth0i M.1; 1/ ! Aut M.1; 1/h0i .1/ is surjective; Auth0i M.1; 1/ is the group generated by ¹ˆ˛;Id ; h;i;ˇ ; y j ˛; ˇ 2 F ; y 2 F x1 [ F x1 @Q2 [ F x12 @2 º: (2) For every y 2 M.1; 1/hki (k > 0) there is y 2 Aut¹kº M.1; 1/ such that y .v/

v

Œy; v 2 M.1; 1/¹kCi C1º

8 v 2 M.1; 1/hi i ; 8 i

2:

Proof. (1) Recall that C.M.1; 1/h0i / D F c where c D x1 @1 C 2x2 @2 , and there is an

isomorphism ƒ W M.1; 1/h0i ! W .1I 1/ ˚ F such that ƒ.@Q1 / D @; ƒ. x1 @1 / D x@; ƒ.2x1 / D x 2 @; ƒ.x1 @Q2 / D x 3 @; ƒ.2x12 @2 / D x 4 @: Œ5 By Lemma 10.5.5 one has @Q1 D .2x1 /Œ5 D .x1 @Q2 /Œ5 D .2x12 @2 /Œ5 D 0. One easily observes that . x1 @1 /Œ5 D . x1 @1 /. Therefore ƒ is a restricted homomorphism when restricted to M.1; 1/h0i .1/ .

92

10


(a) Take ‰ 2 Auth0i M.1; 1/ arbitrary. It is clear that ‰.c/ 2 C.M.1; 1/h0i /. As M.1; 1/h 1i is M.1; 1/h0i -irreducible, both c and ‰.c/ act on M.1; 1/h 1i as a scalar multiple of the identity. Computing traces and observing that dim M.1; 1/h 1i D p 1, one sees that these scalars are equal. Hence ‰.c/ c is an element of M.1; 1/h0i which annihilates M.1; 1/h 1i . But M.1; 1/h0i acts faithfully on M.1; 1/h 1i . This gives ‰.c/ D c. (b) Take ' 2 Aut M.1; 1/h0i .1/ arbitrary. Since M.1; 1/h0i .1/ Š W .1I 1/ is centerless, every such ' is a restricted automorphism. Every element of Auth0i M.1; 1/ gives rise (by restriction) to an element of Aut M.1; 1/h0i .1/ . We write '.@Q1 / D ˛ 1 @Q1 C ˛0 x1 @1 C ˛1 x1 C ˛2 x1 @Q2 C ˛3 x12 @2 : Since '.@Q1 / is not contained in the maximal subalgebra of codimension 1, one has ˛ 1 ¤ 0. Applying ˆ˛ 1 ;Id gives ˛ 1 D 1. Then applying ˛0 x1 2 Auth0i M.1; 1/\ Aut.1/ M.1; 1/ leads to the case ˛0 D 0. Similarly, applying ˛ x @Q 2 Auth0i M.1; 1/ 1 1 2 \ Aut.2/ M.1; 1/ and 3˛2 x 2 @2 2 Auth0i M.1; 1/ \ Aut.3/ M.1; 1/ leads eventually to 1 the case '.@Q1 / D @Q1 C x 2 @2 with 2 F . Apply the restricted homomorphism ƒ 1

and obtain Œ5

0 D ƒ.'.@Q1 // D ƒ.'.@Q1 //Œ5 D .@ C

4 Œ5 x @/ 2

@ 2

.mod W .1I 1/.0/ /:

This gives D 0. As a result, we find ' 0 2 Auth0i M.1; 1/ such that .' 0 1 ı '/.@Q1 / D @Q1 . Since every automorphism of W .1I 1/ acting identically on @ is the identity (Theorem 10.5.1), this gives ' 0 1 ı ' D Id on M.1; 1/h0i .1/ . Then ' is the restriction of the element ' 0 2 Auth0i M.1; 1/. (c) Set Q the group generated by ¹ˆ˛;Id ; h;i;ˇ ; y j ˛; ˇ 2 F ; y 2 F x1 [ F x1 @Q2 [ F x12 @2 º. Clearly, Q Auth0i M.1; 1/. To prove the converse inclusion take ‰ 2 Auth0i M.1; 1/ arbitrary. Arguing as in (b) we find 2 Q such that 1 ı ‰ acts as the identity on M.1; 1/h0i .1/ . As M.1; 1/h 1i is M.1; 1/h0i .1/ -irreducible, 1 ı ‰ acts as a multiple of the identity on M.1; 1/h 1i . Applying a suitable h;i;ˇ 2 Q gives Q WD h;i;ˇ 1 ı 1 ı ‰ acts as the identity on M.1; 1/h 1i . Then ‰ Q acts as the that ‰ identity on ŒM.1; 1/h 1i ; M.1; 1/h 1i D M.1; 1/h 2i also. Assume inductively that Q ‰.w/ D w for all w 2 M.1; 1/hi i . Then Q Q Œu; v D ‰.Œu; v/ D Œu; ‰.v/

8u 2 M.1; 1/h

1i

and 8v 2 M.1; 1/hi C1i :

Q We now recall that the Melikian filtration is a standard filtration. Since ‰.v/ v 2 Q M.1; 1/hi C1i is annihilated by M.1; 1/h 1i , it follows that ‰.v/ D v. Consequently, Q D Id whence ‰ 2 Q. ‰ (2) Let ‰ P 2 Aut¹iº M.1; 1/ n Aut¹i C1º M.1; 1/ for some i 1. Then ‰ D Id C ‰h;i;i C k>i ‰h;i;k where ‰h;i;i 2 M.1; 1/hii . The assignment ‰ 7! ‰h;i;i induces an injective mapping i W Aut¹i º M.1; 1/=Aut¹i C1º M.1; 1/ ! M.1; 1/hii :

10.5

93

Melikian algebras

Due to Theorems 10.5.9 and 10.5.8 one has dim Aut¹0º M.1; 1/ D 3 C dim M.1; 1/.1/ D 3 C .125

9/ D 119:

The first part of this theorem yields 119 D dim Aut¹0º M.1; 1/ D dim Auth0i M.1; 1/ C dim Aut¹1º M.1; 1/ X 5C dim i .Aut¹iº M.1; 1/=Aut¹i C1º M.1; 1// i 1

5 C dim M.1; 1/¹1º D 5 C .125

11/ D 119:

Consequently, every i is surjective. This proves the claim.

In a next step we will determine orbits of toral elements in M.1; 1/ under its automorphism group. Lemma 10.5.11. Let g D g. r/ ¹0º be a finite dimensional Lie algebra with separating filtration, W g.1/ ! Aut.1/ g be an assignment such that .y

Id

ad y/.g.i/ / g.i Cj C1/

8 y 2 g.j /

.j > 0; i

r/;

and G denote the subgroup of Aut.1/ g generated by .g.1/ /. (1) Let a 2 g.i / and V g be a subspace such that gr g D Œgri a; gr g C gr V: For any b 2 g.i/ such that b a 2 g.iCk/ for some k > 0 there is ‰ 2 G \ Aut.k/ g for which ‰.b/ 2 a C V \ g.i Ck/ . (2) If t; t 0 2 g.0/ are ad-semisimple elements such that t conjugate under G .

t 0 2 g.1/ , then t; t 0 are

Proof. (1) By assumption, b a 2 g.i Ck/ D V \g.iCk/ Cg.iCk/ . Suppose inductively b a 2 V \ g.iCk/ C g.i Cj / for some j k. Write b a D v C u where v 2 V \ g.i Ck/ , u 2 g.iCj / . By hypothesis find y 2 g.j / such that gri Cj u Œgri a; grj y 2 gri Cj V . Then u

Œa; y 2 V \ g.i Cj / C g.i Cj C1/ ;

whence y .b/ b C Œy; b D .a C v C u/ C Œy; a C Œy; b a

.mod V \ g.iCk/ C g.iCj C1/ /:

This completes the induction step.

a

94

10


(2) Since t 2 g.0/ and ad t is semisimple, we have g.j / D Œt; g.j / Cg.j / \Cg .t/ for every j . Set in (1) i D 0, a WD t , b WD t 0 , V WD Cg .t/, k D 1. Find ‰ 2 G such that ‰.t 0 / t 2 g.1/ \ Cg .t/. Then t and ‰.t 0 / are commuting ad-semisimple elements. This shows that ‰.t 0 / t is ad-semisimple as well as ad-nilpotent. Consequently ‰.t 0 / D t. Theorem 10.5.12. (1) Every toral element t 2 M.1; 1/ n M.1; 1/. (2) Every toral element t 2 M.1; 1/. with 2 ¹1; 2º.

2/

2/

is conjugate to .1 C x1 /@1 .

n M.1; 1/.

is conjugate to .1 C x1 @Q1 /

1/

(3) Every toral element t 2 M.1; 1/. 1/ n M.1; 1/.0/ is conjugate to .@Q1 C x1 @1 / or to some .@Q1 C x2 @2 / with 2 F5 . (4) Every 1-dimensional torus in M.1; 1/.0/ is conjugate to one and only one of F x1 @1 , F .x1 @1 C x2 @2 /, F .x1 @1 C 2x2 @2 /, or F .x1 @1 x2 @2 /. Proof. Abbreviate M D M.1; 1/. Let t denote a toral element. (1) Let t 2 M n M. 2/ . Since AutŒ0 M acts on MŒ 3 Š .F x1 ˚ F x2 / as the full linear group GL.2/ (Theorem 10.5.4), we may assume t @1 .mod M. 2/ /. Clearly, M D Œ@1 ; M C x14 M, where x14 M WD

4 X X

j

F x14 x2 @l C

4 X

j

F x14 x2 @Ql :

lD1;2 j D0

j D0

lD1;2 j D0

4 X X

j

F x14 x2 C

Therefore we may apply Lemma 10.5.11 with V WD x14 M and assume t D @1 C x14 u, j i Cj 1 u 2 M. Being toral t satisfies the equation t Œ5 D t . Since Œx1i M; x1 M x1 M for i; j > 0, one has Œ5

@1 t D t Œ5 @1 C .ad @1 /4 .x14 u/

u

.mod x1 M/:

This shows that all toral elements in M n M. 2/ are conjugate to @1 x14 @1 . As .1 C x1 /@1 is toral, the claim follows. (2) Let t 2 M. 2/ n M. 1/ . Applying suitable ˆ˛;Id we may assume t 1 Q D D for all D 2 W .2I 1/, one has .mod M. 1/ /. Let us determine Œ1; M. As Œ1; D P j M0 Œ1; M. Next Œ1; W .2I 1/ D div.W .2I 1// D .i;j / 0. It has been mentioned on Volume 1, p. 47 that .x; / is semisimple. As dim g MT.g/ > 0, we now have .x; / 2 t. By Equation (1.5.4), E.x;/ D .p 1/ŠId .p 1/Š ad x C O..ad x/2 /.

114

10


Pdim g MT.g/ 1 Œpi Then f . / D 1 C C O. 2 /. Set q.x/ WD i D1 x . Given 2 .g; t/ and x 2 g set tx WD t .t/ x C q.x/ 8 t 2 t and tx WD ¹tx j t 2 tº: Theorem 1.5.1 proves that E.x;/ is a bijective linear mapping, tx is a torus of maximal dimension, and the roots .g; tx / with respect to tx are given by Equation (1.5.6), namely .x;/ .tx / D .t/ ..x; //.t/ 8 2 .g; t/ n ¹0º: Moreover, E.x;/ .L / D L.x;/ holds for all 2 .g; t/. Theorem 10.7.1. Let t1 ; t2 be tori of maximal dimension in a finite dimensional restricted Lie algebra g and V a finite dimensional restricted g-module. Put .V; ti / the set of weights of V with respect to ti (i D 1; 2). Let F .V; ti / denote the F -span of .V; ti / in ti . There exists an isomorphism of F -spaces W F .V; t1 / ! F .V; t2 / such that ..V; t1 // D .V; t2 / and dimF V D dimF V./ for every 2 .V; t1 /. In particular, the following holds: (1) dim V0 .t1 / D dim V0 .t2 /. (2) If dim V D q for all 2 .V; t1 /, then dim V 0 D q for all 0 2 .V; t2 /. Proof. By Theorem 1.5.5, t2 can be obtained from t1 by a finite chain of elementary switchings. Thus in order to prove the theorem it suffices to assume that there is a root vector x 2 g for some 2 .g; t1 / such that t2 D ¹tx j t 2 t1 º. Choose 2 F such that E.x;/ is an elementary switching. Give gQ WD g ˚ V a restricted Lie algebra structure by letting ŒV; V D V Œp D ¹0º. Obviously, t1 is a torus of maximal dimension in gQ and the ideal V gQ is E.x;/ -stable. Define W t1 ! t2 by the rule ./ WD .t/

..x; //.t/ 8 2 t1 :

Clearly, is F -linear. If ./ D 0 for some 2 t1 , then D ..x; //. But i .x Œp / D 0 for all i > 0, yielding ..x; // D 0, whence D 0. Consequently, is a linear isomorphism. As E.x;/ is invertible, dim V D dim E.x;/ .V / holds for every 2 .V; t1 /. Also, E.x;/ .V / V./ . All claims follow. Our arguments in the following proof rely on the commutator formula Lemma 2.1.5(1) valid in an arbitrary associative algebra A over F : ! ! X sn s1 k1 s1 sn k1 CCkn s1 zx1 xn D . 1/ x xnsn kn k1 kn 1 0ki si

Œxn Œxn Œx1 Œx1 ; z : „ ƒ‚ … „ ƒ‚ … kn

k1

10.7

115

Weights

Theorem 10.7.2. Let g be one of the restricted Cartan type Lie algebras S.3I 1/.1/ , Œp H.4I 1/.1/ , K.3I 1/, or the restricted algebra H.2I .1; 2//.2/ C F @2 , or the restricted Melikian algebra M.1; 1/. Let t be a 2-dimensional torus of g and V a nontrivial restricted g-module. The set .V; t/ of t-weights is a 2-dimensional vector space over Fp . Proof. (a) Let V Vn be a g-composition series. Suppose all composition factors are trivial g-modules. As g.1/ is simple in any case, then V is a trivial g.1/ -module. But then V is a trivial module for the p-envelope g, contradicting our assumption. Therefore it suffices to prove the theorem under the assumption that V is g-irreducible. Since V is a restricted t-module, .t/ 2 Fp holds for all 2 .V; t/ and all toral elements t 2 t. For Fp -independent weights ; ı one therefore has .V; t/ Fp ˚ Fp ı. Let g.k/ denote the k-th component of the natural filtration of g. Let V0 be an irreducible g.0/ -submodule of V . Since g.1/ is a Œp-nilpotent ideal of g.0/ , it annihilates V0 . There is a surjective g-module homomorphism ‰ W u.g/ ˝u.g.0/ / V0 ! V such that ‰.1 ˝ v/ D v for any v 2 V0 . Set VQ WD u.g/ ˝u.g.0/ / V0 . (b) Let g be one of the restricted algebras S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/, M.1; 1/. By Theorem 10.6.3 g has no tori of dimension > 2. Theorem 10.7.1 shows that the claim holds if and only if it is true for any 2-dimensional torus. Therefore it is sufficient to prove the statement for the tori listed below if

g D S.3I 1/.1/ ;

FDH .x1 x3 / ˚ FDH .x2 x4 /

if

g D H.4I 1/.1/ ;

FDK .x1 x2 / ˚ FDK .x3 /

if

g D K.3I 1/;

F x1 @1 ˚ F x2 @2

if

g D M.1; 1/:

F .x1 @1

x2 @2 / ˚ F .x2 @2

x 3 @3 /

These tori are contained in g.0/ . Therefore we may assume that V0 is invariant under t. (c) Suppose g D S.3I 1/.1/ . Then g D g.0/ ˚ F @1 ˚ F @2 ˚ F @3 . Let v 2 V0; be j p 1 p 1 an arbitrary t-weight vector. If @2 @3 ˝ v 62 ker ‰, then all @i2 @3 ˝ v 62 ker ‰ for 0 i; j p 1. In this case the claim follows. So assume that p 1 p 1 @3

@2

˝ v 2 ker ‰:

Using the commutator formula and the fact that ŒDi;j .x1a1 x2a2 x3a3 /; @k D

Di;j .@k .x1a1 x2a2 x3a3 //

for all admissible i; j; k and a1 ; a2 ; a3 , one obtains X p 1 p 1 p 1 p 1 D1;2 .x12 x2 x3 /@2 @3 D D1;2 .x12 / C i2 Ci3 >0

i2 ;i3 @i22 @i33 D1;2 .x12 x2i2 x3i3 /:

116

10


Since g.1/ v D ¹0º, we get p 1 p 1 x3 /

ker ‰ 3 D1;2 .x12 x2

p 1 p 1 @3

.@2

˝ v/ D 1 ˝ D1;2 .x12 / v:

This means that D1;2 .x12 /v D 0. As V0 is a semisimple t-module and v is an arbitrary weight vector of V0 , D1;2 .x12 / annihilates V0 . Since g.0/ =g.1/ Š sl.2/ is a simple Lie algebra, we derive g.0/ V0 D ¹0º. Next we argue similarly to obtain p 1 p 1 x3 /

ker ‰ 3 D1;2 .x1 x2

p 1 p 1 @3

.@2

˝ v/ D D1;2 .x1 / ˝ v;

which gives @2 V0 D D1;2 .x1 / V0 D ¹0º. By symmetry we get g V0 D ¹0º. Since V is assumed to be irreducible and nontrivial, this is aP contradiction. (d) Suppose g D H.4I 1/.1/ . Then g D g.0/ ˚ 4i D1 F @i . Proceed as in the p 1 p 1 former case: assume that @1 @2 ˝ v 2 ker ‰ for every weight vector v 2 V0 . Using the commutator formula, the equality g.1/ V0 D ¹0º, and the fact that ŒDH .x1a1 x2a2 x3a3 /; @k D

DH .@k .x1a1 x2a2 x3a3 //;

one obtains p 1 p 1 2 x2 x3 /

ker ‰ 3 DH .x1

p 1 p 1 @2

.@1

˝ v/ D 1 ˝ DH .x32 / v:

Then DH .x32 / v D 0. As g.0/ =g.1/ Š sp.4/ is a simple Lie algebra and v is an arbitrary weight vector of V0 , we derive g.0/ V0 D .0/. Next we argue similarly to obtain p 1 p 1 x2 x3 /

ker ‰ 3 DH .x1

p 1 p 1 @2

.@1

˝ v/ D DH .x3 / ˝ v;

which gives @1 V0 D DH .x3 / V0 D ¹0º. By symmetry we get @i V0 D ¹0º for i D 1; : : : ; 4. Then g V0 D ¹0º. Since V is assumed to be irreducible and nontrivial, this is a contradiction. (e) Suppose g Š K.3I 1/. Then g D g.0/ ˚ FDK .x1 / ˚ FDK .x2 / ˚ FDK .1/. Note that DK .x1 / and DK .1/ are commuting root vectors relative to t. Moreover, the corresponding roots span t . As before one may assume DK .1/p

1

DK .x1 /p

1

˝ v 2 ker ‰

for each weight vector v 2 V0 . It follows from Equations (4.2.10), (4.2.12), (4.2.13) that .a / .a / .a / .a / .a / .a 1/ ŒDK .x1 1 x2 2 x3 3 /; DK .1/ D 2DK .x1 1 x2 2 x3 3 / and .a1 / .a2 / .a3 / x2 x3 /; DK .x1 /

ŒDK .x1

D

.a1 / .a2 1/ .a3 / x2 x3 /

DK .x1

.a1 C1/ .a2 / .a3 1/ x2 x3 /:

.a1 C 1/DK .x1

10.7

117

Weights

Combining these relations with the commutator formula we obtain .2/ .p 1/ .p 1/

ker ‰ 3 DK .x1 x2 x3 / DK .1/p 1 DK .x1 /p 1 ˝ v X .2/ .m/ .k/ .l/ D DK .x1 / C ˛i;j;k;l;m DK .1/i DK .x1 /j DK .x1 x2 x3 / ˝ v mCkCl>2

D1˝

.2/ DK .x1 /

v: .2/

Now g.0/ =g.1/ Š gl.2/ and the image of DK .x1 / in g.0/ =g.1/ is non-central. In other words, .2/ .2/ FDK .x1 / C FDK .x1 x2 / C FDK .x2 / C g.1/ annihilates V0 . Next compute .p 1/ .p 1/

ker ‰ 3 DK .x1 x2 x3 / DK .1/p 1 DK .x1 /p 1 ˝ v X .m/ .k/ .l/ D DK .x1 / C ˛i;j;k;l;m DK .1/i DK .x1 /j DK .x1 x2 x3 / ˝ v: mCkCl>1

In the latter sum only terms with m 1 occur. Therefore every summand is contained in FDK .x12 / C FDK .x1 x2 / C g.1/ . So DK .x1 / annihilates V0 . The algebra generated .2/ .2/ by FDK .x1 / C FDK .x1 / C FDK .x1 x2 / C FDK .x2 / C g.1/ is all of g, and this implies g V0 D ¹0º. As before this is a contradiction. (f) Suppose g D M.1; 1/ D W .2I 1/ ˚ O.2I 1/ ˚ W .2I 1/. As before we may assume p 1 p 1 @1 @2 ˝ v 2 ker ‰

B

for every weight vector v 2 V0 . Note that x1m x2n @Qj 2 M.1; 1/.1/ whenever mCn 1, j 2 ¹1; 2º. So using the commutator formula and the multiplication formula (4.3.1) of Volume 1, p. 199 it is easy to see that p 1 p 1Q x2 @j /

ker ‰ 3 D .x1

p 1 p 1 @2

@1

˝ v D @Qj ˝ v

.j D 1; 2/:

The subalgebra of M.1; 1/ generated by @Q 1 ; @Q 2 ; M.1; 1/.1/ is the full algebra. Hence M.1; 1/ V0 D ¹0º, and as before this gives a contradiction. (g) Finally we deal with the nonrestricted Lie algebra H.2I .1; 2//.2/ . As in the former proof we may take any 2-dimensional torus for t. Let us take the torus generated by the semisimple element .p 2 1/

t D DH .x1 C x1 x2

/;

t D F t ˚ F t Œp 2

(see Theorem 10.2.7). Due to Lemma 10.2.6 this element satisfies the equation t Œp D 2 t and therefore .t p t/ V D ¹0º .

118

10


Suppose there is a vector v 2 V0 such that the family .t i v/i D0;:::;p2

1

is linearly independent. Then for every ˛ 2 Fp2 the element v˛ WD

2 1 pX

˛

j j

t v

j D1 2

2

t/ v D 0. Similarly, t p 1 v v ¤ 0 is is nonzero. It satisfies .t ˛/ v˛ D .t p a weight vector for the weight 0. Then .V; t/ Š Fp2 and the claim follows. We therefore may assume that for every vector v 2 V0 there is a relation r X

˛j t j ˝ v 2 ker ‰;

r p2

1; ˛r ¤ 0:

j D0

Since t @2 .mod H.2I .1; 2//.0/ / and H.2I .1; 2//.2/ .1/ V0 D ¹0º, one gets .2/ .r/

ker ‰ 3 DH .x1 x2 /

r X

˛j t j ˝ v

j D0

.2/ .r/ .2/ D . 1/r ˛r .ad t/r DH .x1 x2 / ˝ v D . 1/r ˛r ˝ DH .x1 /.v/: .2/

Thus DH .x1 / annihilates V0 . Since g.0/ =g.1/ Š sl.2/ is simple, we obtain g.0/ V0 D ¹0º. But then ker ‰ 3

.r/ DH .x1 x2 /

r X

˛j t j ˝ v D . 1/r ˛r DH .x1 / ˝ v:

j D0

As DH .x1 / and g.0/ generate g as a restricted Lie algebra, we get g V0 D ¹0º. This contradiction proves the theorem. The remaining Cartan type Lie algebras W .2I 1/, W .1I 2/, H.2I 1I ˆ.//.1/ , H.2I 1I ˆ.1// of absolute toral rank 2 do have modules with less than p 2 weights. Namely, the natural modules O.1I 2/=F for W .1I 2/ and O.2I 1/=F for W .2I 1/, H.2I 1I ˆ. //.1/ , H.2I 1I ˆ.1// are examples for these. In these modules only the 0 weight is missing. Theorem 10.7.3. Suppose g is one of the Lie algebras W .2I 1/;

W .1I 2/;

H.2I 1I ˆ.//.1/ ;

H.2I 1I ˆ.1//

and let gŒp denote the minimal p-envelope of g. Let V be a nontrivial restricted gŒp module and t a 2-dimensional torus in gŒp . Then .V; t/ [ ¹0º is a 2-dimensional vector space over Fp .

10.7

119

Weights

Proof. (a) As in the proof of the former theorem one may take any 2-dimensional torus for t. Suppose we have independent Fp -roots ˛, ˇ and root vectors u˛ 2 g˛ , Œp Œp uˇ 2 gˇ for which u˛ 2 t n ¹0º, uˇ 2 t n ¹0º. Then ˛.uŒp ˛ / D 0;

ˇ.uŒp ˛ / ¤ 0;

Œp

Œp

˛.uˇ / ¤ 0;

ˇ.uˇ / D 0:

./

Since V is a restricted t-module, .V; t/ Fp ˛ ˚ Fp ˇ. Let 2 .V; t/ be any Œp nonzero weight, put D r˛ C sˇ. If r ¤ 0 then .uˇ / ¤ 0. Proposition 1.3.2(3) shows that all C iˇ (i 2 Fp ) are weights. Similarly, if s ¤ 0 then all C i˛ are weights. From this one easily deduces the claim. So we have to verify the above assumption. (b) Suppose g Š W .2I 1/. Choose t WD F .1 C x1 /@1 ˚ F .1 C x2 /@2 . Define roots ˛, ˇ by ˛..1Cx1 /@1 / D 1;

˛..1Cx2 /@2 / D 0;

ˇ..1Cx1 /@1 / D 0;

ˇ..1Cx2 /@2 / D 1:

Apply (a) with u˛ WD .1 C x1 /.1 C x2 /@2 , uˇ WD .1 C x1 /.1 C x2 /@1 . (c) Suppose g Š W .1I 2/. Due to Theorem 7.6.3(1) g has basis .ui /i 2Fp2 with Œp

multiplication Œui ; uj D .j i/ui Cj . We take t WD F u0 ˚ F u0 . Set ˛ D 1, choose ˇ 2 Fp2 such that ˛; ˇ form an Fp -basis of Fp2 , and identify these with troots. Apply (a) with u˛ , uˇ . One easily computes Œp Œu1 ; uˇ

D

p Y1

.ˇ

i/uˇ D .ˇ p

ˇ/uˇ ¤ 0;

i D0 Œp Œuˇ ; u1

D

p Y1

.1

iˇ/u1 D .1

ˇp

1

/u1 ¤ 0:

i D0

(d) Suppose g D H.2I 1I ˆ.//.1/ . Theorem 10.3.2 shows that (a) applies for every 2-dimensional torus in H.2I 1I ˆ.//.1/ Œp . (e) Suppose g D H.2I 1I ˆ.1//. We combine the methods of the proof of Theorem 10.7.2 and (a). It suffices to prove the theorem under the assumption that V is girreducible and t is any 2-dimensional torus. Let us take t WD FDH;1 .x2 /Œp ˚ FDH;1 .x1 x2 / D F x2 @2 ˚ F .x2 @2

x1 @1 /:

Define roots ˛.x1 @1 / D 1;

˛.x2 @2 / D 0;

ˇ.x1 @1 / D 0;

ˇ.x2 @2 / D 1:

Since V is a restricted t-module, .V; t/ Fp ˛ ˚ Fp ˇ. Let gŒp;.k/ denote the k-th component of the natural filtration of gŒp and V0 be an irreducible gŒp;.0/ -submodule of V . Observe that V0 is t-invariant.

120

10


(i) Note that DH;1 .x1 / D @2 2 g ˇ . Suppose DH;1 .x1 /p 1 V0 D ¹0º. The commutator formula in combination with the fact that g.1/ V0 D ¹0º yields p 1

¹0º D DH;1 .x12 x2

/DH;1 .x1 /p

1

V0 D DH;1 .x12 / V0 :

Since g.0/ =g.1/ Š sl.2/ we conclude g.0/ V0 D ¹0º. Then p 1

¹0º D DH;1 .x1 x2

/DH;1 .x1 /p

1

V0 D DH;1 .x1 / V0 :

As g=g.0/ is an irreducible g.0/ =g.1/ -module, this gives g V0 D ¹0º, a contradiction. Therefore there is a weight and a weight vector v 2 V0 such that DH;1 .x1 /p 1 v ¤ 0. Then

C Fp ˇ .V; t/: p 1

(ii) Note that DH;1 .x2 / D @1 x1 x2 @2 2 g Arguing as in (a) we derive the implication ı 2 .V; t/; ı.x2 @2 / ¤ 0

)

˛

and DH;1 .x2 /Œp D x2 @2 .

ı C Fp ˛ .V; t/:

(iii) Using (i) and (ii) we obtain Fp ˛ C Fp ˇ .V; t/. Next observe that is an sl.2/-triple, for which 12 DH;1 .x22 / 2

1 1 2 2 2 DH;1 .x2 /; DH;1 .x1 x2 /; 2 DH;1 .x1 / gˇ ˛ , 21 DH;1 .x12 / 2 g˛ ˇ . Put

k WD FDH;1 .x22 / C FDH;1 .x1 x2 / C FDH;1 .x12 / Š sl.2/ and let M be a composition factor of the k-module V which carries the weight ˇ. Note that M is a restricted k-module. If DH;1 .x22 / M ˇ ¤ ¹0º then ˛ is a weight of M . Otherwise the representation theory of sl.2/ (Theorem 5.3.1) shows that .dim M /

1

ˇ.DH;1 .x1 x2 // D

ˇ.x2 @2

x 1 @1 / p

Then DH;1 .x12 /p 1 M ˇ ¤ ¹0º, and this again gives M ¹0º. So we have ˛ 2 .V; t/ in any case. (iv) Finally observe that Q WD

p X1 i D0

FDH;1 .x1i x2 /

D F .@1 C

p 1 x1 x2 @2 /

C

p X1

˛

1

DM

F .x1i @1

.mod .p//: ˇ C.p 1/.˛ ˇ /

¤

ix1i 1 x2 @2 /

i D1

Š W .1I 1/ P is contained in i 2Fp gi ˛ . Let W denote a composition factor of the Q-module V which carries the weight ˛. This is a nontrivial restricted W .1I 1/-module and all t-weights of W are contained in Fp ˛. Theorem 7.6.10 shows that either W has p 1 different weights with only the 0 weight missing or W has p weights. In the first case Fp ˛ .V; t/ and in the second case Fp ˛ .V; t/ holds. As a conclusion, Fp ˛ C Fp ˇ D .V; t/ [ ¹0º. This is the claim.

Chapter 11

1-sections

Applying general results of Volume 1 various structural details for Lie algebras of absolute toral rank 1 and Lie algebras having a CSA of toral rank 1 are collected. Algebras of this type arise in the classification as 1-sections with respect to a torus of maximal dimension or with respect to a CSA.

11.1

Lie algebras of absolute toral rank 1

Let us first extend Corollary 9.2.12 in the spirit of Theorem 10.6.4. Observe the following. Let s be a semisimple Lie algebra and sŒp the p-envelope of s Š ad s in Der s. For every d ˝ x .a/ 2 s ˝ O.mI n/ Š .ad s/ ˝ O.mI n/ Der s ˝ O.mI n/ one has p ads˝O.mIn/ .d ˝ x .a/ / D .ads d /p ˝ .x .a/ /p D ı.a; 0/.ads d /p ˝ 1: This shows that s ˝ O.mI n/ C sŒp ˝ F is the p-envelope of s ˝ O.mI n/ in Der s ˝ O.mI n/ . Theorem 11.1.1. Let g be a non-solvable Lie algebra and h a CSA of toral rank TR.h; g/ D 1 in g. Denote by gŒp a minimal p-envelope of g and by hŒp the penvelope of h in gŒp . Set t0 the maximal torus of the nilpotent restricted Lie algebra hŒp . Let W g ! g= rad g denote the canonical homomorphism. The following holds: (1) g D h C g.1/ , rad g.1/ is nilpotent. (2) s WD .g.1/ / is the unique minimal ideal of .g/. (3) g.1/ = rad g.1/ Š .g.1/ /=.rad g.1/ / 2 ¹sl.2/; W .1I n/; H.2I nI ˆ/.2/ º. (4) There is a realization s Š s0 ˝ O.mI 1/; where s0 is a simple Lie algebra and m 0. This isomorphism induces via the ads -representation a homomorphism of restricted Lie algebras W gŒp ! .Der s0 / ˝ O.mI 1/ Ì Ids0 ˝ W .mI 1/

122

11

1-sections

with ker D rad gŒp , where the projection 2 into the second summand maps .gŒp / onto a transitive subalgebra of W .mI 1/. (5) The isomorphism in (4) can be chosen in such a way that .t0 / D F t0 ˝ 1; where Cs0 .t0 / is a CSA of s0 of toral rank 1 in s0 and t0 is contained in the penvelope of Cs0 .t0 / in Der s0 . (6) If p .h/ acts nilpotently on g, then rad g.1/ D g.1/ \ rad g. Proof. (1) Let t WD ad t0 denote the maximal torus of the p-envelope of adg h in Der g. Since TR.h; g/ D 1, t is 1-dimensional. Corollary 9.2.12 shows that h \ g.1/ is a CSA of g.1/ of toral rank 1. Due to Theorem 1.2.7(1) 1 D TR.h \ g.1/ ; g.1/ / TR.h \ g.1/ ; g/ TR.h; g/ D 1: Therefore t is also the maximal torus of the p-envelope of adg .h \ g.1/ / in Der g, and this gives g D Cg .t/ C g.1/ D h C g.1/ . Set in Theorem 1.2.8(3) L D g.1/ and I D rad g.1/. One obtains TR.rad g.1/ / TR.g.1/ / TR.g.1/ = rad g.1/ / D 0. Therefore rad g.1/ is nilpotent (Theorem 1.2.7). (2) Set gN WD .g/, hN WD .h/ and gN Œp the minimal p-envelope of gN in Der gN . Corollary 1.1.8 shows that gŒp = rad gŒp is the uniquely determined minimal p-envelope of g= rad g, whence gN Œp Š gŒp = rad gŒp DW gŒp as restricted Lie algebras. Note that hN Œp coincides with the image hŒp of hŒp in gN Œp (since both are the reN Then t0 hŒp holds. stricted algebra generated by h). Since TR.h; g/ D 1, the torus t0 acts on g as a 1-dimensional torus. Suppose t0 acts trivially on g= rad g. Then g D h C rad g, whence g would be solvable. As this is not true by assumption, we now have that t0 acts on gN as a 1-dimensional torus. In N gN /=1. other words, dim t0 D 1 and TR.h; r Let ˚j D1 sj denote the sum of all minimal ideals of the semisimple Lie algebra gN . Set hNj WD Csj .t0 / D hN \ sj . If hNj would act nilpotently on sj , then sj would be solvable (by Corollary 1.3.8). As this is not true, .hNj /Œp contains a nonzero torus rj t0 . Since t0 is 1-dimensional, this gives rj D t0 , and from this we conclude .1/ gN D hN C Œt0 ; gN hN C sj . As a result, r D 1 and s1 D s1 D gN .1/ D .g.1/ /. (3) As .g.1/ /=.rad g.1/ / Š g.1/ = rad g.1/ , claim (3) follows from Corollary 9.2.12(3). (4) Due to Corollary 3.3.5 there is a simple Lie algebra s0 and m0 0 such that N s Š s0 ˝ O.m0 I k/. Note that s is not only the unique minimal ideal Pof g but also 0 of gŒp D gŒp Š gŒp = rad gŒp . Then s Š s ˝ O.mI 1/ with m WD kj and, due to Corollary 3.3.5 one has even more that under this isomorphism gŒp injects into

11.1


123

.Der s0 / ˝ O.mI 1/ Ì Ids0 ˝ W .mI 1/ such that the projection 2 W .Der s0 / ˝ O.mI 1/ Ì Ids0 ˝ W .mI 1/ ! W .mI 1/ maps gŒp onto a transitive subalgebra. This is assertion (4). (5) In the course of (2) we have shown that t0 is 1-dimensional and t0 .hN \ s/Œp . Set t0 DW F t with toral element t . Theorem 3.6.1 shows that, up to conjugation by an automorphism of s0 ˝ O.mI 1/, t is mapped onto Id ˝ .1 C x1 /@1 or !.t/ ˝ 1 C Id ˝ .t/; P where !.t / is a toral element in Der s0 and .t/ 2 m i D1 F xi @i also is toral. Recall N \ s/Œp . Then the image of t0 in Der s0 ˝ O.mI 1/ D .Der s0 / ˝ that t0 . h O.mI 1/ Ì Ids0 ˝ W .mI 1/ is contained in the p-envelope of s Š s0 ˝ O.mI 1/. The remark preceding this theorem states that this p-envelope is s0 ˝ O.mI 1/ C s0Œp ˝ F . We conclude that the first case is impossible and in the second case .t/ D 0 holds. Set t0 WD !.t /. We obtain hN \ s Š Cs0 .t0 / ˝ O.mI 1/ and t0 ˝ 1 2 Cs0 .t0 / ˝ O.mI 1/ C Cs0 .t0 / Œp ˝ F \ .Der s0 ˝ F / D Cs0 .t0 / Œp ˝ F: Consequently, t0 is contained in the p-envelope of Cs0 .t0 /. We also have Cs0 .t0 / ˝ F D Cs0 ˝F .t0 ˝ 1/ .h/. Therefore Cs0 .t0 / is nilpotent and self-normalizing, whence a CSA of s0 . Since this CSA acts non-nilpotently on s0 one has (see Theorem 1.2.7(1)) 0 ¤ TR.Cs0 .t0 /; s0 / TR.hN \ s; s/ TR.h; g/ D 1: This proves (5). (6) Theorem 10.6.4(3) applies for gN . Case (b) does not occur, since t0 2 Cs .t0 / and Case (c) is excluded by assumption. Hence s is simple, and this means that s is the unique minimal ideal not only of gN but also of gN .1/ . Then rad .g.1/ / D ¹0º. Therefore rad g.1/ D g.1/ \ .ker / D g.1/ \ .rad g/. Next we partly reformulate the preceding theorem for the special case when g in addition has absolute toral rank 1. Theorem 11.1.2. Let g be a non-solvable Lie algebra of absolute toral rank 1, let t gŒp be a torus of toral rank 1 in gŒp and h WD Cg .t/. The following holds: (1) h is a CSA of g of toral rank 1 in g. (2) g D h C g.1/ . (3) g.1/ = rad g.1/ 2 ¹sl.2/; W .1I 1/; H.2I 1/.2/ º. (4) rad g.1/ is nilpotent and t-invariant.

124

11

1-sections

If p .h/ acts nilpotently on g, then rad g.1/ is an ideal of g and one of the following cases occurs: g= rad g 2 ¹sl.2/; W .1I 1/º

or H.2I 1/.2/ g= rad g H.2I 1/:

Proof. (a) By Remark 1.2.3 we may assume that t is a torus of maximal dimension in gŒp (gŒp is assumed to be any finite dimensional p-envelope of g). Note that adg t is 1-dimensional. If h acts nilpotently on g, then g is solvable (Corollary 1.3.8). This is not true. Let hŒp denote the p-envelope of h in gŒp and t0 be its maximal torus. Since t is maximal, t0 t. The preceding deliberations show dim adg t0 D dim adg t D 1; whence t D t0 C Ct .g/. Then h is selfnormalizing, and this shows that h is a CSA of toral rank dim adg t0 D 1 in g. (b) For a proof of (2), (4) see Theorem 11.1.1(1). Note that .g.1/ / contains 0 s ˝ F . Therefore .rad g.1/ / is invariant under s0 ˝ F . To show that rad g.1/ is t-invariant, we now observe that adg t D adg t0 and apply Theorem 11.1.1(5). (c) Observe that TR.g.1/ = rad g.1/ / TR.g/ D 1 (Theorem 1.2.8(3)). Theorem 11.1.1(3) in combination with Theorem 10.6.1 yields the proof of claim (3). (d) Finally suppose that p .h/ acts nilpotently on g. Theorem 11.1.1(6) shows that rad g.1/ D g.1/ \rad g holds. Therefore g acts on g.1/ = rad g.1/ by derivations. Set K WD ¹D 2 g j ŒD; g.1/ rad g.1/ º. The formal definition gives ŒK .1/ ; K .1/ rad g.1/ , whence K is solvable. As K is an ideal of g and rad g K, this gives K D rad g. Consequently, g= rad g injects into Der.g.1/ = rad g.1/ /. All derivations of sl.2/ and W .1I 1/ are inner. This is the claim for these cases. Suppose g.1/ = rad g.1/ Š H.2I 1/.2/ . Let gN Œp denote the p-envelope of g= rad g in Der.g.1/ = rad g.1/ / Š Der H.2I 1/.2/ D CH.2I 1/ (Theorem 7.1.2). Since gN Œp has absolute toral rank 1 and is centerless, every torus is at most 1-dimensional. Recall that H.2I 1/ is a restricted subalgebra of Der H.2I 1/.2/ (this follows immediately from Lemma 7.2.1(3)). Then it can only be that gN Œp =NgŒp \ H.2I 1/ is Œp-nilpotent (Lemma 1.2.6(2a)). But CH.2I 1/=H.2I 1/ is a torus. Then g= rad g gN Œp H.2I 1/. This proves the final statement. Next we specialize Theorem 11.1.1 for the case that g has a 1-dimensional CSA h of toral rank 1 in g. The following result is due to J. B. Ermolaev [Erm 72] in general and (much earlier) due to I. Kaplansky [Kap 58] for simple algebras. Theorem 11.1.3. Let g be a non-solvable Lie algebra having a 1-dimensional Cartan subalgebra h of toral rank TR.h; g/ D 1. The following holds: (1) g= rad g is isomorphic to sl.2/ or W .1I 1/. (2) g.1/ D g. (3) rad g is abelian.

11.1


125

Proof. (A) In a first step we assume that g is simple. We want to prove that g is isomorphic to sl.2/ or W .1I 1/. In order to do so we assume that the theorem holds for simple Lie algebras of smaller dimension. Choose a maximal subalgebra g.0/ containing h DW F h and a g.0/ -invariant subspace g. 1/ g.0/ such that g. 1/ =g.0/ is g.0/ -irreducible. Consider the standard filtration defined by g.0/ and g. 1/ and the associated graded algebra G WD ˚s r gri g. Note that G satisfies properties (g1)–(g3) of Notation 3.5.2 (by Proposition 3.5.3(1)). The image hN D F hN of h in G0 is a 1-dimensional CSA of G. Let I be any ideal of g.0/ . Lemma 9.2.1(1) states that either I acts nilpotently on g or h I (as h is 1-dimensional). In the latter case g.0/ D I holds. Turning this into information on G (observing that in the first case the image of I in G0 is ¹0º due to (g3)) gives that G0 has no proper ideals. (a) Suppose G0 is 1-dimensional. Then g.0/ is solvable and acts trigonalizably on g. The maximality of g.0/ implies that g.0/ has codimension 1 in g, whence r D 1 and dim Gi D 1 for all nonzero Gi . Adjust hN 2 hN so that it acts as . 1/Id on G 1 . Then hN acts as iId on every Gi (Lemma 3.5.4(2)). Since the CSA in question is 1-dimensional, this gives Gp D ¹0º. As a result, s < p. Then g is at most .p C 1/dimensional. On the other hand, Theorem 11.1.1(3) provides a list of possible simple Lie algebras for g. Comparing dimensions we obtain g Š sl.2/ or g Š W .1I 1/. (b) Suppose G0 is simple of dimension > 1. By induction hypothesis we have G0 Š sl.2/ or G0 Š W .1I 1/. N (b1) Suppose G0 Š sl.2/. Note that the only h-weights on Gi (i ¤ 0) are conN tained in Fp ˛ n ¹0º for some ˛ 2 .h; G/, since hN is contained in G0 and has toral rank 1 in G. Therefore every G0 -composition factor of Gi is a restricted sl.2/-module of dimension less than p (Theorem 5.3.1). The multiplication on g induces an sl.2/invariant pairing G 1 G 1 ! G 2: As G 1 is an irreducible sl.2/-module, Proposition 5.3.6 shows that G 2 is generated as an sl.2/-module by the 0 weight space of G 2 . However, this 0 weight space is ¹0º. Therefore G 2 D ¹0º. Similarly, let G2 ¤ ¹0º and V2 G2 be an irreducible G0 -module. Since 0 is not a weight on V2 , dim V2 is even and less than p. The same is true for dim G 1 , as this is an irreducible G0 -module by definition. Arguing as above one obtains ŒG 1 ; V2 D ¹0º. But this contradicts property (g1) of Notation 3.5.2. Therefore G2 D ¹0º. Next suppose G1 ¤ ¹0º, and let V1 be an irreducible G0 -submodule of G1 . It is easy to see that G 1 ˚ G0 ˚ V1 is a simple Lie algebra, clearly of dimension at most 2.p 1/ C 3. Again, Theorem 11.1.1(3) yields G 1 ˚ G0 ˚ V1 Š W .1I 1/. However, the only non-trivial grading of W .1I 1/ has the property that the degree 0 subalgebra is 1-dimensional (see Theorem 7.4.1). So this case is impossible. If G1 D ¹0º, then we obtain dim g .p 1/ C 3. Theorem 11.1.1(3) yields that g is of the required form.

126

11

1-sections

(b2) Suppose G0 Š W .1I 1/. One proceeds as in the former case. Namely, there is an element hN 2 hN for which all eigenvalues on every Gi (with i ¤ 0) are contained in Fp n ¹0º, since hN has toral rank 1 in G and hN G0 . Therefore every G0 -composition factor of Gi is a restricted W .1I 1/-module of dimension less than p. This gives that every such factor is isomorphic to the .p 1/-dimensional module O.1I 1/=F (Theorem 7.6.10). Note that hN acts as a toral element in G0 . Then it is conjugate to F .1 C x/@ or to F x@ (Theorem 7.5.1). In either case its image fits into F @ C F x@ C F x 2 @ Š sl.2/. Observe that O.1I 1/=F is irreducible even as a module for this sl.2/. Now argue as in the former case. As a result, we obtain that G 2 D G2 D ¹0º, dim G 1 D p 1 and G1 D ¹0º. Then dim g D 2p 1. As in the former cases Theorem 11.1.1(3) yields g Š W .1I 1/. But then the dimensions do not fit. (B) We now consider the general case. Let W g ! g= rad g denote the canonical homomorphism. Theorem 11.1.1(2),(3),(6) state that .g.1/ / is the unique minimal ideal of .g/ and is a simple Lie algebra. Set in Lemma 9.2.1(1) K D K0 D g and I D g.1/ . Since g is not solvable, the lemma gives that h \ g.1/ is a CSA of g.1/ . Then h g.1/ because h is 1-dimensional. Set in Lemma 9.2.1 K D K0 D g and I D rad g. If h I , then g D h C I .1/ would be solvable, a contradiction. Hence h \ rad g D ¹0º because dim h D 1. As a consequence, .h/ is a CSA of .g.1/ /. Therefore we can apply the result of (A) to .g.1/ / and obtain that this algebra is isomorphic to sl.2/ or W .1I 1/. In both cases all derivations are inner. Consequently, .g/ Š .g.1/ / holds. This proves the first statement of the theorem. (C) Observe that 0 is not an h-weight on rad g as h \ rad g D ¹0º. Therefore rad g g.1/ and there is a homomorphism from g= rad g onto g=g.1/ . However, the first algebra is simple by assertion (1) and therefore the image of g= rad g vanishes. This gives g g.1/ C rad g D g.1/ . (D) Let rad g D ˚i 2Fp .rad g/i˛ be the h-weight space decomposition of rad g. Recall that .rad g/0 h \ .rad g/ D ¹0º. The Engel–Jacobson theorem applied to [i 2Fp .rad g/i˛ shows that rad g is nilpotent. Then the multiplication on g induces a surjective .g= rad g/-invariant pairing .rad g/= 2 .rad g/ .rad g/= 2 .rad g/ ! 2 .rad g/= 3 .rad g/: If g= rad g Š sl.2/, then we argue as follows. Since 0 is not a weight on any .g= rad g/composition factor of .rad g/= 2 .rad g/, Theorem 5.3.3(2) yields that .rad g/= 2 .rad g/ is a completely reducible module, while Theorem 5.3.1(2) implies that the dimensions of the irreducible summands are even. Proposition 5.3.6 shows that the image under this pairing is generated as an sl.2/-module by the 0 weight space in

2 .rad g/= 3 .rad g/. However, this 0 weight space is ¹0º.

11.2

127

1-sections

Next suppose g= rad g Š W .1I 1/. Since hN is a 1-dimensional torus, it is conjugate to F .1 C x/@ or to F x@ (Theorem 7.5.1). In either case the image of hN fits into F @ C F x@ C F x 2 @ Š sl.2/. Now argue as in the former case. As a result, 2 .rad g/ 3 .rad g/. Now recall that rad g is nilpotent. Then the former inclusion implies 2 .rad g/ D ¹0º.

11.2

1-sections

In this section we investigate 1-sections in Lie algebras with respect to a (not necessarily maximal) torus. This covers the case of 1-sections with respect to a CSA as well. More exactly, we will consider the following Setting 11.2: (1) Let g be a finite dimensional Lie algebra and gŒp denote a minimal p-envelope. (2) Let t gŒp be a not necessarily maximal torus. (3) Set h WD Cg .t/ and assume that h is nilpotent. (4) Assume that the maximal torus t0 of hŒp gŒp is contained in t. There is a bijection of h-roots and t0 -roots. Recall from §1.3 that every t0 -root ˛ may be regarded an hŒp -root. Namely, for x 2 hŒp there is a Œp-power contained in e e e t0 , x Œp 2 t0 . One puts ˛.x/ WD ˛.x Œp /1=p . Observe that Setting 11.2 also holds if we substitute t by t0 . Hence we treat simultaneously both the cases that t is a torus of maximal toral rank and that t D t0 is the torus of a CSA. For ˛ 2 .g; t/ let X g.˛/ WD gi˛ i 2Fp

denote the respective 1-section. We also set gŒ˛ WD g.˛/= rad g.˛/ the semisimple quotient of such a 1-section. Proposition 11.2.1. Suppose the assumptions of the Setting 11.2 hold. Let ˛ 2 .g; t/ and assume that g.˛/ is non-solvable. Then h is a CSA of g.˛/ of toral rank 1 in g.˛/ and there is h 2 [i 2Fp Œg i˛ ; gi˛ h \ g.˛/.1/ such that t D e F hŒp C t \ ker ˛ for sufficiently large e. P P Proof. The Lie algebra M WD i 2Fp gi˛ C i 2Fp Œg i˛ ; gi˛ carries an Fp -grading defined by setting X Mi WD gi˛ if i 2 Fp and M0 WD Œg i˛ ; gi˛ : i 2Fp

128

11

1-sections

Suppose that every element of [i 2Fp Œg i˛ ; gi˛ acts nilpotently on M . The Engel– Jacobson theorem implies that M0 acts nilpotently on M . Then M would be solvable (Proposition 1.3.7) and g.˛/ D h C M would be solvable, a contradiction. Hence there is i 2 Fp and h 2 Œg i˛ ; gi˛ which acts non-nilpotently. Therefore a suitable e e e power hŒp is contained in t0 and satisfies ˛.hŒp / ¤ 0. Then Cg.˛/ .hŒp / D h and therefore h is selfnormalizing, i.e., it is a CSA of g.˛/. Note that 0 ¤ dim t0 =t0 \ ker ˛ dim t=t \ ker ˛ D 1; e

and this gives TR.h; g.˛// D 1 as well as t D F hŒp C t \ ker ˛.

Let ˛ 2 .g; t/ or ˛ 2 .g; h/ be a t-root (h-root, resp.). Suppose gŒ˛ ¤ ¹0º. Proposition 11.2.1 shows that h is a CSA of g.˛/ of toral rank 1 and that there is e h 2 h \ g.˛/.1/ such that t D F hŒp C t \ ker ˛ for sufficiently large e. Then t D t0 C t \ ker ˛. Theorem 11.1.1 applies to g.˛/ and proves the following: gŒ˛ has the unique minimal ideal s WD gŒ˛.1/ and s Š s0 ˝ O.mI 1/ holds, where s0 is a simple Lie algebra having a CSA of toral rank 1. Since rad s Š s0 ˝ O.mI 1/.1/ , one has s0 Š s= rad s D .gŒ˛.1/ /=.rad gŒ˛.1/ / 2 ¹sl.2/; W .1I n/; H.2I nI ˆ/.2/ º: Moreover, the isomorphism can be taken in such a way that .t/ D .t0 / D F t0 ˝ 1. The following definition is now justified. Definition 11.2.2. Suppose the assumptions of the Setting 11.2 hold, and let ˛ 2 .g; t/ or ˛ 2 .g; h/ be a t-root (h-root, resp.). Set WD .g; t/ and WD .g; h/ in the respective cases. (1) A root ˛ 2 is called solvable classical Witt Hamiltonian

if if if if

gŒ˛ D ¹0º, .gŒ˛.1/ /=.rad gŒ˛.1/ / Š sl.2/, .gŒ˛.1/ /=.rad gŒ˛.1/ / Š W .1I n/, .gŒ˛.1/ /=.rad gŒ˛.1/ / Š H.2I nI ˆ/.2/ .

(2) A root ˛ 2 is called proper if it is solvable or classical or t (h, resp.) stabilizes the unique subalgebra of .gŒ˛.1/ /=.rad gŒ˛.1/ / of maximal dimension, otherwise improper. (3) The torus t is called a proper torus, if all t-roots are proper. We now specialize the setting to tori of maximal toral rank. Then the assumptions of Setting 11.2 are satisfied. Theorem 11.2.3. Let L be a finite dimensional Lie algebra and T LŒp a torus of maximal toral rank, and ˛ 2 .L; T / a nonzero root.

11.2

129

1-sections

(1) If L.˛/ is solvable, then X

N.˛/ WD

Li˛ C

i 2Fp

X

ŒLi˛ ; L

i˛

i 2Fp

is a nilpotent ideal of LŒp .˛/. (2) If L.˛/ is non-solvable, then rad L.˛/ and rad LŒp .˛/ are T -invariant nilpotent ideals of L.˛/ and LŒp .˛/, respectively. Proof. Since T has maximal toral rank, every root on T may be considered a root on HQ WD CLŒp .T / (§1.3). (1) The solvability of ˛ implies that ˛.h/ D 0 whenever [ [ j h2 Li˛ Œp [ ŒLi˛ ; L i˛ : i 2Fp ; j 2N

i 2Fp

P Every such h then acts nilpotently on i 2Fp Li˛ , and then it acts nilpotently on P P i 2Fp Li ˛ C i 2Fp ŒLi˛ ; L i˛ as well. But then every x2

[

Li˛ [

i 2Fp

[

ŒLi˛ ; L

i˛

i 2Fp

also acts nilpotently. The Engel–Jacobson theorem shows that X X Li˛ C ŒLi ˛ ; L i˛ i 2Fp

i2Fp

is nilpotent. This space obviously is an ideal of HQ C L.˛/ D LŒp .˛/. (2) Proposition 11.2.1 shows that TR.H; L.˛// D 1, if ˛ is non-solvable. Let T0 denote the maximal torus of the p-envelope HŒp LŒp of H . Proposition 11.2.1 shows that T D T0 C T \ ker ˛. Consequently, rad L.˛/ is T -invariant. Moreover, we have TR.L.˛/= rad L.˛// ¤ 0 and hence TR.rad L.˛// TR.L.˛// TR.L.˛/= rad L.˛// D 0 (Theorem 1.2.8(3)). This shows that rad L.˛/ is nilpotent. Observe that LŒp also satisfies the assumptions of the theorem. Substituting L by LŒp gives the respective statement for rad LŒp .˛/. Corollary 11.2.4. Let L be a finite dimensional Lie algebra and T LŒp be a torus of maximal toral rank. Suppose H WD CL .T / acts trigonalizably on L. Let ˛ 2 .L; T /. One of the following occurs: (1) LŒ˛ D ¹0º and L.˛/.1/ is nilpotent. (2) LŒ˛ Š sl.2/ and rad L.˛/ is nilpotent. (3) LŒ˛ Š W .1I 1/ and rad L.˛/ is nilpotent.

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(4) H.2I 1/.2/ LŒ˛ H.2I 1/ and rad L.˛/ is nilpotent. In each of these cases rad L.˛/ is T -invariant and LŒ˛ admits a unique Œp-structure. Proof. (a) If ˛ is solvable, then P LŒ˛ D ¹0º byP definition. We have already shown (Theorem 11.2.3) that N.˛/ D i 2Fp Li˛ C i 2Fp ŒLi˛ ; L i˛ is a T -invariant nilpotent ideal of L.˛/. The present assumption gives that H .1/ also acts nilpotently on L.˛/. Then the Engel–Jacobson theorem shows that H .1/ C N.˛/ D L.˛/.1/ is nilpotent. (b) Suppose L.˛/ is non-solvable. Theorem 11.2.3 shows that rad L.˛/ is T invariant and nilpotent. (c) Note that TR.LŒ˛/ 1 (Theorem 1.3.11(3)). Therefore Theorem 11.1.2 shows that LŒ˛ is of the stated form if ˛ is classical, Witt or Hamiltonian. These algebras carry a unique structure as restricted Lie algebras, since they are centerless. Solvable and classical roots are always proper by definition. Let ˛ in the preceding corollary be Witt or Hamiltonian. Then LŒ˛ has the unique minimal ideal LŒ˛.1/ and this minimal ideal is a simple algebra. By definition the question of properness can now be checked in LŒ˛. Namely, we have to decide if T stabilizes W .1I 1/.0/ or LŒ˛ \ H.2I 1/.0/ in the respective cases. Theorem 11.2.5. Let L be a finite dimensional Lie algebra and T LŒp a torus of maximal toral rank. Suppose H WD CL .T / acts trigonalizably on L. (1) Every solvable or classical root is proper. (2) A Witt root is proper if and only if T acts on W .1I 1/ as a 1-dimensional torus conjugate under an automorphism of W .1I 1/ to F x@, and improper if and only if T acts on W .1I 1/ as a 1-dimensional torus conjugate to F .1 C x/@. (3) A Hamiltonian root ˛ is proper if and only if T acts on H.2I 1/.2/ as a 1-dimensional torus conjugate under an automorphism of H.2I 1/.2/ to FDH .x1 x2 /, and improper if and only if T acts on H.2I 1/.2/ as a 1-dimensional torus conjugate to FDH ..1 C x1 /x2 /. Proof. Solvable and classical roots are proper by definition. Let ˛ be a Witt root. Then T acts on W .1I 1/ as a 1-dimensional torus, and as such it is conjugate to F x@ or to F .1 C x/@ (Theorem 7.5.1). By definition ˛ is proper if and only if T stabilizes W .1I 1/.0/ . This is true if and only if T is conjugate to F x@. Let ˛ be a Hamiltonian root. Then T acts on LŒ˛ \ H.2I 1/ as a 1-dimensional torus. More exactly, Proposition 11.2.1 yields the existence of h 2 L.˛/.1/ for e which F .adL.˛/ h/p D adL.˛/ T . Then there is t 2 LŒ˛.1/ Š H.2I 1/.2/ such that F adLŒ˛ t D adL˛ T . Theorem 7.5.8 shows that F t it is conjugate to FDH .x1 x2 / D F .x2 @2 x1 @1 / or to FDH ..1 C x1 /x2 / D F .x2 @2 .1 C x1 /@1 /. By definition ˛ is proper if and only if T stabilizes LŒ˛ \ H.2I 1/.0/ . This is true if and only if F t is conjugate to FDH .x1 x2 /.

11.3

Representations of dimension < p 2

131

For further reference we derive a result which allows to detect CSAs of toral rank 1. Proposition 11.2.6. Let L be finite dimensional and simple of absolute toral rank 2, let LŒp be the semisimple p-envelope of L in Der L and T LŒp a 2-dimensional torus. Every nilpotent 1-section L.˛/ with respect to T is a trigonalizable CSA of L of toral rank 1. Proof. We may regard ˛ a root on HQ WD CLŒp .T /. Let L.˛/Œp denote the penvelope of L.˛/ in LŒp . Recall that j

L.˛/Œp D L.˛/ C span ¹x Œp j x 2 [i 2Fp Li˛ ; j > 0º (Volume 1, p. 19). Therefore j

T \ L.˛/Œp CL.˛/Œp .T / H C span ¹x Œp j x 2 [i 2Fp Li˛ ; j > 0º: r

The nilpotency of L.˛/ yields that H acts nilpotently on L.˛/. Hence every x Œp e (x 2 H , r 0) acts nilpotently on L.˛/. Let x 2 [i 2Fp Li˛ and xs D x Œp be its semisimple part. As T has maximal toral rank, one has xs 2 T C C.LŒp /. But then ˛.xs / D 0 as 0 D Œxs ; x D ˛.xs /x. Observe that HQ .1/ H . Remark 1.3.3 (with j N D HQ ) implies that ˛ is linear on HQ H C span ¹x Œp j x 2 [i 2Fp Li˛ ; j > 0º. Consequently, T \ L.˛/Œp T \ ker ˛. Theorem 1.2.9 yields (as dim T D 2) TR.L.˛/; L/ TR.L/

dim T =T \ L.˛/Œp D dim T \ L.˛/Œp

dim T \ ker ˛ 1: Suppose L.˛/ acts nilpotently on L. Choose a root ˇ which is Fp -independent of ˛ and observe that .L; T / Fp ˛ P C Fp ˇ (as L has absolute toral rank 2). Impose an Fp -grading on L by setting Lj WD i 2Fp Li˛Cjˇ . By assumption, L0 D L.˛/ acts nilpotently on L. Proposition 1.3.7 implies that L would be solvable, a contradiction. Hence L.˛/ acts non-nilpotently on L. Then TR.L.˛/; L/ ¤ 0. The above gives TR.L.˛/; L/ D 1 and T \ L.˛/Œp D T \ ker ˛. Choose a root vector x 2 Li˛ for r which ˇ.x Œp / ¤ 0. Then x acts invertibly on L=L.˛/. This shows that L.˛/ is selfnormalizing, whence a CSA of L. Since it has toral rank 1 in L it acts trigonalizably on L (Theorem 9.2.11).

11.3

Representations of dimension < p2

Let L be a simple Lie algebra of absolute toral rank 2 and T a 2-dimensional torus in the semisimple p-envelope P of L. For independent T -roots ˛ and ˇ the 1-section M WD L.˛/ acts on V WD i 2Fp Lˇ Ci˛ . In a rather natural setting H WD CL .T / acts trigonalizably on V and dim V < p 2 holds. This will become the situation when the following results are to apply in the Classification Theory.

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11

1-sections

Theorem 11.3.1. Let H be a CSA of an arbitrary Lie algebra M . Assume that TR.H; M / D 1, and that M has a faithful irreducible module V of dimension < p 2 , such that H acts trigonalizably on V . One of the following holds: (1) dim M D 1; (2) rad M D C.M / and .M=C.M //.1/ is simple; (3) M Š H1 Ì sl.2/ is the 6-dimensional split extension of sl.2/ by the Heisenberg algebra H1 ; (4) M has an abelian ideal I 6 C.M /. Proof. (a) Observe that H acts trigonalizably on M as it does so on V and V is a faithful module. If M is abelian, then it is 1-dimensional (since it has a faithful irreducible finite dimensional module). If M is solvable but not abelian, then it has an abelian ideal I 6 C.M / ([S-F, Corollary 5.8.2]). We now assume that M is non-solvable and C.M / is the only maximal abelian ideal. Suppose rad M D C.M /. Theorem 11.1.1(6) shows that rad M .1/ D M .1/ \ rad M . Therefore in the notation of Theorem 11.1.1 .rad M .1/ / D ¹0º and part (3) of that theorem then states that .M=C.M //.1/ D .M /.1/ D .M .1/ / is simple. This is Case (2) of the theorem. If rad M ¤ ¹0º and C.M / D ¹0º, then M does have a non-central abelian ideal. (b) We now may assume rad M © C.M / ¤ ¹0º. Let I be an ideal of M maximal with the property I .1/ C.M /. Due to our assumption I © C.M /. Since V is an irreducible module, C.M / Š F is 1-dimensional. Consider the bilinear pairing I I ! C.M / Š F;

.x; y/ 7! Œx; y:

Note that J WD ¹x 2 I j Œx; I D ¹0ºº is an abelian ideal of M . Due to our assumption, J D C.M /. Therefore the induced pairing I =C.M / I =C.M / ! C.M / Š F;

.x C C.M /; y C C.M // 7! Œx; y

is nondegenerate. As a consequence, I Š Hm is a Heisenberg algebra of dimension 2m C 1 for some m > 0. Moreover, C.M / acts invertibly on V . The representation theory of the Heisenberg algebra (cf. Volume 1, p. 149) gives dim V p m , whence m D 1. (c) The multiplication of M induces an action of M on I =C.M /. The kernel of this action is K WD ¹x 2 M j Œx; I C.M /º: The formerly mentioned pairing extends to a pairing K=C.M / I =C.M / ! C.M /. Since the restriction I =C.M / I =C.M / ! C.M / Š F is nondegenerate, there is a decomposition K D Q C I , where Q WD ¹x 2 M j Œx; I D ¹0ºº;

Q \ I D C.M /:

11.3


133

Clearly, Q is an ideal of M . Suppose Q is not solvable. Lemma 9.2.1 (with K D K0 D M and Q instead of I ) implies that M D H C Q.1/ and H \ Q is a CSA of Q. Let x 2 I be arbitrary and decompose x D h C q with h 2 H , q 2 Q.1/ . Then for every t 2 H \ Q 0 D Œt; x D Œt; h C Œt; q 2 H C Œt; q: Therefore ŒH \ Q; q H , and this implies q 2 H \ Q as H \ Q is a CSA of Q. As a consequence, I H . But H acts trigonalizably on V , a contradiction. Therefore Q is solvable. If C.M / ¨ Q, then there would be an ideal Q0 ¤ C.M / of M in Q, which satisfies Q0.1/ C.M /. The ideal I C Q0 then would be bigger than I and its derived algebra is contained in C.M /. This contradicts the maximality of I . As a consequence, Q D C.M /, whence K D I . (d) As a result, M=I injects into gl.I =C.M // Š gl.2/. If M=I Š gl.2/, then the image of H in gl.2/ is a CSA, necessarily of dimension 2. However, since C.M / I and TR.H; M / D 1, the image of H can only contain a 1-dimensional torus. Therefore the image of M=I cannot be the full gl.2/. As M is assumed to be non-solvable, M=I Š sl.2/. Choose elements e; h; f 2 M , such that the images in M=I constitute an sl.2/-triple. Since I =C.M / is a faithful 2-dimensional .M=I /-module, the eigenvalues of .ad h/jI are ˙1 each of multiplicity 1. The eigenvalues of .ad h/jM=I are ˙2; 0 again of multiplicity 1. Consequently, the eigenvalues of ad h are ˙1; ˙2; 0 and all .ad h/-weight spaces of M for nonzero eigenvalues are 1-dimensional. Choose e 0 ; f 0 2 M such that Œh; e 0 D 2e 0 , Œh; f 0 D 2f 0 , set h0 WD Œe 0 ; f 0 2 F h C C.M /. Then F e 0 C F h0 C Ff 0 Š sl.2/, and the extension splits. The simple algebras occurring in Case (2) of the preceding theorem are ruled by Theorem 11.1.1(3). Not all the algebras listed in Theorem 11.1.1(3) have a faithful irreducible representation of dimension < p 2 . However, it is not really important for our purposes to determine these exactly. In the context of our applications there will occur additional conditions which will limit the list of possible algebras even more. We will later on need information for the particular case that M D W .1I 1/. Theorem 11.3.2. Let V be a faithful irreducible W .1I 1/-module of dimension less than p 2 . Then all x i @ (i > 1) act nilpotently on V and V is one of the following: (a) O.1I 1/=F ; p 1

(b) ˚i D0 F @i u, where F u is a 1-dimensional W .1I 1/.0/ -submodule and @p u 2 F u. Proof. (a) Set Ei WD x i @. Note that W .1I 1/.1/ acts nilpotently on W .1I 1/. The irreducibility of V implies that there is an eigenvalue function W W .1I 1/.1/ ! F such that E .E/IdV is nilpotent for every E 2 W .1I 1/.1/ . Suppose there is i0 with 1 i0 p 2 such that .Ei0 / ¤ 0. Choose i0 maximal with this property.

134

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1-sections

If i0 p 3, then apply Proposition 3.2.11 with M WD W .1I 1/.0/ , I WD W .1I 1/.2/ and e WD E 1 , f WD Ei0 C1 . This shows that there is an irreducible M -submodule V0 such that dim V p dim V0 . Next apply Proposition 3.2.11 with G WD M D W .1I 1/.0/ , M 0 D I 0 WD W .1I 1/.1/ and e WD E0 , f WD Ei0 . Then dim V0 p. Therefore this case cannot happen. So assume i0 D p 2. Apply Proposition 3.2.11 with G WD W .1I 1/.0/ , M D I WD W .1I 1/..p 1/=2/ W .1I 1/.p 3/ and ej WD Ej , fj WD Ep 2 j for j D 0; 1. Then dim V p 2 . This proves the first statement. (b) As in (a) there are a; b 2 F such that @p D aId and .x@/p x@ D bId. As a consequence of (a) W .1I 1/.0/ acts trigonalizably on V . Choose a 1-dimensional Pp 1 i submodule F u. Then V D i D0 F @ u. If a D b D 0, then the module is a restricted one. Apply Theorem 7.6.10. Otherwise there are p distinct x@-eigenvalues p 1 on V , and therefore dim V p. Then V D ˚i D0 F @i u. Case (4) of Theorem 11.3.1 needs further investigation. Proposition 11.3.3. Let H be a CSA of an arbitrary Lie algebra M . Assume that TR.H; M / D 1, and that M has a faithful irreducible module V of dimension < p 2 , such that H acts trigonalizably on V . Suppose M contains an abelian ideal I 6 C.M /. Regard M as a subalgebra of gl.V / and let MŒp denote the p-envelope of M in gl.V /. Then there is a homomorphism W MŒp ! W .1I 1/ such that .MŒp / 6 W .1I 1/.0/ and J WD ker is abelian. Moreover, I J and Œ 1 .W .1I 1/.0/ /; J consists of nilpotent transformations. Proof. (a) Since V is M -irreducible and I is abelian, there is a linear eigenvalue function W I ! F such that every x .x/Id is a nilpotent endomorphism of V (Lemma 3.2.7). Let V0 WD ¹v 2 V j x.v/ D .x/v 8x 2 I º denote the weight space. Set B the associative subalgebra of End V generated by I and K WD ¹x 2 MŒp j .Œx; I / D 0º. By definition of MŒp , V is a restricted module. Corollary 3.2.5 implies that V Š u.MŒp / ˝u.K/ V0 is induced by a subrepresentation of the restricted subalgebra K of MŒp . If K D MŒp , then ŒM; I is an ideal of M acting nilpotently on V . Engel’s theorem yields ŒM; I D ¹0º, whence I D C.M /. But this is not true, and therefore K ¤ MŒp . Note that p 2 > dim V D p dim MŒp =K dim V0 . This gives dim MŒp =K D 1;

b

b b

dim V0 < p:

b b

We now apply Theorem 2.3.2 in the following way. Observe that the universal penvelope MŒp of MŒp satisfies MŒp D MŒp C C.MŒp / and that K 0 WD K C C.MŒp / is a restricted subalgebra of MŒp satisfying dim MŒp =K 0 D dim MŒp =K D 1. The flag E.K 0 / is trivial, because every element x 2 MŒp satisfies x p x Œp 2 K 0 . Theorem 2.3.2 gives rise to a special homomorphism W MŒp ! W .1I 1/

b

11.3


135

with respect to K 0 and Theorem 2.3.4(3) gives that this homomorphism is transitive (Definition 2.3.1). Note that K D K 0 \ MŒp . Then Theorem 2.3.4(1) shows that KD

1

.W .1I 1/.0/ /:

1 .W .1I 1/ / D K. Theorem 2.3.4(1) shows that (b) Recall that J D ker .0/ I J (as I K). Suppose J is not solvable. Lemma 9.2.1 (with K D K0 D M and J for I ) shows that M D H C J .1/ , and H \ J is Pa CSA of J . Since I and J are ideals, there is a decomposition I D .I \ H / ˚ ˛¤0 I \ J˛ where ˛ runs through all nonzero .H \ J /-roots. Then I˛ ŒI; H \ J ker . Since H .1/ acts nilpotently on V , this yields (as is linear) X I \ J˛ C ŒI; K ker ; ŒI; M ŒI; H C ŒI; J .1/ I \ H .1/ C ˛¤0

hence M K and then MŒp K, a contradiction. Therefore J is solvable. Since both .K/ W .1I 1/.0/ and ker D J are solvable, so is K. Recall that V0 is a K-submodule of dimension < p. As every irreducible module for a solvable Lie algebra has dimension a p-power, there is a 1-dimensional K-submodule. This is annihilated by K .1/ . We now have that J .1/ K .1/ is an ideal of M , which has a nonzero annihilator V1 WD ¹v 2 V j x.v/ D 0 8x 2 J .1/ º. But V1 is an M -submodule of V . The M -irreducibility of V gives V1 D V , which means J .1/ D ¹0º. Next we recall that the Fitting-0-component V2 of ŒK; J K .1/ on V is nonzero (because even more the 0 weight space is nonzero). Since ŒK; J J acts nilpotently on M , V2 is M -invariant. Then it is all of V and ŒK; J acts nilpotently on V . Theorem 11.3.4. Let H be a CSA of an arbitrary Lie algebra M . Assume that TR.H; M / D 1, and that M has a faithful irreducible module V of dimension < p 2 , such that H acts trigonalizably on V . Suppose M contains an abelian ideal I 6 C.M /. Then M is one of the following: (1) M D F t ˚i 2SFp F yi , S n ¹0º ¤ ;, Œt; yi D iyi ;

Œyi ; yj D 0;

and yi acts invertibly for all i . (2) M D F t ˚ F u ˚kiD0 F yi , k p Œt; u D

u;

1,

Œt; yi D iyi ;

Œu; yi D yi

1;

Œyi ; yj D 0;

and yi acts nilpotently for 1 i k. (3) M Š O.1I 1/ÌW .1I 1/ is the split extension of W .1I 1/ by the restricted W .1I 1/module O.1I 1/ and O.1I 1/.1/ Ì W .1I 1/.1/ acts nilpotently on V .

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(4) M Š O.1I 1/ Ì sl.2/ is the split extension of sl.2/ (considered as a subalgebra of W .1I 1/) by the restricted sl.2/-module O.1I 1/ and O.1I 1/.1/ Ì .sl.2/ \ W .1I 1/.1/ / acts nilpotently on V . Proof. We consider M as a Lie subalgebra of gl.V / and denote by MŒp the penvelope of M in gl.V /. In a first step we will prove that MŒp (instead of M ) is of the listed form, and then show that M is of the required form as well. We adopt the notations of the preceding proposition. Recall that .MŒp / is not contained in W .1I 1/.0/ and I J WD ker . Proposition 7.6.8 shows that MŒp =J Š .MŒp / is isomorphic to one of F @;

F .1 C x/@;

F @ ˚ F x@;

F @ ˚ F x@ ˚ F x 2 @;

or

W .1I 1/:

Adjusting we may assume that .MŒp / is one of these. (a) Suppose .MŒp / is 1-dimensional and there is u 2 MŒp with .u/ ¤ 0 such that ad u is nilpotent. Since J is abelian, this implies that MŒp is nilpotent. But then M D H acts trigonalizably on V . This gives M D C.M /, which is not true in the present case. (b) Suppose .MŒp / is 1-dimensional and there is u 2 MŒp with .u/ ¤ 0 such that ad u is not nilpotent. Let JŒp be the p-envelope of J in MŒp . If J ¨ JŒp , then JŒp D MŒp (by dimension reasons) and MŒp were abelian. We argued in (a) that this is impossible. Then J is a restricted ideal, and there exists a toral element t 2 MŒp P n J . This element t acts semisimply on J with eigenvalues in Fp . Decompose J D i 2Fp Ji into eigenspaces with respect to ad t . Note that MŒpPD F t ˚J . Since J is abelian, J admits a linear eigenvalue function 2 J and i 2Fp Ji \ ker is an ideal of MŒp . This ideal acts nilpotently on V , and therefore is ¹0º by Engel’s theorem. As a result, dim Ji 1 for all i. Since J is not central, there is an index i ¤ 0 with Ji ¤ ¹0º. This is Case (1) of the theorem. (c) Next consider the remaining three cases. Then .MŒp / is centerless, and this proves that is a homomorphism of restricted algebras. Recall that K D 1 .W .1I 1/ / and that ŒK; J acts nilpotently on V (Proposition 11.3.3). In the .0/ present cases .K/ contains F x@. Since K is a restricted Lie algebra, we find a toral element t 2 K for which .t/ D x@, and then an element u 2 MŒp for which .u/ D @, Œt; u D u. Consider J 0 WD ¹y 2 J j .ad u/j .y/ 2 ker 8j > 0º. Since ŒK; J ker and MŒp D F u C K, J 0 is an ideal of MŒp satisfying ŒMŒp ; J 0 ker . Engel’s theorem gives ŒMŒp ; J 0 D ¹0º, whence J 0 D C.MŒp /. Note that uŒp 2 J . Since J is abelian, this gives ŒuŒp ; J D ¹0º. Consider the nilpotent subalgebra N WD F u C J of MŒp . Let VN denote an irreducible N submodule of V and R the kernel of this representation. Note that R\J J 0 , whence R \J C.MŒp /\ker D ¹0º. The assumption R ¤ ¹0º would imply N D R CJ , and hence Œu; J ŒR; J R \ J D ¹0º. But then I J J 0 D C.MŒp /, a contradiction. The N -module VN is therefore faithful.

11.3


137

Decompose J D ˚Ji into indecomposable F u-modules. As J is abelian, every Ji is an ideal of N and as such it intersects C.N / nontrivially (recall that N is nilpotent). Since VN is faithful and irreducible, C.N / is 1-dimensional. Then J itself is indecomposable. We mentioned above that ŒuŒp ; J D ¹0º, whence .ad u/p jJ D 0. The Jordan form of .ad u/jJ provides a basis .y0 ; : : : ; yk / for J with Œu; yi D yi 1 if i > 0 and Œu; y0 D 0. Note that y0 2 J 0 D C.MŒp /, and therefore C.MŒp / D F y0 ¤ ¹0º. As .ad u/p jJ D 0, we have k p 1. P There is a t -eigenvector in J of the form yk C ki D01 i yi . Obviously this element also generates the F u-module J . Therefore we may take yk a t -eigenvector. Then any yi is a t-eigenvector. As Œt; y0 D 0, this yields Œt; yi D iyi for all i . We also have for i ¤ 0 that yi 2 ŒK; J ker . (d) If .MŒp / D F @˚F x@, then (c) shows that we are in Case (2) of the theorem. So assume that .MŒp / is one of F @˚F x@˚F x 2 @ Š sl.2/ or W .1I 1/. Applying the results of (c) we obtain that C.MŒp / ¤ ¹0º and dim J =C.MŒp / p 1. Computing traces one gets k X k.k C 1/ ; 0 D trace.ad tjJ / D iD 2 i D1

i.e., dim J D k C 1 D p. Note that J =F y0 is a restricted .MŒp /-module of dimension p 1, .MŒp / Pp 1 contains the sl.2/-triple .@; 2x@; x 2 @/, and J =F y0 Š i D1 F yi contains the highest weight vector y1 C F y0 of weight 2. Theorem 5.3.1 shows that the irreducible sl.2/-composition factor containing the image of this highest weight vector has dimension p 1. But then J =F y0 is .MŒp /-irreducible and the lowest weight space is spanned by yp 1 C F y0 . we recall that ŒK; J acts nilpotently on V . Therefore ŒK; J J \ ker D PpNext 1 F yi . Since .K/.1/ W .1I 1/.1/ is Œp-nilpotent, K .1/ acts nilpotently on i D1 Pp 1 ŒK; J . Therefore there is a 1-dimensional K-submodule in i D1 F yi . But the only space invariant even under x 2 @ alone is F yp 1 . Therefore F yp 1 is K-invariant, and the .MŒp /-module J is isomorphic to J Š u. .MŒp // ˝u.

.K//

F yp

1

Š O.1I 1/

under isomorphisms yp

1 i

7! @i ˝ yp

1

7! @i .x p

1

/:

This gives a simultaneous realization J Š O.1I 1/ and .MŒp / D F @˚F x@˚F x 2 @ or .MŒp / D W .1I 1/. (e) Next we prove that the extension 0 ! J ! MŒp ! MŒp =J ! 0 splits.

138

11

1-sections

Consider first the case that MŒp =J Š .MŒp / D F @ ˚ F x@ ˚ F x 2 @. Since ad t is semisimple, there is a t-eigenvector y 2 MŒp n J such that .y/ D x 2 @. Then v WD Œu; y 2t 2 ker D J , Œt; y D y. Therefore Œt; v D 0, and this proves that v 2 F y0 D C.MŒp /. Put t 0 WD 21 Œu; y, whence .t 0 / D .t/ D x@. Then F u C F t 0 C F y is isomorphic to sl.2/, and the extension splits. Next consider the case that MŒp =J Š W .1I 1/. Using the former result we put e 1 WD u, e0 WD t 0 , e1 WD y. Since t 0 is semisimple, there are elements e2 ; : : : ; ep 2 2 MŒp n J with the property Œe0 ; ei D iei . Note that the .ad e0 /-eigenspace for the eigenvalue i 2 Fp is 2-dimensional and is spanned by ei ; yi . Then Œe 1 ; ei .i C 1/ei 1 D ri 1 yi 1 for some ri 1 2 F . Substituting inductively ei by ei ri 1 yi for i D 2; : : : ; ep 2 one obtains Œe 1 ; ei D .i C 1/ei

1;

Œe0 ; ei D iei

8 i 2 Fp :

Next observe that Œei ; ej .j i/ei Cj D si;j yi Cj for some si;j 2 F and si;j D 0 if i 0 or j 0. Then one obtains inductively (for 1 i; j; i C j p 2) si;j yi Cj

1

D Œe 1 ; Œei ; ej D .i C 1/Œei

.j

1 ; ej

i/ei Cj

C .j C 1/Œei ; ej

1

.j

i/.i C j C 1/ei Cj

1

D 0: Pp 2 One similarly proceeds if 1 i; j p 2 < i C j . Then i D 1 F ei Š W .1I 1/, and the extension splits. (f) Note that K Š O.1I 1/ Ì .MŒp \ W .1I 1/.0/ / as restricted algebras, whence .1/ K Š O.1I 1/.1/ Ì .MŒp \ W .1I 1/.1/ / is a Œp-nilpotent subalgebra of gl.V /. Then K .1/ acts nilpotently on V . (g) We now derive the structure of M . Note that MŒp .1/ M . Case (1): Since I ¤ C.M /, M is not abelian. Then there exist ˛; ˇ 2 F , ˛ ¤ 0, such that t 0 WD ˛t C ˇy0 2 M . Then either MŒp .1/ D M or M D F t 0 C P i 2Sn¹0º F yi . P Case (2): One has MŒp .1/ D F u C kiD0 F yi . Then MŒp .1/ M is nilpotent of codimension 1 in MŒp . Since M ¤ H , M is not nilpotent and one obtains M D MŒp . In Cases (3) and (4) one has MŒp D MŒp .1/ and this shows MŒp D M . We explicitly mention that the element t in Theorem 11.3.4 is not necessarily contained in the CSA H .

11.4

More on H.2I 1/.2/

The Z-gradings of the simple Cartan type Lie algebras are ruled by Theorem 7.4.1. We are particularly interested in the case H.2I 1/.2/ . This algebra is defined by use of the mapping DH , in which definition fixed generators x1 ; x2 of O.2I 1/ are involved.

More on H.2I 1/.2/

11.4

139 .u/

Using different generators u1 ; u2 gives different mappings DH W f 7! @u1 .f /@u2 @u2 .f /@u1 and isomorphic but not necessarily identical subalgebras of W .2I 1/. Now let 2 Autc O.2I 1/ be such that ı H.2I 1/.2/ ı 1 D H.2I 1/.2/ . Put u0i WD .xi / (i D 1; 2). Then u01 ; u02 are generators of O.2I 1/. Theorem 7.3.6 shows that ¹u01 ; u02 º D ˛ for some ˛ 2 F and ı DH .f / ı 1 D DH .˛ 1 .f // holds for all f 2 O.2I 1/ (with the Poisson bracket ¹ ; º introduced in Volume 1, p. 188). If one adjusts u1 WD u01 , u2 WD ˛ 1 u02 , then ¹u1 ; u2 º D 1 and j

DH .ui1 u2 /.uk / D ˛

j

¹u01 i u02 j ; uk º D ˛

j

iık;2 u01 i

1 0j u2

˛1

j

j ık;1 u01 i u02 j

1

:

This gives j

j

DH .ui1 u2 / D iui1 1 u2 @u2

j 1

j ui1 u2

.u/

j

@u1 D DH .ui1 u2 /;

.u/

i.e., DH D DH . It is clear that p 1

p 1

Der H.2I 1/.2/ D DH .O.2I 1// C F u1 @u2 C F u2 @u1 C F u2 @u2 X j .u/ p 1 p 1 D FDH .ui1 u2 / C F u1 @u2 C F u2 @u1 C F u2 @u2 : 0i;j p 1 02 p 1

C F u2

@u1 C F u2 @u2 ;

.1/

and hence M0 acts nilpotently on M . Moreover, there are at least 2 different indices i1 ; i2 < 0 with Mik ¤ ¹0º, k D 1; 2. (5) Suppose M H.2I 1/. (a) If the grading is as in (2) or (3), then C.M0 / D ¹0º. (b) In Case (2) any torus F h0 M0 is proper in M if and only if it is proper in M0 . In Cases (3) and (4) any torus F h0 M0 is proper in M . Proof. According to Theorem 7.4.1 and Corollary 7.4.2 there is an automorphism 2 Autc O.2I 1/ and a1 ; a2 2 Z such that ı H.2I 1/.2/ ı 1 D H.2I 1/.2/ and the grading of M is given by a degree deg .x1 / DW a1 , deg .x2 / DW a2 . Applying the isomorphism H.2I 1/.2/ ! H.2I 1/.2/ , .x1 / 7! .x2 /, .x2 / 7! .x1 / (cf. Theorem 7.3.2) if necessary we may assume ja1 j ja2 j. Set u1 WD .x1 / and u2 WD ˛ .x2 /, where ˛ 2 F is chosen such that ¹u1 ; u2 º D 1. The introductory .u/ remark shows that DH D DH , whence Der H.2I 1/ D DH .O.2I 1// C F u1

.u/

p 1

@u2 C F u2

.u/

p 1

@u2 C F u2

H.2I 1/ D DH .O.2I 1// C F u1

p 1

@u1 C F u2 @u2 ;

p 1

@u1 ;

.u/

H.2I 1/.2/ D DH .O.2I 1//.1/ : (1)–(4) In Case 2 one has (if i C j > 0) .u/

j

j

deg .DH .ui1 u2 // D deg .iui1 1 u2 @u2 p 1

@u2 D

p 1

@u1 D .p

deg u1 deg u2

a2 ; 1/a2 ;

deg u1 @u1 D deg u2 @u2 D 0: An easy computation gives the result.

j 1

j ui1 u2

@u1 / D .j

1/a2 ;

More on H.2I 1/.2/

11.4

141

In Case 3 one has (if i C j > 0) .u/

j

j

j 1


deg u1

p 1

@u2 D deg u2

j ui1 u2

@u1 D .p

@u1 / D .i C j

2/a2 ;

2/a2 ;

deg u1 @u1 D deg u2 @u2 D 0: An easy computation gives the result. In Case 4 set q D a1 =a2 and observe that q ¤ 0; 1. Note that (if i C j > 0) j

.u/

j

j 1


@u2 D .q.p

p 1

@u1 D .p

deg u1 deg u2

1/ 1

j ui1 u2

@u1 / D .q.i

1/ C .j

1//a2 ;

1/a2 ;

q/a2 ;

deg u1 @u1 D deg u2 @u2 D 0: .u/

j

Thus deg .DH .ui1 u2 // ¤ 0 for .i; j / 2 ¹.1; 0/; .2; 0/; .0; 1/; .0; 2/º, and hence P .u/ .u/ j p 1 p 1 M0 FDH .u1 u2 / C i Cj >2 FDH .ui1 u2 / C F u1 @u2 C F u2 @u1 C F u2 @u2 . Since .kuk1 1 u2 @u2 uk1 @u1 / 2 M.k 1/a1 for k D 2; 3, the final claim follows if a1 < 0. The case a2 < 0 is symmetric. So let us assume a1 ; a2 > 0. As @u2 2 M a2 and @u1 2 M a1 and a1 ¤ a2 , this proves the final claim for all cases. (5) Suppose M H.2I 1/. The statement on C.M0 / is obtained by a dimension argument since u2 @u2 62 H.2I 1/. Note that every maximal torus is 1-dimensional and the root lattice with respect to any maximal torus is spanned by one element. Let F h0 M0 be any 1-dimensional torus and ˛ denote a spanning root. Then M Œ˛ D M . Let M.0/ be the uniquely determined maximal subalgebra of codimension 2 in M for which M.0/ =M.1/ Š sl.2/. It follows from Definition 11.2.2(2) that F h0 is proper in M if and only if ˛ is proper, and this is true if and only if F h0 M.0/ . Assume that the grading of M is as in Case (2). Note that F h0 is proper in M if and only if F h0 M.0/ \ M0 . Since M.0/

XX

.u/

j

FDH .ui1 u2 / C

j 2 i 0

X

.u/

FDH .ui1 u2 / C

i 1

C

p 1 F u1 @u2

X

.u/

FDH .ui1 /

i 2

C

p 1 F u2 @u1 ;

Pp 1 .u/ one computes in the present case that M.0/ \ M0 D i D1 FDH .ui1 u2 /, and this is the unique subalgebra of codimension 1 in M0 Š W .1I 1/. The result follows. If the grading of M is as in Cases (3) or (4), then F h0 M0 M.0/ . So F h0 is proper.

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1-sections

Theorem 11.4.2. Let M , u1 ; u2 and a1 ; a2 be as in Theorem 11.4.1(2) and T M0 be a 2-dimensional torus. There is an automorphism of Der H.2I 1/.2/ such that Mi \ H.2I 1/.2/ D Mi \ H.2I 1/.2/ for all i and .T / D F .ı1 C u1 /@u1 C F u2 @u2 ;

ı1 2 ¹0; 1º:

Proof. By assumption the grading of M is given by generators u1 , u2 having degrees deg u1 D 0, deg u2 D a2 ¤ 0 and satisfying ¹u1 ; u2 º D 1. Put t0 WD u2 @u2 2 Der H.2I 1/.2/ and observe that t0 is the mod.p/-degree derivation, Œt0 ; d D jd

8 d 2 Mja2 :

Due to Theorem 7.1.2(3c) one has Der H.2I 1/.2/ D CH.2I 1/. Since T is a 2dimensional torus, it is a maximal torus in CH.2I 1/ (Theorem 7.5.5). Moreover, this theorem states that T is conjugate under an automorphism of CH.2I 1/ to one of F .ı1 C u1 /@u1 ˚ F .ı2 C u2 /@u2 ; ı1 ; ı2 2 ¹0; 1º (we may substitute x1 ; x2 by u1 ; u2 in that theorem according to the introductory remark of this section). As T M0 , one has ŒT; t0 D ¹0º, whence t0 2 T by the maximality of T . Set .t0 / D 1 .ı1 C u1 /@u1 C 2 .ı2 C u2 /@u2 : Recall that Der H.2I 1/.2/ .0/ is the subalgebra of all derivations which stabilize the natural maximal subalgebra H.2I 1/.2/ .0/ . This subalgebra is invariant under all automorphisms. As t0 2 Der H.2I 1/.2/ .0/ , then 1 ı1 D 2 ı2 D 0: Since t0 62 H.2I 1/.2/ , we have .t0 / 62 H.2I 1/.2/ , and this gives 1 C 2 ¤ 0. Consequently, Œ .t0 /; .ı1 C u1 /p Since M D

Pp

1 iD 1

1

@u2 ¤ 0;

Œ.t0 /; .ı2 C u2 /p

1

@u1 ¤ 0:

Mia2 ,

.M0 / ¹d 0 2 CH.2I 1/ j Œ.t0 /; d 0 D 0º D F .ı1 C u1 /@u1 ˚ F .ı2 C u2 /@u2 .u/ ˚ span ¹DH .ı1 C u1 /i .ı2 C u2 /j j 1 .i

1/ C 2 .j

1/ 0 .p/º:

If 1 2 ¤ 0, then ı1 D ı2 D 0 by an earlier remark and we obtain .M0 / Der H.2I 1/.2/ .0/ . But as Der H.2I 1/.2/ .0/ = Der H.2I 1/.2/ .1/ Š gl.2/, there is no room for .M0 / W .1I 1/ in Der H.2I 1/.2/ .0/ . Consequently, 1 2 D 0.

11.4

More on H.2I 1/.2/

143

If 2 D 0, then 1 ¤ 0 and .t0 / D 1 u1 @u1 ;

.T / D F u1 @u1 ˚ F .ı2 C u2 /@u2 :

The automorphism 0 induced by the mapping u1 7! u2 , u2 7! u1 brings us to the case . 0 ı /.t0 / 2 F u2 @u2 and . 0 ı /.T / D F .ı2 C u1 /@u1 ˚ F u2 @u2 . We suppress the notion of 0 and assume in either case 1 D 0, 2 ¤ 0, whence .t0 / D 2 u2 @u2 ;

.T / D F .ı1 C u1 /@u1 ˚ F u2 @u2 :

Set .@u2 / 1 @u1 C 2 @u2

.mod Der H.2I 1/.2/

.0/

/;

i 2 F:

As @u2 2 M 1 , 1 @u1 C 2 @u2 .@u2 / D .Œ@u2 ; t0 / D Œ.@u2 /; .t0 / 2 2 @u2 .mod Der H.2I 1/.2/ .0/ /; and this gives 1 D 0, 2 .2 1/ D 0. Since .@u2 / 62 H.2I 1/.2/ .0/ we have 2 ¤ 0. Then 2 D 1. Consequently, .t0 / D u2 @u2 D t0 . Theorem 11.4.1(2a) shows the following: if Mia2 ¤ ¹0º and Mja2 ¤ ¹0º and i j.p/ but i ¤ j , then i j 1.p/. Therefore if i 6 1.p/, then we have Mi a2 D ¹d 2 M j Œt0 ; d D id º. As .t0 / D t0 , this gives .Mia2 / D Mia2 . p 1 Next assume i 1.p/. Note that M.p 1/a2 F u2 @u1 and therefore we have M.p 1/a2 \ H.2I 1/.2/ D ¹0º. Then there is only one choice for such i for which Mi a2 \ H.2I 1/.2/ ¤ ¹0º, namely i D 1. Now argue as before. Next we will describe the central extensions of H.2; 1/.2/ . It is a standard fact, that central extensions 0 ! F ! g ! gN ! 0 of a Lie algebra gN are described by the second cohomology group H 2 .Ng; F /. Central extensions g1 and g2 are called equivalent if there is an isomorphism W g1 ! g2 such that the following diagram commutes 0! F k 0! F

! !

g1 # g2

! gN ! 0 k ! gN ! 0:

For completeness we will include the proof of the following theorem. For u 2 g write uN WD u C F 2 g=F Š gN . Theorem 11.4.3. Let gN be a Lie algebra admitting a nondegenerate associative bilinear form . The central extensions 0 ! F ! g ! gN ! 0

144

11

1-sections

are given by cocycles fD .x; y/ WD .D.x/; y/ for some D 2 Der gN satisfying .D.u/; N v/ N C .D.v/; N u/ N D0

8u; N vN 2 gN :

If D1 D2 is an inner derivation of gN , then the cocycles fD1 and fD2 describe equivalent extensions. N D uN for all Proof. (a) Choose a linear endomorphism W gN ,! g satisfying .u/ u 2 g. Then we have a vector space decomposition g D .Ng/ ˚ F z;

Œz; g D ¹0º:

Let Œ ; and Œ ; denote the multiplications in gN and g, respectively. Define a skewsymmetric bilinear form f by Œu; v D .Œu; N v/ N C f .u; N v/z N

8u; v 2 g:

Note that ŒŒu; v ; w D Œ.Œu; N v/; N w D Œ.Œu; N v/; N w N C f .Œu; N v/; N wN N C f .Œu; N v; N w/; N D .ŒŒu; N v; N w/ and therefore the Jacobi identity gives the cocycle property f .Œu; N v; N w/ N C f .Œv; N w; N u/ N C f .Œw; N u; N v/ N D 0: Since is a nondegenerate form, there is a linear transformation D 2 EndF .Ng/, such that f .u; N v/ N D .D.u/; N v/: N The skew-symmetry of f gives .D.u/; N v/ N C .D.v/; N u/ N D0

8u; N vN 2 gN :

Then .D.Œu; N v/; N w/ N D f .Œu; N v; N w/ N D f .u; N Œv; N w/ N C f .v; N Œw; N u/ N D .D.u/; N Œv; N w/ N

.D.v/; N Œu; N w/ N

D .ŒD.u/; N v; N w/ N

.ŒD.v/; N u; N w/ N

for all u; N v; N wN 2 gN , and this shows that D is a derivation of gN . (b) Suppose g1 and g2 are extensions given by cocycles fD1 .u; N v/ N D .D1 .u/; N v/ N and

fD2 .u; N v/ N D .D2 .u/; N v/; N

respectively;

11.4

More on H.2I 1/.2/

145

such that D1 D2 D ad x is an inner derivation. As in (a) identify gi D i .Ng/ ˚ F z N D uN for all uN 2 gN , and write the multiplication in gi as vector spaces such that i .u/ as Œu; vi D i .Œu; N v/ N C fDi .u; N v/z N 8u; v 2 gi ; i D 1; 2: Define a linear mapping W g1 ! g2 ;

.1 .u/ N C ˛z/ WD 2 .u/ N C .˛

.x; u//z N

8uN 2 gN :

Then induces the identity mapping on F z and gN . Note that fD1 .u; N v/ N

fD2 .u; N v/ N D .D1 .u/; N v/ N

.D2 .u/; N v/ N

D .Œx; u; N v/ N D .x; Œu; N v/ N and u

i .u/ N 2 F z;

.u/

2 .u/ N 2 F z:

Then N 2 .v// N C fD2 .2 .u/; N 2 .v//z N Œ .u/; .v/2 D Œ2 .u/; N 2 .v/ N 2 D 2 .Œ2 .u/; D 2 .Œu; N v/ N C fD2 .u; N v/z N D 2 .Œu; N v/ N

.x; Œu; N v/z N C fD1 .u; N v/z N D 1 .Œu; N v/ N C fD1 .u; N v/z N D .Œu; v1 /:

The extensions g1 and g2 are equivalent. In the setting of Theorem 11.4.3 the central extensions are described by N W g ! g=C.g/ Š gN the canonical homomorphism; W gN ,! g a linear endomorphism satisfying N ı D Id; Œu; v D .Œu; N v/ N C .D.u/; N v/; N

(11.4.1)

D 2 Der gN :

The concepts of equivalent extensions and isomorphic algebras are different. Lemma 11.4.4. Let gN be simple and a nondegenerate associative bilinear form on gN . For every ' 2 Aut gN the central extensions g, g0 described by (11.4.1) and derivations D and ' ı D ı ' 1 , respectively, are isomorphic algebras. Proof. Note that .u; N v/ N 7! .'.u/; N '.v// N defines a nondegenerate associative bilinear form on gN . Hence there is 2 GL.Ng/ such that .'.u/; N '.v// N D ..u/; N v/ N for all u; N vN 2 gN . The associativity of the form yields that is gN -invariant. The simplicity of gN and Schur’s lemma gives rise to 2 F such that .'.u/; N '.v// N D .u; N v/ N 8u; N vN 2 gN :

146

11

1-sections

Then ..' ı D ı ' 1 /.'.u//; N '.v// N D .D.u/; N v/ N for all u; N vN 2 gN . Define a vector space automorphism of gN ˚ F by .uN C ˛/ WD '.u/ N C ˛. Denote the multiplications of the central extensions given by the derivations D and ' ı D ı ' 1 by Œ ; D and Œ ; ' , respectively. One obtains Œ .uN C ˛/; .vN C ˇ/' D Œ'.u/; N '.v/ N ' D Œ'.u/; N '.v/ N C .'.D.u//; N '.v// N D '.Œu; N v/ N C .D.u/; N v/ N D D

.Œu; N v N D/

.ŒuN C ˛; vN C ˇD /:

Lemma 11.4.5. Let n D .n1 ; n2 / be arbitrary. Define f 2 Autc O.2I n/ for any f 2 O.2I n/ by setting f .xk / WD

p X1 i D0

i 1 DH .f x22 / .xk /; iŠ

k D 1; 2:

Then f .u/ D exp DH .f x22 / .u/ for u 2 O.2I 1/ and f .¹u; vº/ D ¹f .u/; f .v/º

8u; v 2 O.2/:

In particular, ˆf .DH .u// D DH .f .u// holds for all u 2 O.2/. Proof. (a) Let g1 ; g2 2 O.2/. One proves DH .f x22 /r .gi / D Pr;gi x2r , where Pr;gi 2 O.2/, by induction on r. Then p p DH .f x22 / .g1 / 2 O.2/x2 D ¹0º; and if r C s p DH .f x22 /r .g1 / DH .f x22 /s .g2 / 2 O.2/x2rCs D ¹0º: From these equations one derives that exp DH .f x22 / is an automorphism of O.2/. This automorphism coincides with f on x1 ; x2 and hence on every element of the commutative algebra generated by these 2 elements. (b) As in (a) one computes for r C s p p

¹DH .f x22 /r .x1 /; DH .f x22 /.x2 /º D ¹Pr;1 x2r ; Ps;2 x2sC1 º 2 O.2/x2 D ¹0º and derives from this exp DH .f x22 /.¹x1 ; x2 º/ D ¹exp DH .f x22 /.x1 /; exp DH .f x22 /.x2 /º: Then 1 D ¹f .x1 /; f .x2 /º. Theorem 7.3.6 now yields the claim.

More on H.2I 1/.2/

11.4

.p/

147

.p 1/ .p 1/

.p/

Lemma 11.4.6. Let D WD DH .˛x1 C ˇx1 x2 C x2 / and ˛ ¤ 0. Then .p/ there exists 2 Autc O.2I 1/ such that ˆ .D/ DH .˛x1 / .mod H.2I 1/.2/ / and ˆ DH ..1 C x1 /x2 / D DH ..1 C x1 /x2 /. Q C x1 /2 and 1 WD f Proof. Choose Q 2 F satisfying 2˛ Q p D and set f WD .1 as in the previous lemma. Lemma 11.4.5 shows that 1 is, for arbitrary n, a divided power automorphism of O.2I n/ satisfying ˆ1 .DH .u// D DH .1 .u// for all u 2 O.2I n/. Note that DH . .1 Q C x1 /2 x22 / .1 C x1 /x2 D 0, whence 1 ..1 C x1 /x2 / D .1 C x1 /x2 . Applying Equations (2.1.4) and (2.1.5) we obtain .p/

1 ˛x1

.p 1/ .p 1/ x2

C ˇx1

.p/

C x2

˛1 .x1 /.p/ C 1 .x2 /.p/ ˛.x1 .p/

˛x1

.p/

2 x Q 2 /.p/ C x2 C .

.p/

˛x1 .p/

.p/

2˛ Q p /x2

.mod O.2I 1//:

.p 1/ .p 1/

x2 / C D 0 , where ˇ1 2 F and We obtain ˆ1 .D/ D DH .˛x1 / C ˇ1 DH .x1 0 .2/ D 2 H.2I 1/ . Choose ˇQ 2 F satisfying ˇ1 C ˛ ˇQ D 0 and define 2 2 Autc O.2/ by Q .p 2 .x1 / WD x1 C ˇx 2

1/

;

2 .x2 / WD x2 :

Note that 2 .O.2I 1// D O.2I 1/ holds. Computing the Poisson bracket one obtains ¹2 .x1 /; 2 .x2 /º D ¹x1 ; x2 º D 1 and therefore (see Theorem 7.3.6) ˆ2 .DH .u// D DH .2 .u// 8u 2 O.2/: Applying Equations (2.1.4) and (2.1.5) we obtain .p/

2 ˛x1

.p 1/ .p 1/ x2

C ˇ1 x1

.p 1/ .p/

Q D ˛.x1 C ˇx 2 .p/

D ˛x1

/

.p 1/ .p 1/ .p 1/ / x2

Q C ˇ1 .x1 C ˇx 2

Q .p C .ˇ1 C ˛ ˇ/x 1

1/ .p 1/ x2

.p/

D ˛x1 :

One also has Q .p 2 ..1 C x1 /x2 / D x2 C .x1 C ˇx 2

1/

/x2 D x2 C x1 x2 D .1 C x1 /x2 :

According to [S-F, Theorem 4.6.5] the algebra H.mI n/.2/ possesses a nondegenerate associative form given by the following: put .n/ the m-tuple .p n1 1; : : : ; p nm 1/, define ˛ 2 HomF .O.mI n/; F / by ˛ .x .a/ / WD ı..n/; a/ 8a .n/:

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Then the setting .DH .x .a/ /; DH .x .b/ // WD ˛ .x .a/ x .b/ /

8a; b .n/

gives the required nondegenerate associative bilinear form on H.mI n/.2/ . In the particular case m D 2 and n D 1 one obtains ´ p 1 p 1 a1 Ca2 if a C b D p 1; i i .a/ .b/ a1 a2 D . 1/ .DH .x /; DH .x // D (11.4.2) 0 otherwise: Let g.0/ denote the inverse image of H.2I 1/.2/ .0/ H.2I 1/.2/ Š g=C.g/. Theorem 11.4.7. There are exactly 3 isomorphism classes of Lie algebras g having a 1-dimensional center such that g=C.g/ Š H.2I 1/.2/ . (1) The split extension. It is characterized by the property C.g/ \ g.1/ D ¹0º. (2) The extension g has a basis j

.x1i x2 j .0; 0/ .i; j / < .p

1; p

1//

and the multiplication is given by the Poisson bracket. This case is characterized by the property C.g/ g.1/ ; C.g/ \ g.0/ .1/ D ¹0º: (3) The extension g has a basis j

.zI x1i x2 j .0; 0/ < .i; j / < .p and the multiplication is given by 8 ˆ 4, one has p 1 bC1 p 1 x2 º; ¹x1 ; ¹x1 ; x1 x2bC1 ºº

¹x1 ; x1

2 g.0/ :

By choice of .a; b/ the Engel–Jacobson theorem shows that ¹M; I º g0.pCb p 1 1/x1 x2b .

2/

acts

p 1 .x1 x2b /

¤ 0, this comAs nilpotently on V . Note that ¹e; f º D .b C mutator acts invertibly on V . The proposition now yields that there is an irreducible g.0/ -submodule V0 of V such that dim V p dim V0 . The present assumption gives dim V0 < p 3 . Next we apply Proposition 3.2.11 to the Lie algebra g.0/ and the g.0/ -module V0 . Set M WD g.1/ , I WD g0.pCb 3/ , e1 WD x12 ;

e2 WD x1 x2 ;

p 2 bC1 x2 ;

f1 WD x1

p 1 b x2 :

f2 WD x1

One has ¹ei ; fj º; ¹ei ; ¹ek ; fj ºº 2 ¹g.0/ ; g.pCb 3/ º g.1/ D M . The Lie algebra P p 2 p 1 R generated by F ¹ei ; fj º D F x1 x2bC1 C F x1 x2b is abelian, and we have P p 1 2 x2b º D 0. Therefore i <j F ¹ei ; fj º D F ¹x1 ; x1 X F ¹ei ; fj º D ¹g.1/ ; g0.pCb 3/ º g0.pCb 2/ ¹g.1/ ; g0.pCb 3/ º C R.1/ C i<j

acts nilpotently on V0 . Moreover, the matrix .¹ei ; fj º/ 1i;j 2 is a nonsingular triangular matrix. This means that .¹ei ; fj º/ is nilpotent if i < j and invertible if i D j . By Proposition 3.2.11 there is an irreducible g.1/ -submodule V1 of V0 such that dim V0 p 2 dim V1 . The present assumption on V yields dim V1 < p. As g.1/ is solvable, dim V1 is a p-power, i.e., dim V1 D 1. This shows that every E 2 g.1/ .1/ p 3 has eigenvalue 0 on V , whence .g.1/ .1/ / D 0. However, .¹x13 ; x1 x2bC1 º/ D p 1 3.b C 1/.x1 x2b / ¤ 0. Therefore this case cannot occur. (c) Suppose () holds with a D p 1 and b p 2. As a Cb 2p 3, this gives b D p 2. We apply Proposition 3.2.11 to the Lie algebra g.0/ and any irreducible g.0/ -submodule V0 of V . Set M WD g.1/ , I WD g0.2p 5/ , e1 WD x12 ;

e2 WD x1 x2 ;

p 2 p 1 x2 ;

f1 WD x1

p 1 p 2 x2 :

f2 WD x1

As above, there is an irreducible g.1/ -submodule V1 of V0 with dim V0 p 2 dim V1 . The present assumption on V yields dim V1 < p 2 . We now set M WD g.2/ , I WD g0.2p 6/ and e WD x13 ;

p 3 p 1 x2 :

f WD x1

By Proposition 3.2.11 there is an irreducible g.2/ -submodule V2 of V1 with dim V1 p dim V2 . The present assumption on V1 yields dim V2 < p. As g.2/ is solvable,

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dim V2 is a p-power, i.e., dim V2 D 1. This shows that every E 2 g.2/ .1/ has eigenp 4 p 1 value 0 on V , whence .g.2/ .1/ / D 0. However, we have .¹x14 ; x1 x2 º/ D p 1 p 2 4.x1 x2 / ¤ 0. Therefore this case cannot occur. (d) Suppose () holds with b D p 1. This case is reduced to the former one by applying the automorphism of g induced by the mapping x1 7! x2 , x2 7! x1 . (e) Suppose () holds with a; b a. We proceed as in (c). Set M WD g.0/ , I WD g0.aCb 1/ , e1 WD x1 ;

f1 WD x1a x2bC1 ;

e2 WD x2 ;

f2 WD x1aC1 x2b :

Then we have ¹ei ; fj º; ¹ei ; ¹ek ; fj ºº 2 g.aCb 3/ g.0/ D M , and (as a C b 3) ¹¹ei ; fj º; ¹ek ; fl ºº 2 g0.2.aCb 2// I . By choice of .a; b/ and the Engel–Jacobson theorem, I C F ¹e1 ; f2 º acts nilpotently on V . Also, .¹ei ; fj º/ 1i;j 2 is a nonsingular triangular matrix. By Proposition 3.2.11, V has a g.0/ -submodule V0 of dimension dim V0 < p 2 . Next put M WD g.1/ , I WD g0.aCb 2/ and arrange e1 WD x12 ;

e2 WD x1 x2 ;

f1 WD x1a

1 bC1 x2 ;

f2 WD x1a x2b

if a ¤ 0, a ¤ b; e1 WD x12 ;

e2 WD x22 ;

f1 WD x1a

1 bC1 x2 ;

f2 WD x1aC1 x2b

1

if a ¤ 0, a D b; e1 WD x1 x2 ;

f1 WD x2b ;

e2 WD x22 ;

f2 WD x1 x2b

1

if a D 0. As in the former cases we obtain dim V0 p 2 , a contradiction.

Theorem 11.5.2. Suppose g=C.g/ Š H.2I 1/.2/ . Let W g ! gl.V / be an irreducible representation of dimension dim V < p 4 . Assume in addition that every Cartan subalgebra of g acts trigonalizably on V . Then the subalgebra Œg.0/ ; g.1/ C Œg; g.2/ acts nilpotently on V . Proof. (a) Note that either ker C.g/ or g D ker C C.g/. In the first case .g/=C..g// Š H.2I 1/.2/ holds, while in the second case the statement is trivial. Therefore we may assume that the first case is true. Suppose .g/ is as in Case 3 of Theorem 11.4.7. The multiplication is given by Equation (11.4.3). Expanding the equation in polynomials .1 C /3 .1 C /p

3

D .1 C /p D 1 C p

one derives the equations on binomial coefficients ! ! p 3 X X3 3 p 3 D 0; 1 l p i k iD0 iCkDl

kD0

1;

11.5

Low dimensional representations of H.2I 1/.2/

153

which yield ¹.1 C x1 /3 x23 ; .1 C x1 /p D

3 p X X3 i D0 kD0

D

3z C

3 p 3 x2 º

3 i

!

p

3

!

k

p 3 X X3 iD0 i Ck¤0

kD0

3 i

p 3

¹x1i x23 ; x1k x2 !

! 3

p k

º

. 3i

3k/x1i Ck

1 p 1 x2

D

3z:

This shows that the Cartan subalgebra h WD C.g/ ..1 C x1 /x2 / has the property z 2 h.1/ . But then 1 .h/ is a CSA of g which acts non-trigonalizably on V . This contradicts the present assumption. Since is irreducible, dim C..g// 1. Then either C..g// D ¹0º or .g/ is as described in Theorem 11.4.7. The above reasoning shows that in the latter case .g/ is split or a Poisson Lie algebra. In order to treat all cases simultaneously suppose C..g// D ¹0º or .g/ is split. In these cases one has to prove the proposition only for g.1/ . Choose a central extension gQ of .g.1/ / Š H.2I 1/.2/ which is isomorphic to the Poisson Lie algebra. Define the corresponding representation Q W gQ ! .g.1/ / gl.V /: Then .Qg; / Q satisfies the assumptions of the proposition. It is sufficient to prove the proposition for .Qg; / Q where ker Q C.Qg/. As a consequence, it suffices to prove the theorem under the additional assumption that g is the Poisson extension of H.2I 1/.2/ and ker C.g/. Lemma 11.5.1 j shows that .x1i x2 / consists of nilpotent transformations if i C j 3. Note that these elements span Œg.0/ ; g.1/ C Œg; g.2/ . The Engel–Jacobson theorem concludes the proof. In a next step we allow g to have a noncentral radical. Lemma 11.5.3. Every proper subalgebra of H.2I 1/.2/ of codimension 3 is contained in H.2I 1/.2/ .0/ . Proof. Let M be a subalgebra of codimension 3. Note that H.2I 1/.2/ has a natural grading which is given by attaching degree 1 to the indeterminates x1 ; x2 . Consider the graded subalgebra gr M associated with this grading. It is clear that dim H.2I 1/.2/ = gr M 3, and M H.2I 1/.2/ .0/ holds if and only if gr M H.2I 1/.2/ .0/ . Therefore we may assume that M is a graded subalgebra of H.2I 1/.2/ which is not contained in H.2I 1/.2/ .0/ . Then there is a nonzero element u1 2 F x1 C F x2 satisfying FDH .u1 / 2 M 1 . A suitable linear combination u2 D 1 x1 C 2 x2

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11

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exists with the properties that u1 ; u2 span F x1 C F x2 and ¹u1 ; u2 º D 1 holds. Theorem 7.3.6 then shows that the mapping ui 7! xi defines an automorphism ˆ of H.2I 1/.2/ . Since H.2I 1/.2/ .0/ is invariant under all automorphisms, it suffices to show the lemma for ˆ.M /, i.e., DH .x1 / 2 M . Next we observe P4 one may assume i that by dimension reasons i D1 FDH .x2 / \ M ¤ ¹0º. Applying DH .x1 / to a nonzero element of this space several times gives DH .x2 / 2 M as well. Since H.2I 1/.2/ 2p 5 ¤ ¹0º and 2p 5 5, a dimension argument shows that there is a full homogeneous component H.2I 1/.2/ k contained in M , where k 2. Then P .2/ M . [S-F, Theorem 4.7.1] shows that M D H.2I 1/.2/ . i k H.2I 1/ Theorem 11.5.4. Suppose g= rad g Š H.2I 1/.2/ , g.1/ D g and TR.g/ D 1. Let W g ! gl.V / be an irreducible representation of g such that every Cartan subalgebra of g acts trigonalizably on V . The following holds: (1) If dim V < p 4 , then V contains an irreducible g.0/ -submodule of dimension < p2. (2) If dim V < p 3 , then Œg.0/ ; g.1/ acts nilpotently on V . Proof. (a) Note that either ker rad g or g D ker C rad g. In the first case all assumptions hold for .g/ as well, and if the result is true for .g/ then it is true for g. In the second case there is an element t 2 ker which acts non-nilpotently on g. The semisimple part ts of t then spans a torus of maximal toral rank (by the assumption TR.g/ D 1) and as t 2 Cg .ts /, then Cg .ts / is a CSA. In this case g D ker C Cg .ts / holds, and the assumption on CSAs shows that g acts trigonalizably on V . Then the statement is trivial. Therefore we may assume that is faithful. Identify g and .g/. (b) Let gŒp denote the p-envelope of g in gl.V /. Corollary 1.1.8 shows that gŒp = rad gŒp is a p-envelope of g= rad g. It is semisimple, hence is a minimal p-envelope. Since g= rad g Š H.2I 1/.2/ is restricted, one gets gŒp = rad gŒp Š H.2I 1/.2/ as restricted Lie algebras. Then there is a restricted homomorphism W gŒp ! gŒp = rad gŒp Š H.2I 1/.2/ : For r 0 set gŒp;.r/ WD

1

.H.2I 1/.2/ .r/ /:

Due to Theorem 1.2.8(3) 0 ¤ TR.gŒp = rad gŒp / TR.gŒp /

TR.rad gŒp / D 1

TR.rad gŒp /:

One obtains TR.rad gŒp / D 0, whence rad gŒp is nilpotent (Theorem 1.2.7(2)). Set r WD rad gŒp . As r is nilpotent, there is n 0 such that nC1 .r/ F IdV . Choose n minimal with this condition. (c) Suppose n D 0, i.e., r F IdV . Theorem 11.5.2 applies and yields that Œg.0/ ; g.1/ acts nilpotently on V . Let V 0 be an irreducible g0 -submodule of V with

11.5


155

representation 0 . Then Œg.0/ ; g.1/ ker 0 by Engel’s theorem. Since Œg.0/ ; g.1/ Š sl.2/ is simple, it can only be that g.0/ D g.1/ C ker 0

or

ker 0 D g.1/

or

ker 0 ¨ g.1/ :

In the respective cases 0 .g.0/ / is abelian or 0 .g.0/ / Š sl.2/ or 0 .g.0/ / Š sl.2/ ˚ F . In all cases V 0 is at most p-dimensional (Theorem 5.3.1). The theorem holds in this case. (d) Consider the case n > 0, whence r 6 F IdV . Choose an ideal A of gŒp minimal subject to the conditions A.1/ F IdV ¨ A r: Suppose A.1/ ¤ ¹0º. There is a pairing f W A A ! F;

Œa; b D f .a; b/IdV

8a; b 2 A:

Note that f is gŒp -invariant. As the radical of f is gŒp -invariant as well, the minimality of A=F IdV implies rad f D F IdV . Then A is a Heisenberg algebra acting faithfully on V . As dim V < p 4 , one has dim A=F IdV DW 2s 6 (Volume 1, p. 149). The nilpotency of r in combination with the minimality of A=F IdV and Engel’s theorem gives Œr; A F IdV , and therefore A=F IdV is a restricted gŒp =r-module. Observe that f induces a nondegenerate skew-symmetric bilinear form on A=F IdV and this form is invariant under gŒp =r. Therefore gŒp =r maps into the classical algebra C3 . If the image is nonzero, then, as gŒp =r Š H.2I 1/.2/ is simple, the preceding reasoning yields p 2 2 D dim gŒp =r dim C3 D 21: This contradiction proves Œg; A ŒgŒp ; A F IdV . Since g D g.1/ we get Œg; A D ¹0º and A F IdV , a contradiction. Hence A.1/ D ¹0º. Next observe that there is a gŒp -invariant pairing g W r A ! F;

Œa; b D g.a; b/IdV

8a 2 r; b 2 A:

Suppose that this pairing is nonzero. Note that g.A; A/ D 0 by the former result. Obviously, A.?;g/ WD ¹x 2 r j g.x; A/ D 0º is a restricted subalgebra of r and an ideal of gŒp . Let Vr denote an irreducible r-submodule of V . Proposition 3.2.11 (setting .?;g/ G D r, M D A.?;g/ and I D A) shows that p dim r=A dim Vr 9 dim gl.r=A.?;g/ / and gŒp =r is simple, gŒp acts trivially on every composition factor as well. Since g is perfect, this gives Œg; r A.?;g/ . As a consequence, 0 D g.Œg; r; A/ D g.r; Œg; A/. The minimality of A=F IdV gives the alternative A D Œg; A C F IdV or Œg; A F IdV . As g is assumed to be nonzero, the first case is impossible. The second alternative gives (since g is perfect) Œg; A D ¹0º. As before we conclude A IdV , a contradiction. Consequently, g D 0, which means Œr; A D ¹0º.

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As V is gŒp -irreducible, there is a linear function 2 A such that x is nilpotent for any x 2 A. Let

.x/IdV

V0 WD ¹v 2 V j y.v/ D .y/v for all y 2 Aº and gŒp WD ¹x 2 gŒp j .Œx; A/ D 0º: Obviously gŒp is a restricted subalgebra of gŒp and it is easy to see that gŒp D ¹x 2 gŒp j x.V0 / V0 º. By Corollary 3.2.5 (setting B D u.A/), V0 is gŒp irreducible and V Š u.gŒp / ˝u.gŒp / V0 as gŒp -modules. If gŒp D gŒp , then the ideal ŒgŒp ; A acts nilpotently on V , hence annihilates this irreducible module. We obtain the contradiction A F IdV . Therefore gŒp ¤ gŒp . Also,

dim V D p dim gŒp =gŒp dim V0 : As dim V < p 4 one gets 0 ¤ dim gŒp =gŒp 3. We have shown that Œr; A D ¹0º, and this implies r gŒp . Since ker D r, .gŒp / is a proper subalgebra of H.2I 1/.2/ of codimension 3. Then .gŒp / H.2I 1/.2/ .0/ by Lemma 11.5.3, whence gŒp is a subalgebra of gŒp;.0/ . Note that gŒp;.0/ is the unique subalgebra containing r of maximal dimension with this property. Therefore gŒp;.0/ is a restricted subalgebra. Set V1 WD u.gŒp;.0/ / ˝u.gŒp / V0 u.gŒp / ˝u.gŒp / V0 Š V: Note that dim V D p dim gŒp =gŒp;.0/ dim V1 D p 2 dim V1 , whence dim V1 < p 2 . This proves assertion 1. (e) Now consider the case dim V < p 3 . The above gives the stronger estimate 0 ¤ dim gŒp =gŒp 2 and this gives gŒp D gŒp;.0/ and V1 D V0 , dim V0 < p. Let x be an arbitrary element of gŒp;.1/ and xs the semisimple part of x in gŒp . Suppose x is not Œp-nilpotent. Then F xs generates a nonzero torus t. As TR.g/ D 1 by assumption, then F xs is a torus of maximal toral rank, whence CgŒp .xs / is a CSA. Since H.2I 1/.2/ .1/ acts nilpotently on H.2I 1/.2/ , one has Œxs ; gŒp r. As ad xs is semisimple, gŒp D CgŒp .xs / ˚ Œxs ; gŒp D CgŒp .xs / C r: But then gŒp would be solvable. Consequently, gŒp;.1/ is Œp-nilpotent. This result shows that the eigenvalue function W A ! F extends to gŒp;.1/ , i.e., x .x/IdV is nilpotent for any x 2 gŒp;.1/ . For u 2 gŒp;.0/ and v 2 gŒp;.1/ one has 0 D traceŒu; vjV0 D .dim V0 /.Œu; v/: As dim V0 < p, we derive that vanishes on ŒgŒp;.0/ ; gŒp;.1/ . In other words, ŒgŒp;.0/ ; gŒp;.1/ acts nilpotently on V . The second assertion follows.

11.5


157

Next we describe lower bounds for the dimensions of modules of central extensions of H.2I 1/.2/ . Lemma 11.5.5. Let g be the Poisson central extension of H.2I 1/.2/ as described in j Theorem 11.4.7(2). Suppose W g ! gl.V / is a representation such that all .x1i x2 / for i C j 3 are nilpotent transformations. Assume, moreover, that no g-submodule of V is 1-dimensional. Let V0 denote an irreducible g.0/ -module. (1) Suppose .x22 / is invertible on V0 . Then we have dim V0 D p and for every basis .v0 ; : : : ; vp 1 / of V0 the family .x2 /i vj ; 0 i; j p 1 is linearly independent. (2) Suppose V0 D F v0 is 1-dimensional. The family .x1 /i .x2 /j v0 ; 0 i; j p 1; 0 i C j 2p

3;

is linearly independent. (3) Suppose .x22 / is not invertible on V0 and dim V0 2 holds. There is v0 2 V0 such that .x1 x2 /v0 2 F v0 ;

v1 WD .x22 /v0 ¤ 0;

.x22 /v1 D 0:

For any such choice of v0 the family .x1 /i .x2 /j v1 [ .x1 /k v0 ; 0 i p 1; 0 j; k p 2; i Cj 2p 4; is linearly independent. Proof. By assumption, j

g D span ¹x1i x2 j 0 i C j 2p

3; 0 i; j p

1º

and the Lie multiplication of g is given by j

¹x1i x2 ; x1k x2` º D .i`

j k/x1i Ck

1 j C` 1 x2 :

Note that ker C.g/ D F . P j Recall that g.0/ D F C i Cj 2 F x1i x2 is a subalgebra containing the ideal P j g0.1/ WD i Cj 3 F x1i x2 , and g0.1/ acts nilpotently on V . Choose an irreducible g.0/ module V0 of V . Engel’s theorem shows that j

.x1i x2 /V0 D ¹0º

if i C j 3:

./

In particular, V0 is an irreducible module for g.0/ =g0.1/ Š gl.2/. Then it is an irreducible module for the algebra F.x22 / ˚ F.x1 x2 / ˚ F.x12 / Š sl.2/.

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P For every w 2 i Cj k .x1 /i .x2 /j V0 we write w D 0.k/. (1) Suppose .x22 / is invertible on V0 . As V0 is an irreducible sl.2/-module, its dimension is p (Theorem 5.3.1). Assume that the assertion is not true. There is a nontrivial relation (for some k p 1) p X1 ˛i .x2 /k vi D O.k 1/: i D0

Since

j .x1i x2 /V0

D ¹0º if 3 i C j 2p pX1

.x1k x22 /

0D

k

3, one obtains

˛i .x2 / vi D

pX1

kŠ.x22 /

i D0

˛i vi ;

iD0

whence all ˛i vanish, a contradiction. (2) Suppose V0 D F v0 is 1-dimensional, and assume that the assertion is not true. Then there is a nontrivial relation X ˛i .x1 /i .x2 /j v0 D O.k 1/; k 2p 3: i Cj Dk

If ˛p 1 ¤ 0, then multiply with .x22 / (which annihilates V0 ) to obtain a new relation in which ˛p 1 D 0. Then multiply with suitable powers of .x1 / and .x2 / and obtain a relation .x1 /p 2 .x2 /p 1 v0 D O.2p 4/: j

Applying Lemma 2.1.5(1) and observing .x1i x2 /V0 D ¹0º if i C j 2 we derive p 1 p 2 x2 /.x1 /p 2 .x2 /p 1 v0

O.1/ D .x1

D . 1/2p „

4

4

.p

p 1 p 2 x2 // v0

..x1

…

1/.x2 / .ad .x2 //p 2 .ad .x1 //p ƒ‚

2

p 1 p 2 x2 // v0

..x1

b1 Dp 2; b2 Dp 2

2 .p

2/Š. 1/p

.p

2/Š. 1/p

D .p

3

b1 Dp 3; b2 Dp 1

C . 1/2p „ D

2/.x1 / .ad .x2 //p 1 .ad .x1 //p ƒ‚

.p

1 2

.p

1/Š.x1 /.x2 /v0

.p

1/Š.x2 /.x1 /v0

2/Š.x1 /.x2 /v0 C .p

2/Š.1/v0

and then 0 D .x12 x2 /.x1 /.x2 /v0 D

2.x1 /v0 :

But then .g/v0 D 0, and F v0 would be a 1-dimensional g-submodule of V .

…

11.5


159

(3) Since V0 is an irreducible module for F.x22 / ˚ F.x1 x2 / ˚ F.x12 / Š sl.2/ and dim V0 ¤ 1 and .x22 / is nilpotent on V0 , there is v0 as required. Next proceed as in the former cases and assume that there is a nontrivial relation X ˛i .x1 /i .x2 /j v1 C ˇ.x1 /k v0 D O.k 1/: i Cj Dk

Here we have i p 1, j p 2, i C j 2p 4, and k p 2 if ˇ ¤ 0. Let .x1 x2 /v0 D v0 . Then .x1 x2 /v1 D . C 2/v1 . Considering eigenvectors with respect to .x1 x2 / we find even more such a relation ˛.x1 /i .x2 /k i v1 C ˇ.x1 /k v0 D O.k

1/;

where i C .k i/ C 2 k.p/ if ˇ ¤ 0. Consider first the case that ˇ ¤ 0. Then 2.k i/ 2.p/, and this gives k i D p 1. However, we assume that the power of .x2 / does not exceed p 2. Then ˛ D 0. We get using () (recall that x1 x2kC1 2 g) 0 D .x12 x2k /.x1 /k v0 D . 1/k kŠ.x12 /v0 ; 0 D .x1 x2kC1 /.x1 /k v0 D . 1/k .k C 1/Š.x1 x2 /v0 : But then F v1 is a 1-dimensional module for g.0/ , contradicting the assumption that V0 is g.0/ -irreducible and not 1-dimensional. As a consequence, there is a nontrivial relation .x1 /i .x2 /k i v1 D O.k 1/. Multiplying with suitable powers of .x1 / and .x2 / one obtains a nontrivial relation .x1 /r .x2 /s v1 D O.2p Note

5/;

.r; s/ 2 ¹.p

that x1sC1 x2r 2 g. Applying Lemma V0 3 .x1sC1 x2r /.x1 /r .x2 /s v1

1; p

3/; .p

2; p

2/º:

2.1.5(1) and observing ./ we derive

D . 1/rCs .ad .x2 //s .ad .x1 //r ..x1sC1 x2r // v1 ƒ‚ … „ b1 Dr; b2 Ds

C . 1/r „

1Cs

C . 1/rCs „

r.x1 / .ad .x2 //s .ad .x1 //r ƒ‚ b1 Dr 1; b2 Ds

1

..x1sC1 x2r // v1 …

1 s.x2 / .ad .x2 //s 1 .ad .x1 //r ..x1sC1 x2r // v1 ƒ‚ … b1 Dr; b2 Ds 1

D . 1/r .rŠ/.s C 1/Š.x1 /v1 C . 1/r C . 1/r s.rŠ/

1

r.rŠ/.s C 1/Š.x1 /.x1 x2 /v1

.s C 1/Š .x2 /.x12 /v1 2Š

D . 1/r .rŠ/.s C 1/Š .x1 /

s r.x1 /.x1 x2 / C .x2 /.x12 / v1 : 2

160

11

1-sections

Applying .x13 / brings us to the equation .x12 /2 v1 D 0 and applying .x12 x2 / gives s r.x1 /.x1 x2 / C .x2 /.x12 / v1 2 2 2 .x1 / C r.x1 /.x1 x2 / C s.x1 x2 /.x12 / v1 :

0 D .x12 x2 / .x1 / D

Next observe that . 21 .x22 /; .x1 x2 /; 21 .x12 // is an sl.2/-triple and V0 is an irreducible module for this algebra of dimension dim V0 2. Moreover, v1 is a highest weight vector and .x12 /2 v1 D 0. Then dim V0 D 2 and .x1 x2 /v1 D v1 . The final equation then yields 1 C r s D 0. However, in both possible cases we have r s 2 ¹2; 0º, a contradiction. Theorem 11.5.6. Suppose g=C.g/ Š H.2I 1/.2/ . Let W g ! gl.V / be a faithful representation. Then dim V p 2 2. Proof. Let V ¹0º be a g-composition series, and let Ni denote the induced representation on Vi =Vi C1 . If ker Ni is non-solvable, then ker Ni C C.g/=C.g/ Š H.2I 1/.2/ . In this case g= ker Ni is solvable. Consequently, the faithfulness of the representation shows that there is a composition factor for which ker Ni is solvable. Then ker Ni C.g/. It suffices to prove the claim for Ni .g/. So we may assume that V is an irreducible module. Suppose the central extension F ,! g H.2I 1/.2/ is as in Case (3)P of Theorem p 1 11.4.7. Then g contains the p-dimensional Heisenberg algebra F z C i D1 F x2i . The representation theory of Heisenberg algebras (see Volume 1, p. 149) shows that dim V p .p 1/=2 p 2 . So we may assume that the central extension is not of that type. Then g is the split central extension (with the possibility that C.g/ D ¹0º) or the Poisson extension. In the first case we put gQ the Poisson extension of g.1/ and Q the representation defined by .C.Q Q g// D ¹0º, and Q D on g.1/ . Skipping the assumption that is faithful but assuming that ker C.g/ we may assume that g is the Poisson extension of H.2I 1/.2/ in any case. Due to Lemma 11.5.1 we may assume that all j .x1i x2 / for i C j 3 are nilpotent. Therefore Lemma 11.5.5 applies and shows the following: if .x22 / is invertible on V0 , then dim V p 2 ; if V0 is 1-dimensional, then dim V p 2 1; if none of these cases holds, then dim V is bounded from below by ˇ ˇ ˇ ˇ ˇ¹.i; j / j 0 i p 1; 0 j p 2; i C j 2p 4ºˇ C ˇ¹k j 0 k p 2ºˇ ˇ ˇ D ˇ¹.i; j / j 0 i p 1; 0 j p 2º n ¹.p 1; p 2/ºˇ C .p 1/ D p.p This is the claim.

1/

1Cp

1 D p2

2:

Theorem 11.5.7. Suppose g D F x2 @2 ˚ H.2I 1/.2/ and t g is a 2-dimensional torus. Let W g ! gl.V / be a faithful irreducible restricted representation. Then j.V; t/j p 2 2.

11.5


161

Proof. Theorem 10.7.1 shows that it suffices to prove the theorem for t D F x1 @1 ˚ F x 2 @2 . sum of all trivial H.2I 1/.2/ -submodules. If W ¤ ¹0º, then V D P Let W be the .2/ i i 0 .x2 @2 / W , and, obviously, V is annihilated by H.2I 1/ . As this is not true by assumption, we have W D ¹0º. Let gQ denote the Poisson central extension of H.2I 1/.2/ with induced representation Q W gQ H.2I 1/.2/ ! .H.2I 1/.2/ /: j

Since is a restricted representation, .DH .x1i x2 // is nilpotent if i C j 3. Then j .x Q 1i x2 / is nilpotent if i C j 3. Since every 1-dimensional gQ -submodule gives rise to a trivial H.2I 1/.2/ -submodule, no such submodules exist. Then the assumptions of Lemma 11.5.5 are satisfied. Choose an irreducible F x2 @2 ˚ H.2I 1/.2/ .0/ -submodule V0 of V . Then V0 is j annihilated by the Lie algebra spanned by the elements .DH .x1i x2 // for i C j 3 .2/ (which is H.2I 1/ .1/ ). Therefore V0 is an irreducible module for the Lie algebra F x2 @2 ˚ H.2I 1/.2/ .0/ =H.2I 1/.2/ .1/ Š gl.2/. But then .x1 @1 C x2 @2 /jV0 2 F IdV0 and V0 is irreducible even as an H.2I 1/.2/ .0/ -submodule. We consider in detail the 3 j j possibilities mentioned in Lemma 11.5.5. Recall that .x Q 1i x2 / D .DH .x1i x2 // for all i; j . (1) Since is a restricted representation, .x Q 22 / D .DH .x22 // is nilpotent. Therefore this case does not occur. (2) Suppose V0 D F v0 is 1-dimensional. Then v0 is a t-weight vector. Also, DH .x1 / and DH .x2 / are t-eigenvectors for independent roots. Therefore there are at least p 2 1 different weights on V . (3) Define ˛ and ˇ dual to x1 @1 and x2 @2 . Note that v0 is an eigenvector for both .x1 @1 C x2 @2 / and .DH .x1 x2 //, whence it is a t-weight vector. Say v0 2 V . Then v1 D .DH .x22 //v0 2 VCˇ ˛ and .DH .x1 //i .DH .x2 //j v1 2 VCˇ

˛ j˛ iˇ ;

.DH .x1 //k v0 2 V

kˇ :

Lemma 11.5.5 shows that all these vectors are nonzero if 0i p

1;

0 j; k p

2;

.i; j / ¤ .p

1; p

2/:

Therefore all ¹ C j˛ C iˇ j 0 i p [ ¹

kˇ j 0 k p

1; 1 j p

1; .i; j / ¤ .2; 1/º

2º

are weights. The only missing weights are C ˛ C 2ˇ and C ˇ.

Chapter 12

Sandwich elements and rigid tori

The purpose of Chapters 13 and 14 is to classify the simple Lie algebras over F having absolute toral rank 2. The idea of a proof is to construct a filtration such that the Recognition Theorem 5.6.2 applies. Filtrations are closely connected with sandwich elements. In this chapter we give sufficient conditions for the existence of sandwich elements which are, in addition, eigenvectors with respect to a given 2-dimensional torus. The methods applied in this Chapter 12 mainly go back to ideas of A. A. Premet [Pre 87, Pre 94] and A. I. Kostrikin as a predecessor.

12.1

Deriving identities

For an arbitrary finite dimensional Lie algebra g and a torus t Der g set Nk .g; t/ WD ¹x 2 g n ¹0º j .ad x/k D 0; t.x/ F xº: We will need some subtle identities in the following. In order to simplify notations write X for ad x. Recall that ! m X m m m j m ad.X .y// D .ad X/ .Y / D X j YX m j ; (12.1.1) . 1/ j j D0

which is true for arbitrary elements x; y in any Lie algebra g (cf. [S-F, Proposition 1.1.3] with A WD End g). Proposition 12.1.1. Let x; y 2 g and assume X k D 0. k 1 (1) If 4 k p 1, then ad.X k 1 .y// D 0. In particular, if Np then N3 .g; t/ ¤ ; holds. (2) If k D 3, then the following holds: X 2 YX D XYX 2 ; X 2 YX 2 D 0; 2 ad.X 2 .y// D X 2 Y 2 X 2 ; 3 ad X 2 .y// D 0:

1 .g; t/

¤ ;,

(12.1.2) (12.1.3) (12.1.4) (12.1.5)

12.1

163

Deriving identities

(3) If x 2 N2 .g; t/ and y is a t-eigenvector, then Œx; y 2 N3 .g; t/ [ ¹0º. (4) If N2 .g; t/ ¤ ;, then there is c 2 N2 .g; t/ satisfying ŒC; ŒC; Der g D ¹0º: Proof. Suppose X k D 0. We quote the following identities from [S-F, Lemma 3.1.2]: XYX

(a)

k 1

k X1

X

(b) (c)

Xk

1

YX k

2

YX

YX k

1;

D0

if 2 k p

1;

1

2

D0

if 4 k p

1:

1 .y// k 1

(1) Expanding ad.X k one gets an equation ad.X k

k 1

k .y//

1

j

if 3 k p

D

j D2 k 1

! 1 k X j YX k . 1/ k j j

D .ad X/k

X

D

1 .Y / k 1

and observing X k D 0

˛.i1 ; : : : ; ik /X i1 YX i2 YX ik

.i1 ;:::;ik /

with suitable ˛.i1 ; : : : ; ik / 2 F and (d)

k X

il D .k

1/2 ;

lD1

(e)

il k

1

for 1 l k:

Let .i1 ; : : : ; ik / be such that there is l with il D 0. By (d) and (e) we obtain .i1 ; : : : ; ik / D .k 1; : : : ; k 1; 0; k 1; : : : ; k 1/, and the summand corresponding to .i1 ; : : : ; ik / contains a factor X k 1 YX k 1 as k 4. Such a summand vanishes according to (b). Consequently, we may assume that il ¤ 0 for all l. Using (a) we may assume that il k 2 for 2 l k. Then (d) and (e) give .k

1/2 D i1 C

k X

il i1 C .k

1/.k

2/:

lD2

Hence i1 D k 1, as well as il D k 2 for all l 2. Now (c) proves the equation .ad X k 1 .y//k 1 D 0. Suppose inductively that Nk .g; t/ ¤ ; for some k 4. Let x 2 Nk .g; t/. If k X 1 D 0, then x 2 Nk 1 .g; t/. Otherwise, there is y 2 g for which X k 1 .y/ ¤ 0. Since x is a t-eigenvector, there is a t-eigenvector y with this property. Then X k 1 .y/ 2 Nk 1 .g; t/.

164

12


(2) Suppose X 3 D 0. Equation (a) for k D 3 is Equation (12.1.2). Multiplying this by X gives Equation (12.1.3). Using these results one obtains (see Equation (12.1.1)) 2 ad.X 2 .y// D .X 2 Y 2XYX C YX 2 /2 D

2X 2 YXYX C X 2 Y 2 X 2 C 4XYX 2 YX

2XYXYX 2

D X 2Y 2X 2; which is Equation (12.1.4). Finally, 3 2 ad.X 2 .y// D ad.X 2 .y// ı ad.X 2 .y// D .X 2 Y

2XYX C YX 2 /X 2 Y 2 X 2 D 0:

(3) Take x 2 N2 .g; t/, let y be a t-eigenvector and set z WD Œx; y. Equation (b) above for k D 2 yields XYX D 0. As X 2 D 0, one obtains Z 3 D .XY

YX/3 D .XY

YX/. XY 2 X/ D 0:

Therefore z 2 N3 .g; t/ [ ¹0º. (4) Take any x 2 N2 .g; t/. If ŒX; ŒX; Der g D ¹0º, then put c WD x. Otherwise there is d 2 Der g for which adŒx; d.x/ D

Œad x; Œad x; d ¤ 0:

Set c WD Œx; d.x/ 2 g, which means C D Œad x; Œad x; d D .adDer g X/2 .d /. Observe that ŒX; ŒX; ŒX; Der g ad.Œx; Œx; g/ D ¹0º, whence .adDer g X/3 D 0. Now apply Equation (12.1.4) for X instead of x and d instead of y (which are elements in Der g). One obtains that .adDer g C /2 D .adDer g X/2 .adDer g d /2 .adDer g X/2 ; and this gives ŒC; ŒC; Der g D .adDer g X/2 .Œd; Œd; Œad x; Œad x; Der g/ ŒX; ŒX; ad g D ad.Œx; Œx; g/ D ¹0º:

We need some more identities if k D 2. Proposition 12.1.2. Let c 2 N2 .g; t/ and x; y 2 g. The following identities hold: CXC D 0; 2

.adŒc; x/ D

(12.1.6) 2

CX C;

(12.1.7)

CXY C D C YXC; 2

CX CXY C D

(12.1.8) 2

CXY CX C;

(12.1.9)

CX 2 CX 2 C D 0;

(12.1.10)

.ad.CX 3 .c///2 D 0;

(12.1.11)

12.1

165

Deriving identities

ŒCX 3 .c/; C Y 3 .c/ D 0; 2

2

(12.1.12) 2

2

CX CX Y C Y C D C Y CXY CX C; .adŒŒc; x; y/m ı C D .CXY /m C; 4CX Y C YXC D 3

2

2

CX C Y C 2

(12.1.13)

m > 0; 2

(12.1.14) 2

C Y CX C;

2

2

(12.1.15) 2

2.CXY / C D CX C Y CXY C D C Y CX CXY C:

(12.1.16)

Proof. Equation (12.1.6) is the statement of [S-F, Lemma 3.1.2(2)] for k D 2. Then .adŒc; x/2 D .CX XC /2 D CX 2 C . This is (12.1.7). Substitute in (12.1.6) x by Œx; y to obtain 0 D C ı .adŒx; y/ ı C D CXY C

C YXC;

whence (12.1.8). Next, (12.1.6) gives C ı .adŒy; c/ D .adŒy; c/ ı C D 0. Using (12.1.1) one gets 0 D C ı .ad.X 3 .Œy; c/// ı C D

C ı .adŒy; c/ ı X 3 C C 3CX ı .adŒy; c/ ı X 2 C 3CX 2 ı .adŒy; c/ ı XC C CX 3 ı .adŒy; c/ ı C

D 0 C 3CXY CX 2 C C 3CX 2 C YXC C 0: Observing (12.1.8) this gives (12.1.9), and specializing y D x gives (12.1.10). Next (12.1.1) and (12.1.6) imply ad.CX 3 .c// D C ı ad.X 3 .c// ad.X 3 .c// ı C D C. CX 3 C 3XCX 2

3X 2 CX C X 3 C /

. CX 3 C 3XCX 2 D 2CX 3 C

3CX 2 CX

3X 2 CX C X 3 C /C

3XCX 2 C:

From this one gets (applying (12.1.6) and (12.1.10)) 2 ad.CX 3 .c// D .2CX 3 C 3CX 2 CX 3XCX 2 C /2 D 4.CX 3 C /.CX 3 C /

6.CX 3 C /.CX 2 CX/

6.CX 3 C /.XCX 2 C /

6.CX 2 CX/.CX 3 C /

C 9.CX 2 CX/.CX 2 CX/ C 9.CX 2 CX/.XCX 2 C / 6.XCX 2 C /.CX 3 C / C 9.XCX 2 C /.CX 2 CX/ C 9.XCX 2 C /.XCX 2 C / D 0:

166

12


This is (12.1.11). Next we obtain (using (12.1.6)) ŒCX 3 .c/; C Y 3 .c/ D ad.CX 3 .c// .C Y 3 .c// D .2CX 3 C

3CX 2 CX

3XCX 2 C /C Y 3 .c/

D 0: This is Equation (12.1.12). Again using the former equation one gets ad.CX 3 .c// ı ad.C Y 3 .c// D .2CX 3 C

3CX 2 CX

3XCX 2 C /.2C Y 3 C

3C Y 2 C Y

3Y C Y 2 C /

D 9CX 2 CXY C Y 2 C: From (12.1.12) we conclude ad.CX 3 .c// ı ad.C Y 3 .c// D ad.C Y 3 .c// ı ad.CX 3 .c// ; and this in combination with the preceding equation gives CX 2 CXY C Y 2 C D C Y 2 C YXCX 2 C: Now (12.1.13) follows from (12.1.8). Equation (12.1.14) is easily proved by induction on m, namely .adŒŒc; x; y/ ı C D CXY C holds by (12.1.6) and then .adŒŒc; x; y/m ı C D .adŒŒc; x; y/m 1 ı CXY C D .CXY /m 1 CXY C follows. Expanding C ı ad.X 2 Y 2 .c// ı C D 0 (which is (12.1.6)) one obtains 0 D C ı ad.X 2 Y 2 .c// ı C D C X 2 ı ad.Y 2 .c// 2X ı ad.Y 2 .c// ı X C ad.Y 2 .c// ı X 2 C D CX 2 C Y 2 C C 4CXY C YXC C C Y 2 CX 2 C: This is (12.1.15). Applying (12.1.8) gives 4CXY CXY C D

CX 2 C Y 2 C

C Y 2 CX 2 C;

and then applying (12.1.8), (12.1.9) and (12.1.13) to this equation 4.CXY /3 C D 4.CXY CXY C /XY C D . CX 2 C Y 2 C

C Y 2 CX 2 C /XY C

D CX 2 CXY C Y 2 C C C Y 2 CXY CX 2 C D 2CX 2 CXY C Y 2 C D 2C Y 2 CXY CX 2 C; which is the remaining Equation (12.1.16).

Corollary 12.1.3. Suppose C.g/ D ¹0º and Cg .t/ is nilpotent. If N2 .g; t/ ¤ ;, then there is a nonzero root 2 .g; t/ for which N2 .g; t/ \ g ¤ ;.

12.1

167

Deriving identities

Proof. (a) Suppose the statement is not true, whence N2 .g; t/ Cg .t/ DW h. The nilpotency of h gives rise to a natural number d such that N2 .g; t/ \ d .h/ ¤ ;, N2 .g; t/ \ d C1 .h/ D ;. Choose c 2 N2 .g; t/ \ d .h/. Let ˛ be a nonzero root and ˇ be an arbitrary root, let x 2 g˛ and y 2 gˇ be arbitrary elements. 2 (b) By Equation (12.1.11), ad.CX 3 .c// D 0. Note that CX 3 .c/ 2 g3˛ . The general assumption then yields CX 3 .c/ D 0. Using Equation (12.1.1) this gives 0 D ad.CX 3 .c// D ŒC; ad.X 3 .c// D ŒC; X 3 C D 2CX 3 C

3CX 2 CX

3X 2 CX C 3XCX 2

CX 3

3XCX 2 C:

Multiplying the above equation from the left by C Y and observing Equation (12.1.6) one gets C YXCX 2 C D 0: Equation (12.1.15) in conjunction with (12.1.10) then gives CX 2 C Y 2 CX 2 C D . 4CXY C YXC

C Y 2 CX 2 C /X 2 C D 0:

./

Next we observe that u WD Œc; x satisfies .ad u/3 D 0 (by Proposition 12.1.1(3)), and this gives (see Equations (12.1.2) and (12.1.3))) .adŒc; x/2 ı Y ı .adŒc; x/ D .adŒc; x/ ı Y ı .adŒc; x/2 ; .adŒc; x/2 ı Y ı .adŒc; x/2 D 0: From these equations we conclude by Equations (12.1.1), (12.1.7) and ./ 2 ad .adŒc; x/2 .y/ 2 D .adŒc; x/2 ı Y 2.adŒc; x/ ı Y ı .adŒc; x/ C Y ı .adŒc; x/2 D .adŒc; x/2 ı Y ı .adŒc; x/2 ı Y 2.adŒc; x/2 ı Y ı .adŒc; x/ ı Y ı .adŒc; x/ C .adŒc; x/2 ı Y 2 ı .adŒc; x/2 2.adŒc; x/ ı Y ı .adŒc; x/3 ı Y C 4.adŒc; x/ ı Y ı .adŒc; x/2 ı Y ı .adŒc; x/ 2.adŒc; x/ ı Y ı .adŒc; x/ ı Y ı .adŒc; x/2 C Y ı .adŒc; x/4 ı Y

2Y ı .adŒc; x/3 ı Y ı .adŒc; x/

C Y ı .adŒc; x/2 ı Y ı .adŒc; x/2 D. 2C4

2/.adŒc; x/ ı Y ı .adŒc; x/2 ı Y ı .adŒc; x/

C .adŒc; x/2 ı Y 2 ı .adŒc; x/2 D CX 2 C Y 2 CX 2 C D 0:

168

12


(c) If there are such x; y for which .adŒc; x/2 .y/ ¤ 0, then, by the preceding computation, .adŒc; x/2 .y/ 2 N2 .g; t/ \ g2˛Cˇ . The general assumption implies ˇ D 2˛. Therefore .adŒc; x/2 .y/ D

CX 2 C.y/ 2 Œc; h d C1 .h/;

whence .adŒc; x/2 .y/ 2 N2 .g; t/ \ d C1 .h/. This contradicts the choice of d . Consequently, .adŒc; x/2 D 0 for every x 2 g˛ where ˛ is any nonzero root. If there is such x satisfying Œc; x ¤ 0, then P Œc; x 2 N2 .g; t/ \ g˛ . This Pagain contradicts our general assumption. Hence Œc; ˛¤0 g˛ D ¹0º. Put I WD ˛¤0 g˛ C P Œg ˛¤0 ˛ ; g ˛ and Z WD Cg .I /. Since I is a t-invariant ideal of g, so are Z and Z \ I . Note that Z \ I is an abelian ideal, whence every t-root vector of Z \ I is contained in N2 .g; t/. The definition of I and the general assumption imply Z \ g D Z \ I \ g D ¹0º for all nonzero . Consequently, Z h. Note that c 2 Z, whence Z is a nonzero ideal of h. As h is nilpotent, there is a nonzero z 2 Z satisfying Œz; h D ¹0º. But then Œz; g D Œz; h C I D ¹0º. However, we assumed C.g/ D ¹0º, and this contradiction proves the corollary. There is the following important subset of N3 .g; t/: E.g; t/ WD ¹x 2 g n ¹0º j .ad x/3 D 0; t.x/ F x; x 2 .ad x/2 .g/º: Every element e of E.g; t/ fits into an sl.2/-triple .e; h; f / (Proposition 5.3.5). We are interested in sl.2/-triples which consist of t-eigenvectors. Such an sl.2/-triple is called a t sl.2/-triple. Note that necessarily Œt; h D ¹0º holds. Proposition 12.1.4. Suppose Np and the following holds:

1 .g; t/

¤ ; and N2 .g; t/ D ;. Then E.g; t/ ¤ ;

(1) Every element e 2 E.g; t/ fits into a t

sl.2/-triple .e; h; f /.

(2) There is a decomposition into t-invariant spaces gDg where gi D ¹x 2 g j .ad h

2

˚g

1

˚ g0 ˚ g1 ˚ g2 ;

i Id/dim g .x/ D 0º.

(3) The mappings .ad e/2 W g

2

! g2 ;

.ad f /2 W g2 ! g 2 ;

ad e W g

1

! g1 ;

ad f W g1 ! g

are bijections. In particular, .ker.ad e//\g hold.

2

1

D ¹0º and .ker.ad f //\g2 D ¹0º

(4) Œg2 ; g2 D Œg2 ; g1 D ¹0º, .ad v/2 .g/ g2 for every v 2 g2 and .ad e/2 .g/ D g2 .

12.1

169

Deriving identities

Proof. According to Proposition 12.1.1 N3 .g; t/ ¤ ;. Choose x 2 N3 .g; t/ ¤ ; such that dim X 2 .g/ is minimal. As N2 .g; t/ D ;, there is a t-eigenvector y 2 g satisfying X 2 .y/ ¤ 0. Equation (12.1.5) gives X 2 .y/ 2 N3 .g; t/ and Equation (12.1.4) shows 2 ad.X 2 .y// .g/ X 2 .g/. The minimality assumption implies 2 ad.X 2 .y// .g/ D X 2 .g/ 3 X 2 .y/: This means X 2 .y/ 2 E.g; t/ and E.g; t/ ¤ ;. (1) Let e 2 E.g; recall that it fits into an sl.2/-triple .e; h; f /. P t/ be arbitrary and P Decompose h D ˛2.g;t/ h˛ , f D ˛2.g;t/ f˛ into the sum of t-eigenvectors. Since e is a t-eigenvector by definition, say e 2 g , one has that Œh0 ; e D 2e, h0 D Œe; f . The proof of Proposition 5.3.5 essentially shows that there is y 2 ker.ad e/ such that (with u D f ) Œh0 ; y C 2y D .Œh0 ; u C 2u/. As h0 and u D f are t-homogeneous, one may take y D y homogeneous. Setting f 0 WD y C u one obtains that .e; h0 ; f 0 / is an sl.2/-triple as well. This means that we may take e; h; f as t-root vectors. (2) Set s WD F e C F h C Ff Š sl.2/ and let U be an s-composition factor of any composition series of g and U the induced representation. Then U .e/3 D 0 D U .e/dim U and Theorem 5.3.1(2) shows the following1 : U D ˚kiD0 F ui and U .f /.ui / D ui C1 U .h/.ui / D .˛ U .e/.ui / D i.˛

.0 i k .0 i k/;

2i/ui i C 1/ui

1/;

1

.1 i k/:

Moreover, if dim U < p then ˛ k D dim U 1. If k 2, then dim U < p and therefore ˛ k. The only eigenvalues of U .h/ are contained in ¹ k; : : : ; kº D ¹˙2; ˙1; 0º. Next consider the case that k 3 (one always has k C 1 D dim U p). For 3 i k one gets 0 D U .e/3 .ui / D i.˛

i C 1/.i

1/.˛

i C 2/.i

2/.˛

i C 3/ui

3:

In particular, ˛ 2 Fp . Choose an inverse image ` 2 ¹0; : : : ; p 1º for which ˛ `. Taking i D 3 gives ` 2. Taking i D k then gives k 2 ¹` C 1; ` C 2; ` C 3º, whence k ¤ ` and k 5. But then ˛ 6 k and this gives dim U D p. In this case 6 k C1 D dim U D p, i.e., p D 5. Since ˛ 2 Fp , all eigenvalues of U .h/ are contained in F5 D ¹˙2; ˙1; 0º. We mention for the next step that 4 D k 2 ¹` C 1; ` C 2; ` C 3º and this implies ` ¤ 0. Consequently, only ` D 1; 2 are possible. (3) We proved in (2) that every composition factor U is at most 3-dimensional provided p > 5 or dim U < p. Assertion (3) then holds for composition factors (Theorem 5.3.1). Now suppose dim U D p D 5. Recall that only ` D 1; 2 are 1 see

also the chapter “Supplements to Volume 1”

170

12


possible. Next one computes that ker.ad e/ D F u0 C F u`C1 . Since u0 2 g` and u`C1 2 g ` 2 we get that ker.ad e/ g2 C g1 . Similarly, ker.ad f / F u4 g`C2 , while ` C 2 ¤ 1; 2. Assertion (3) holds for composition factors also in this case. Let g0 g00 be proper s-submodules, and assume inductively that assertion (3) is true for g0 and g00 =g0 . Since .g00 =g0 /k Š g00 k =g0 k for all k 2 ¹˙2; ˙1; 0º, the assertion follows by induction on the composition length. (4) By (3), g1 D .ad e/.g 1 / and g2 D .ad e/2 .g 2 /. Since .ad e/3 D 0, Equations (12.1.3), (12.1.2) yield for arbitrary d; d 0 2 g Œ.ad e/2 .d /; .ad e/2 .d 0 / D .ad e/2 ı .ad d / ı .ad e/2 .d 0 / D 0; Œ.ad e/2 .d /; .ad e/.d 0 / D .ad e/2 ı .ad d / ı .ad e/.d 0 / 2.ad e/ ı .ad d / ı .ad e/2 .d 0 / D

.ad e/ ı .ad d / ı .ad e/2 .d 0 /:

The first equation gives Œg2 ; g2 D ¹0º. To prove Œg2 ; g1 D ¹0º we make the following observation. Suppose that we have .ad e/.v1 / ¤ ¹0º for some v1 2 g1 . As .ad e/.v1 / 2 g3 , necessarily p D 5 and .ad e/.v1 / 2 g 2 hold. But then .ad e/2 ..ad e/.v1 // ¤ 0 by (3), a contradiction. This shows Œe; g1 D ¹0º. Next take d 2 g 2 , d 0 2 g 1 arbitrary and apply the second of the above equations. We obtain Œ.ad e/2 .d /; .ad e/.d 0 / 2 Œe; g1 D ¹0º: This proves Œg2 ; g1 D ¹0º. Finally, the former results show .ad v/2 .g/ D .ad v/2 .g 2 / g2 for v 2 g2 and .ad e/2 .g/ D .ad e/2 .g 2 / D g2 .

12.2

Sandwich elements

The set N2 .g; t/ is of particular interest. Definition 12.2.1. The elements of N2 .g; t/ D ¹x 2 g n ¹0º j .ad x/2 D 0; t.x/ F xº are called t-sandwiches of g. Sandwiches give rise to filtrations of height bigger than its depth. Namely, let c 2 g be a nonzero element satisfying .ad c/2 D 0. Then ker.ad c/ is a subalgebra of g containing Œc; g C F c. Moreover, one has Œg; Œc; g ker.ad c/ (Equation (12.1.6)). Choose a maximal subalgebra g.0/ containing ker.ad c/ and construct the

12.2

Sandwich elements

171

filtration with respect to g. 1/ WD g and g.0/ (cf. Definition 3.5.1, we do without the assumption of g. 1/ =g.0/ being g.0/ -irreducible). Then Œc; g C F c g.1/ ;

F c g.2/ ;

and this gives a filtration of depth 1 and height at least 2. This is the reason, why sandwiches play an important role in the Classification Theory. A. I. Kostrikin ([Kos 67]) showed that the existence of sandwiches distinguishes the simple restricted nonclassical Lie algebras from the classical ones, provided p > 5 and there exists a root vector x ¤ 0 with respect to a suitable torus such that .ad x/p 1 D 0. A. A. Premet ([Pre 86], [Pre 87]) extended this result by showing that a simple Lie algebra g is classical if and only if N2 .g; ¹0º/ D ;. Theorem 12.2.6 below is formulated in the spirit of Kostrikin’s result. Corollary 12.4.7 is an analogue of Premet’s result. The main difficulties for the proofs of these results indeed arise for the case p D 5. In this section we consider the following Setting 12.2: (1) Let L be simple and LŒp denote its semisimple p-envelope. (2) Let T LŒp be a 2-dimensional torus. (3) Set H WD CL .T / and assume that H is nilpotent. (4) Assume that the maximal torus of HŒp LŒp is contained in T . Setting 12.2(4) implies that every T -root extends via Equation (1.3.1) on p. 31 of Volume 1 to a root on HŒp , which though needs not be linear any more. In this section we want to prove the alternative Np 1 .L; T / D ; or N2 .L; T / ¤ ; or L is classical. In order to do so we assume from now on that Np

1 .L; T

/ ¤ ; and N2 .L; T / D ;:

Then E.L; T / ¤ ; by Proposition 12.1.4. Set Emin .g; t/ WD ¹e 2 N3 .g; t/ j dim.ad e/2 .g/ is minimalº \ E.L; T /: Lemma 12.2.2. One has Emin .L; T / ¤ ;. Take e 2 Emin .L; T / and a T sl.2/triple .e; h; f /. Let L D L 2 ˚ L 1 ˚ L0 ˚ L1 ˚ L2 be the decomposition according to Proposition 12.1.4. The following holds: (1) L2 \ L n ¹0º Emin .L; T / for all 2 .L; T /. (2) If v 2 L2 \ L n ¹0º, then .ad v/2 .L/ D .ad v/2 .L 2 / D L2 . (3) dim L2 \ L 1 for all 2 .L; T /. (4) dim L2 2 ¹1; pº; if dim L2 D p, then there are independent roots ˛; ˇ 2 .L; T / such that ˛.h/ D 2, ˇ.h/ D 0 and .L2 ; T / D ˛ C Fp ˇ.

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12


Proof. Take e 2 N3 .L; T / such that dim E 2 .L/ is minimal and a T -eigenvector v 2 E 2 .L/ n ¹0º. There is a T -eigenvector y 2 L satisfying E 2 .y/ D v. Equation 2 (12.1.5) gives v 2 N3 .L; T /. Equation (12.1.4) shows ad.E 2 .y// .L/ E 2 .L/. The minimality assumption implies 2 .ad v/2 .L/ D ad.E 2 .y// .L/ D E 2 .L/ 3 E 2 .y/ D v: This means .ad v/2 .L/ D .ad e/2 .L/ and v 2 E.L; T /. But then v 2 Emin .L; T /. (1), (2) Take arbitrary v 2 L2 n ¹0º. Arguing as before one gets v 2 Emin .L; T /. By Proposition 12.1.4(4) one has .ad v/2 .L/ D .ad v/2 .L 2 / and .ad e/2 .L/ D L2 . (3) We intend to prove that dim L2 \ L 1. So assume the contrary and let v; w be independent elements of L2 \ L . Recall that .ad f /2 establishes a vector space isomorphism from L2 to L 2 by Proposition 12.1.4. Next we observe that the determinant of the linear endomorphism .ad.v C w//2 ı .ad f /2 W L2 ! L2 ;

2 F;

is a polynomial in with highest coefficient det..ad v/2 ı .ad f /2 /. By the above this coefficient is not zero. Let 0 2 F be a zero of this nonconstant polynomial. Then .ad.0 v C w//2 W L 2 ! L2 is not surjective. But this contradicts the minimality of .ad e/2 .L/. (4) Suppose dim L2 > 1. Then there are different roots ˛; with L2 \ L˛ ¤ ¹0º, L2 \ L ¤ ¹0º by (3). Choose v˛ 2 L2 \ L˛ , v 2 L2 \ L and T sl.2/triples .v˛ ; h˛ ; u˛ /, .v ; h ; u /. Let h D hs C hn denote the Jordan–Chevalley q decomposition of h where hs D hŒp (q 0) is the semisimple part. Setting 12.2(4) implies hs 2 T . Therefore Œhs ; h˛ D 0, whence h˛ 2 L0 . Then we may choose u˛ 2 L 2 \ L ˛ . As a result, .ad u˛ /2 .L2 / L 2 . Recall from (2) that L2 D .ad v˛ /2 .L/. Then Proposition 12.1.4 applied to the triple .v˛ ; h˛ ; u˛ / shows that the mapping .ad v˛ /2 ı .ad u˛ /2 is a linear isomorphism of L2 . As dim L 2 D dim L2 , this gives .ad u˛ /2 .L2 / D L 2 . Applying these arguments for both triples shows that .ad v /2 ı .ad u˛ /2 is a linear isomorphism of L2 . It maps any root space L2 \ L onto L2 \ LC2. ˛/ . In particular, ˛ 2 .L; T / and all ˛ C Fp . ˛/ are roots on L2 . Put ˇ WD ˛. On the other hand, hs has the single eigenvalue 2 on L2 . As the full root space is spanned by two elements, at most p roots can occur on L2 . Due to (3) the claim follows. Choose e 2 Emin .L; T / and a T sl.2/-triple .e; h; f /. Let according to Proposition 12.1.4 L D L 2 ˚ L 1 ˚ L0 ˚ L1 ˚ L2 be the primary decomposition with respect to F h and recall that all Li are T -invariant. Then L0 C L1 C L2 is a T invariant subalgebra of L. Choose a maximal T -invariant subalgebra Q of L containing L0 C L1 C L2 and find an associated T -invariant standard filtration of L (see Definition 3.5.1), L D L.

r/

L.

1/

L.0/ WD Q L.s/ ¹0º;

12.2

173

Sandwich elements

where L. 1/ =Q is .T C Q/-irreducible. Note that this filtration is exhaustive because we have chosen Q maximal T -invariant. It is also separating (i.e. Q contains no proper ideals of L), because L is assumed to be simple. Put G WD gr L D ˚skD

r Gk :

Since h 2 H and H is nilpotent, one has H L0 Q. Therefore there are natural actions of T and H on G and all Gk are T -invariant as well as H -invariant. Write TN for the isomorphic image of T in Der G. For x 2 L.k/ n L.kC1/ we denote the image of x in Gk by x. N Lemma 12.2.3. The following holds: (1) 1 r D s 2; (2) if r D s D 2, then L.k/ D Lk ˚ L.kC1/ for all k D

2; : : : ; 2.

Proof. One has by construction Œe; L L0 C L1 C L2 Q, whence e 2 L.1/ and hence 1 s. If s > r, P then there is a T -eigenvector u 2 L.s/ which by choice of s satisfies Œu; Œu; L k>s L.k/ D ¹0º. This element would be a T -sandwich, a contradiction. Hence one has 1 s r. Set R WD ¹x 2 L. 1/ j Œx; L. 1/ L. 1/ º. Clearly, R is .T C Q/-invariant and contains Q. The .T C Q/-irreducibility of L. 1/ =Q forces R D L. 1/ or R D Q. In the first case the definition of a standard filtration yields r D 1, whence s D r D 1. So suppose R D Q. The .ad h/-invariance of L. 1/ gives L.

1/

D .L.

1/

\ L 2 / C .L.

1/

\ L 1 / C Q:

Since ŒL 2 ; L 2 C L 1 D ¹0º if p > 5, and this space is contained in L1 C L2 Q if p D 5, one has L. 1/ \ L 2 R. Then L. 1/ D .L. 1/ \ L 1 / C Q, and therefore hN acts on G 1 with the single eigenvalue 1. Since G0 acts faithfully on G 1 , it can only be that hN acts on G0 with the single eigenvalue 0. This means that adG0 hN is nilpotent. According to Proposition 3.5.3(1) the grading of T C G satisfies properties (g1)–(g3) of Notation 3.5.2. Then Lemma 3.5.4(2) implies that hN acts on every Gk with single eigenvalue k. Since L0 C L1 C L2 Q, therefore only G 1 and G 2 can be nonzero negative parts of the grading. This shows r 2. Moreover, ¹0º ¤ L2 Q. Then G2 ¤ ¹0º, and this gives s 2. Consequently, r D s D 2. Suppose r D s D 2. The preceding deliberations show that hN acts on every Gk with the single eigenvalue k. Then L.k/ D Lk C L.kC1/ for all k. Moreover, Lk \ L.kC1/ L.kCp/ D ¹0º since s D 2. Next we want apply Weisfeiler’s Theorem 3.5.7. By construction, TN C G satisfies conditions (g1)–(g3) of Notation 3.5.2 (see Proposition 3.5.3).

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12


If r D 1, then G1 ¤ ¹0º and the irreducibility of G 1 implies that also (g4) is N fulfilled. ideal A.TN C P Theorem 3.5.6 shows that T C G contains a unique minimal N G/ j 1, then r D s D 2. Proof. Suppose L1 ¤ ¹0º, dim L2 > 1 and r D s D 1. (a) Suppose m ¤ 0. Q p 1 is TN -invariant. Choose a ‰.TN /If 2 .‰.TN // W .mI 1/.0/ , then S1 ˝ m i D1 xi Qm p 1 eigenvector v D u ˝ i D1 xi , u 2 S1 . Then Œv; ‰.G/ S ˝ O.mI 1/.1/ and N D hence Œv; Œv; ‰.G/ D ¹0º. Choose a T -eigenvector w 2 L.1/ for which ‰.w/ v. As G is mapped injectively into .Der S/ ˝ O.mI 1/ C Id ˝ W .mI 1/, one has N G D ¹0º. But then .ad w/2 .L.i/ / L.i C3/ for all i . The latter vanishes as Œw; N Œw; i r D 1, s D 1. Then w 2 N2 .L; T /. This contradicts our assumption. Suppose 2 .‰.TN // 6 W .mI 1/.0/ . Recall that hN 2 TN . According to Theorem 3.6.1 there is a realization ‰ such that N D !.h/ N ˝ 1 C Id ˝ .h/; N ‰.t1 / D Id ˝ .1 C x1 /@1 ; TN D F hN ˚ F t1 ; ‰.h/

12.2

175

Sandwich elements

N is a toral element of Der S and .h/ N N 2 W .mI 1/.0/ . In addition, ‰.h/ where !.h/ N N is contained in S ˝ O.mI 1/, and this implies .h/ D 0 and !.h/ 2 S . Arguing as Q p 1 before, every ‰.TN /-eigenvector of S1 ˝ m x would give rise to a T -sandwich i D2 i of L. No such sandwiches exist. Therefore m D 1. As a consequence, HN \A.TN CG/ N ˝ 1. Then CS .!.h// N is nilpotent. As it contains !.h/, N it is is mapped onto CS .!.h// a CSA of S . Since H Q, there is a surjective restricted homomorphism HŒp ! HN Œp ! ‰.HN Œp / D ‰.HN /Œp : N in Der S. Then t ˝ 1 Let t denote the maximal torus of the p-envelope of CS .!.h// ‰.HN /Œp . Hence there is a torus T0 in HŒp mapped onto t ˝ 1 ([S-F, Theorem 2.4.5]). Since the maximal torus of HŒp is contained in T , we now obtain T0 T . On the other hand, since 2 .‰.TN // ¤ ¹0º D 2 .‰.T0 //, one has T0 ¤ T . As T is 2N is a CSA of S of dimensional, this gives dim t D 1, which means that CS .!.h// toral rank 1. Moreover, S is generated as an algebra by S 1 [ S1 (as it is simple) and every element x 2 S 1 [ S1 satisfies .ad x/3 D 0. Theorem 4.1.4 shows that N has only S is classical. Then S Š sl.2/ by Theorem 9.2.11. This implies that !.h/ 2 different nonzero eigenvalues, not only on S but also on ‰.G/. But then h has only 2 different nonzero eigenvalues on L. This contradicts our assumption. As a consequence, m D 0 and A.TN C G/ D S is a simple Lie algebra. (b) Note that S is generated as an algebra by S 1 [ S1 and every element x 2 S 1 [ S1 satisfies .ad x/3 D 0. Theorem 4.1.4 shows that S is classical. In particular, N D S is a restricted Lie algebra. By Lemma 12.2.2(4) there is a root ˇ such that CS .h/ P P N N N 2 Sj \ Siˇ i 2Fp CS .h/ \ Siˇ . Note that CS .h/ j 0 Sj as L0 Q. Let x (j 0, i ¤ 0). As S is restricted, there is an element yN 2 S \ HN such that xN Œp D y. N We may take inverse images x 2 L.0/ \ Liˇ ;

y 2 H L.0/ ; q

.ad x/p

ad y 2 Der.1/ L: qC1

Setting 12.2(4) implies that t WD y Œp 2 T for sufficiently large q. Then .ad x/p qC1 ad t 2 Der.1/ L. Observe that d WD .ad x/p ad t is centralized by ad T , whence qC1 p .ad x/ and ad t commute. Since d is p-nilpotent, we may increase q to obtain qC1 qC1 .ad x/p D ad t . Note that ˇ.x Œp / D 0. N Then the Engel–Jacobson Now suppose that S \ HN acts nilpotently on CS .h/. N theorem implies that CS .h/ is nilpotent. But then it is a CSA of S . The preceding arguments show that N Œp TN \ CS .h/ j j D TN \ span ¹x Œp j x 2 [i 2Fp Siˇ ; j > 0º C span ¹hŒp j h 2 S \ HN ; j 0º ker ˇ: N has toral rank 1 in S. The classification of these algebras gives S Š sl.2/ Then CS .h/

176

12


as S is classical. But then G Der S Š sl.2/, a contradiction. Hence S \ HN does N Therefore the p-envelope of S \ HN contains TN . But not act nilpotently on CS .h/. N N S \ H D CS .T / is closed under Œp-powers. Then TN S \ HN . This shows that S \ HN is a CSA of S . Due to [Sel 67, Theorem IV.1.2, Note] all CSAs are conjugate. Then S \ HN satisfies the requirements of Theorem 4.1.2. In particular, S \ HN acts as a torus on S . This gives S \ HN S0 . Find a torus T0 HŒp which is mapped onto S \ HN ([S-F, Theorem 2.4.5]). Then Setting 12.2(4) implies T0 T , whence 2 D dim TN dim S \ HN dim T0 dim T D 2: Consequently, S \ HN D TN is a 2-dimensional CSA of L. Lemma 10.6.2 [Sel 67, Lemma II.3.2] shows that for every 2 .S; TN / n ¹0º only ˙ ; 0 are roots in Fp and dim S˙ D 1. Let us turn this into information on root spaces of L. The above shows that for every 2 .L; T / n ¹0º only ˙ ; 0 are roots in Fp and dim L˙ D 1. Lemma 12.2.2(4) implies that L2 \ Lk is nonzero for some k 2 Fp . Then we have L 2 \ L k ¤ ¹0º. But then L. / D L 2 \ L k C L \ H C L2 \ Lk . As a result, L D L 2 C H C L2 L 2 C L0 C L2 . This contradicts the assumption L1 ¤ ¹0º. Lemma 12.2.5. If L1 ¤ ¹0º, then dim L2 D 1. Proof. (a) Suppose the contrary. Lemma 12.2.4 gives r D s D 2. Then Lemma 12.2.3(2) states L.k/ D Lk ˚ L.kC1/ for all k. In particular we have G

2

D L 2;

G

1

D L 1;

G0 D L0 ;

G1 D L1 ;

G2 D L2 :

We conclude ŒfN; v N D Œf; v for v 2 L2 [ L1 [ L0 and Œe; N D Œe; v for v 2 N v N 2 W G 2 ! G2 are bijections L0 [ L 1 [ L 2 . Then .ad fN/2 W G2 ! G 2 and .ad e/ (Proposition 12.1.4(3)). Take roots ˛; ˇ 2 .L; T / as in Lemma 12.2.2(4) such that .L2 ; T / D ˛ C Fp ˇ and ˇ.h/ D 0. Every root vector v 2 L2 \ L is contained in Emin .L; T / (Lemma 12.2.2(1)). Choose a T sl.2/-triple .v ; h ; u /. Then L2 \L D F v and L 2 \L D F u (as the respective root spaces are 1-dimensional by Lemma 12.2.2(3)). We mentioned N is contained in a remark preceding Lemma 12.2.4 that every such h (substituting h) in TN and acts as 2Id on .ad v /2 .L/ D G2 . Then hN h vanishes on G2 . But then ˛.hN h / D ˇ.hN h / D 0 and this gives .hN h / D 0 for all roots . Consequently, hN D h . This shows that N ŒG 2 ; G2 \ HN D F h: (b) Set M WD G

2

˚ G0 ˚ G2 :

Let I ¤ ¹0º be a minimal TN -invariant ideal of M . As hN 2 M , this ideal is a graded subspace of M , I D .I \ G 2 / ˚ .I \ G0 / ˚ .I \ G2 /:

12.2

Sandwich elements

177

Suppose I \ G 2 D ¹0º D I \ G2 . Then I G0 and ŒI; G 2 D ŒI; G2 D ¹0º. In particular, ˛ and ˇ vanish on I \ HN . As these roots span the full lattice, this means that I \ HN acts nilpotently on G. Note that TN acts on I as a 1-dimensional torus N I D ¹0º). Corollary 1.3.8 then shows that I is solvable. The minimality of (as Œh; I yields that I is abelian. Since G 1 is .TN C G0 /-irreducible by construction of the filtration and I is .TN C G0 /-invariant, G 1 is an eigenvalue module for I and some eigenvalue function (Lemma 3.2.7). But then G 2 D ŒG 1 ; G 1 is an eigenvalue module for I and the eigenvalue function 2. Since I annihilates G 2 , this gives D 0. This means that I acts nilpotently on G 1 . Engel’s theorem shows that I annihilates G 1 . The definition of the standard filtration of L gives I D ¹0º. This contradiction yields I \G

2

¤ ¹0º or I \ G2 ¤ ¹0º:

Next we recall that .ker.ad fN/2 / \ G2 D ¹0º. Therefore I \ G 2 ¤ ¹0º in any case. Theorem 3.5.6(3) applied for TN C G yields that G 2 is .TN C G0 /-irreducible. Therefore we now have G 2 I . Recall that .ad e/ N 2 .G 2 / D G2 . Therefore G 2 C G2 I , and the minimality of I yields I DG

2

˚ ŒG 2 ; G2 ˚ G2 :

In particular, I is non-abelian. Corollary 3.3.5 gives rise to a graded simple Lie algebra S 0 and m0 0 such that ‰ W I Š S 0 ˝ O.m0 I 1/; Sk0 ˝ O.m0 I 1/ Š I \ Gk ;

k D ˙2; 0;

S 0 D S 0 2 ˚ S00 ˚ S20 ; TN C M ! .Der S 0 / ˝ O.m0 I 1/ Ì Id ˝ W .m0 I 1/ : Corollary 4.1.5 shows that S 0 is classical. In particular, it is restricted. (c) Suppose m0 D 0. Then I D S 0 is classical simple. Recall that S 0 \ HN D 0 N N Recall that CS 0 .h/ N D N N ŒG CS 0 .h/. P 2 ; G2 \ NH D F h. Then SN \ H annihilates Œp 0 N N 2 S \ HN D F h. As ˇ.h/ D 0, the i 2Fp CS 0 .h/iˇ . Take x 2 CS 0 .h/iˇ . Then x N Engel–Jacobson theorem implies that CS 0 .h/ is nilpotent. Then it is a CSA of S 0 of toral rank 1. We conclude I D S 0 Š sl.2/. But then dim G2 D 1, a contradiction. As a result, m0 ¤ 0. (d) We argue word-by-word as in (a) of the proof of Lemma 12.2.4 (only substitute S1 by S20 and S 1 by S 0 2 ) to show that m0 D 1, that there is a realization N ˝ 1 C F Id ˝ .1 C x1 /@1 ; ‰.TN / D F !.h/ N 2 S 0 , and that S 0 Š sl.2/. In particular, Der S 0 D S 0 holds. where !.h/

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12


Next we consider the .TN CM /-module G 1 ˚G1 . Let V G 1 ˚G1 be a nonzero N .T C M /-submodule. Applying fN if necessary we obtain G 1 \ V ¤ ¹0º. Since G 1 is .TN C G0 /-irreducible, this gives G 1 V . Applying eN one gets G1 V . Hence G 1 ˚ G1 is .TN C M /-irreducible. k For every ‰.u/ 2 S 0 ˝O.1I 1/.1/ there is a suitable power such that .ad u/p .TN C k As G 1 is .TN C G0 /-irreducible, M / D ¹0º. In particular, .ad u/p .TN C G0 / D ¹0º. kˇ p there is an eigenvalue 2 F such that .ad u/ ˇG 1 D IdG 1 . But then we have kˇ .ad u/p ˇG 2 D 2IdG 2 D 0. This shows that D 0. Using property (g1) of Notation 3.5.2 one easily shows that u acts nilpotently on G. Therefore ‰ 1 S 0 ˝ O.1I 1/.1/ acts nilpotently on G 1 ˚ G1 . Theorems 3.3.10 and 3.6.1 combined with the fact that Der S 0 D S 0 yield a simultaneous realization (from now we suppress the notion of ‰) M S 0 ˝ O.1I 1/ Ì Id ˝ W .1I 1/ ; G

1

˚ G1 D U ˝ O.1I 1/; dim U 2; N ˝ 1 ˚ F Id ˝ .1 C x/@ : TN D F !.h/

As S 0 ˝ O.1I 1/ M , there is a subalgebra D0 W .1I 1/ such that M D S 0 ˝ O.1I 1/ ˚ Id ˝ D0 : The multiplication of G gives rise to a .TN C M /-invariant bilinear mapping ƒ W .G

1

˚ G1 / .G

1

˚ G1 / ! M;

which induces a .TN C M /-invariant bilinear mapping ƒ2 WD 2 ı ƒ W U ˝ O.1I 1/ U ˝ O.1I 1/ ! M=.S 0 ˝ O.1I 1// Š D0 : Theorem 8.2.5 yields that either ƒ2 D 0 or the image under ƒ2 is a nonsolvable N ˝ x 2 M several times gives subalgebra W of D0 . Suppose ƒ2 ¤ 0. Applying !.h/ ƒ2 .U ˝ O.1I 1/; U ˝ O.1I 1// D ƒ2 .U ˝ F; U ˝ O.1I 1//: The TN -invariance of ƒ2 yields that ƒ2 .u ˝ 1; u0 ˝ .1 C x/i / is a .1 C x/@-eigenvector in W .1I 1/ of eigenvalue i for every u; u0 2 U , whence ƒ2 .U ˝ 1; U ˝ .1 C x/i / F .1 C x/i C1 @: Since W is not solvable, there is an exponent j ¤ 0 such that .1 C x/j C1 @ 2 W .

12.2

Sandwich elements

179

Then Id ˝ .1 C x/j C1 @ 2 M and ƒ2 .U ˝ 1; U ˝ O.1I 1// p X1 k D ƒ2 U ˝ 1; Id ˝ .1 C x/j C1 @ U ˝ F .1 C x/j kD0

D

p X1

ad.1 C x/j C1 @

k

ƒ2 .U ˝ 1; U ˝ F .1 C x/j /

kD0

D F .1 C x/j C1 @: But then the image under ƒ2 is solvable, a contradiction. Hence ƒ2 D 0. N D Si if i ¤ 0. (e) We now turn this into information on S. Note that Gi D ŒGi ; h 0 From (d) we get ŒG 1 ; G1 S ˝ O.1I 1/ \ G0 D I0 , and then ŒS 1 ; S1 D ŒG 1 ; G1 I0 D ŒG 2 ; G2 D ŒS 2 ; S2 : The simplicity of S gives S0 D ŒS 1 ; S1 C ŒS 2 ; S2 D ŒS 2 ; S2 D I0 D S00 ˝ O.1I 1/ D F hN ˝ O.1I 1/. But then S0 is abelian. The simplicity of S implies that S2 is S0 -irreducible ([S-F, Theorem 3.3.7]). This gives dim L2 D dim G2 D dim S2 D 1, the final contradiction. The result of this section is Theorem 12.2.6. Under the assumption of the Setting 12.2 one and only one of the following cases occurs: (1) L is classical; in this case L is of type A2 , B2 , or G2 and T D CL .T / is a CSA which meets the requirements of Theorem 4.1.2; (2) Np

1 .L; T

/ D ;;

(3) N2 .L; T / ¤ ;. Proof. Suppose that Np 1 .L; T / ¤ ; and N2 .L; T / D ;. (a) Proposition 12.1.4 shows that E.L; T / ¤ ;, whence Emin .L; T / ¤ ;. Choose according to that proposition e 2 Emin .L; T / and a T sl.2/-triple .e; h; f / and decompose L into h-invariant spaces L D L 2 ˚ L 1 ˚ L0 ˚ L1 ˚ L2 . Suppose L1 D ¹0º. Then L 1 D Œf; L1 D ¹0º as well. We obtain L D L 2 ˚ L0 ˚ L2 , whence ŒL 2 ; L 2 D ¹0º. Therefore every x 2 L˙2 satisfies .ad x/3 D 0. This implies that exp.ad x/ is an automorphism of L. Now note that the simple Lie algebra L is generated as an algebra by this set (Proposition 1.3.5). Theorem 4.1.4 shows that L is classical. So assume L1 ¤ ¹0º. Suppose p 7. Then L˙3 D L˙4 D ¹0º and therefore every x 2 L˙2 satisfies .ad x/3 D 0. This implies that exp.ad x/ is an automorphism of L. Next consider any

180

12


x 2 L1 . Then .ad x/3 .L/ L1 C L2 , .ad x/4 .L/ L2 , .ad L/5 D 0. This gives Œ.ad x/i .L/; .ad x/p i .L/ ŒL1 C L2 ; L2 D ¹0º, whence Œ.ad x/i .L/; .ad x/j .L/ D ¹0º if i C j p: Again this implies that exp.ad x/ is an automorphism of L. By symmetry we obtain the same result S if x 2 L 1 . Consequently, exp.ad x/ is an automorphism of L for every x 2 i 2¹˙1;˙2º Li . Now note that the simple Lie algebra L is generated as an algebra by this set (Proposition 1.3.5). The proof of Theorem 4.1.4 applies word-byword2 to the present setting and shows that L is classical. So assume p D 5. Lemma 12.2.5 shows that L2 D F e. Call the respective root 2˛. Since 2˛.h/ D 2, we obtain L2˛ L2 and hence L2 D L2˛ D F e;

L

2

DL

2˛

D Ff:

Take any root ˇ independent of ˛, choose ` 2 F5 for which .`˛ C ˇ/.h/ D 0 and set ˇQ WD `˛ C ˇ. Possibly ˇQ is not a root, but the space F5 ˛ C F5 ˇQ contains the full root space. (b) Suppose ŒL 1 ; L 2 ¤ ¹0º. Choose a root vector v 2 L 1 for which Œv; f ¤ 0. Since ad f W L1 ! L 1 is bijective, there is a root vector w 2 L1; for which v D Œf; w. Then .ad f /2 .w/ D Œf; v is a nonzero element in L2; 4˛ , whence D ˛. Adjusting w 2 L1;˛ we get .ad f /4 .w/ D f . Next we determine the ˛-chain through kˇ. Let j˛ C k ˇQ be a root for k 2 F5 . Q This chain does not meet LP 2 C L 2 , and this means ˙2 ¤ .j˛ C k ˇ/.h/ D j . 2 Therefore the L.˛/-module j 2F5 Lj˛Ck ˇQ is annihilated by .ad f / . Suppose L1 carries roots only in the ˛-direction. Then so does Œf; L P1 D L 1 and we obtain L D L C L.˛/. But then the simplicity of L implies L D 0 62F5 ˛ L C P ŒL ; L L , a contradiction. Hence there is k 2 F 0 ;62F5 ˛ 5 and nonzero u 2 L˛Ck ˇQ L1 . By Equation (12.1.1) Œf; u D ad.ad f /4 .w/ .u/ ! 4 X 4 D . 1/4 j .ad f /j ı .ad w/ ı .ad f /4 j

j

.u/ D 0;

j D0

contradicting the fact that ad f W L1 ! L

1

is bijective. We now have

ŒL 1 ; L 2 D ¹0º and ŒL 2 ; L 2 D F Œf; f D ¹0º: As a consequence, .ad f /3 D .ad e/3 D 0. Then L is completely reducible as a module for Ff C F h C F e. Therefore h acts semisimply on L. 2 see


12.2

181

Sandwich elements

(c) Let u 2 L1; be an arbitrary root vector in L1 . Obviously, .ad u/4 .L/ L2 . This implies .ad u/4 D 0 unless 2˛ C 4 D 2˛, i.e., ˛ D . Suppose .ad u/4 .L/ ¤ ¹0º. Adjusting u by a nonzero scalar we may assume .ad u/4 .f / D e. Set M WD L.˛/.1/ = rad L.˛/.1/ , and for x 2 L.˛/.1/ set xQ 2 M the image of x. Then F eQ ˚ F hQ ˚ F fQ Š sl.2/;

Œu; Q e Q D 0;

.ad u/ Q 4 .fQ/ D e: Q

Since L.˛/ is not solvable, Proposition 11.2.1 shows that H is a CSA of toral rank 1 in L.˛/. Corollary 9.2.12 then states that M is one of sl.2/, W .1I n/, or H.2I nI ˆ/.2/ . As dim M 4, one has M 6Š sl.2/. Suppose M Š W .1I n/. Every element g sticking out of W .1I n/.0/ satisfies .ad g/3 ¤ 0. Since .adL e/3 D .adL f /3 D 0, eQ and fQ do not have the former property and we obtain e; Q fQ 2 W .1I n/.0/ . This shows that W .1I n/.0/ contains an sl.2/. But W .1I n/.0/ is solvable, a contradiction. Suppose M Š H.2I nI ˆ/.2/ . As above we obtain that e; Q fQ; hQ 2 H.2I nI ˆ/.2/ .0/ . Consider the graded algebra associated with the natural filtration, H.2I n/.2/ gr H.2I nI ˆ/.2/ H.2I n/: The homomorphism H.2I nI ˆ/.2/ .0/ ! H.2I nI ˆ/.2/ .0/ =H.2I nI ˆ/.2/ .1/ D H.2I n/.2/ 0 Š sl.2/ Q maps F eQ CF hCF fQ onto H.2I n/.2/ 0 D FDH .x12 /CFDH .x1 x2 /CFDH .x22 /. Let e ; h ; f denote the respective images. We consider this subalgebra as contained in H.2I 1/.2/ H.2I n/.2/ gr H.2I nI ˆ/.2/ . Theorem 7.5.8 proves that h is conjugate under an automorphism of H.2I 1/.2/ to DH .x1 x2 /, where 2 F . Since both h and DH .x1 x2 / are toral, 2 Fp . From this we conclude that all eigenspaces of gr H.2I nI ˆ/.2/ for ad h for nonzero eigenvalues are at least 5-dimensional. This in turn implies that dim L2 5, a contradiction. As a consequence, every u 2 L1; for arbitrary 2 .L; T / satisfies .ad u/4 D 0. (d) Next consider the action of any such u 2 L1; in more detail. Set U D adL u. As U 4 .ŒU.x/; y/ D 0, it follows that 4ŒU 2 .x/; U 3 .y/ C 6ŒU 3 .x/; U 2 .y/ D 0, whence ŒU 3 .x/; U 2 .y/ D ŒU 2 .x/; U 3 .y/ and hence 1 ŒU 2 .x/; U 3 .y/ D U.ŒU 2 .x/; U 2 .y// 8x; y 2 L: 2 If the elements x; y are homogeneous and ŒU 2 .x/; U 3 .y/ ¤ 0, then x; y 2 L 1 [ L 2 and not both are contained in L 1 . Since L 2 D Ff is 1-dimensional, the

182

12


symmetry mentioned above tells us that we may take x D f , y 2 L 1 . But then U 3 .y/ D e for some 2 F and ŒU 2 .x/; U 3 .y/ D ŒŒu; Œu; f ; e D

Œu; Œu; h D 0:

From this we conclude that ŒU i .x/; U j .y/ D 0 for i C j 5 and for all x; y 2 L. Then exp.ad u/ is an automorphism of L. Note that L is generated by elements u 2 .[2.L;T / L1; / [ L˙2 . We have now proved that for every such u the mapping exp.ad u/ is an automorphism of L. Arguing as in the proof of Theorem 4.1.43 one obtains that L is classical. (e) Since L is classical, it is restricted. Then T L, hence T H and H is a CSA of L. [Sel 67, Theorem IV.1.2, Note] shows that H satisfies the requirements of Theorem 4.1.2. In particular, H is a torus. Setting 12.2(4) implies that H D T is 2-dimensional. The corresponding Dynkin diagram has rank 2. Then L is one of A2 , B2 , G2 . (f) In order to show that only one of the listed cases can occur let L be one of A2 , B2 , G2 . For all these algebras the Killing form is nonzero ([Sel 67, p. 47]). Then the ˛-root string through ˇ has gaps, which gives Np 1 .L; T / ¤ ;, and every root vector fits into an sl.2/, whence N2 .L; T / D ;. From this it is clear that not two of the cases in the theorem can occur simultaneously.

12.3

Rigid roots

In this section we intend to show that Np 1 .L; T / ¤ ; for relevant Lie algebras L and tori T . However, there are examples which show that additional assumptions are necessary. Let g be an arbitrary Lie algebra, gŒp a minimal p-envelope and t a torus in gŒp . The following notation is essentially independent from the choice of the p-envelope, because p-envelopes differ only by their centers. For our purposes it is sufficient to consider semisimple Lie algebras, and for those algebras the minimal p-envelope is uniquely determined. For ˛ 2 .gŒp ; t/ n ¹0º set N .˛/ .g; t/ WD ¹x 2 gŒp .˛/ n ¹0º j .adg x/p D 0; t.x/ F x; .ad x/2 .g.˛// t \ ker ˛º: Different applications of the main result Theorem 12.3.7 will require subtle assumptions. We need the following Setting 12.3: (1) Let L be a semisimple Lie algebra; denote the minimal p-envelope of L by LŒp . (2) Assume that L has a unique minimal ideal L which is a simple Lie algebra in its own right; denote the p-envelope of L in LŒp by LŒp . 3 see


12.3

183

Rigid roots

(3) Let T LŒp be a 2-dimensional torus. (4) Set H WD CL .T / and assume that H is a nilpotent restricted subalgebra of LŒp . (5) Assume that the maximal torus of H is contained in T . (6) Assume L D H C L. In the following we presuppose the notations and assumptions of the Setting 12.3. In order to derive a contradiction we also assume that N .˛/ .L; T / ¤ ; for some ˛ 2 .L; T / n ¹0º but Np

1 .L; T

/ D ;:

Note that LŒp is semisimple. There is an injective restricted homomorphism LŒp ,! Der L. In particular, L is the unique minimal ideal of LŒp , LŒp is the minimal p-envelope of L, and T acts as a 2-dimensional torus on L. Lemma 12.3.1. If x 2 LŒp is a nonzero T -eigenvector, then .ad x/p

1 .L/

¤ 0.

Proof. Suppose .ad x/p 1 .L/ D ¹0º. Choose k minimal with .ad x/k .L/ D ¹0º. Since LŒp acts faithfully on L, one has k ¤ 1. Choose a T -eigenvector y 2 L such that z WD .ad x/k 1 .y/ ¤ 0. Note that z 2 L. If k 4, then z 2 Nk 1 .L; T / (Proposition 12.1.1(1)). If k D 3; 2, then Proposition 12.1.1 shows that z 2 N3 .L; T /. But in all of these cases one would have Np 1 .L; T / ¤ ;. We now fix a 2 N .˛/ .L; T /. Let t˛ 2 T be a nonzero toral element for which T \ ker ˛ D F t˛ . Define an Fp -grading on L by Lj WD ¹x 2 L j t˛ .x/ D jxº: Since LŒp .˛/ D ¹x 2 LŒp j Œt˛ ; x D 0º, it is clear that ŒLŒp .˛/; Lj Lj for all j . For x 2 LŒp write X WD adL x. Note that L0 D L.˛/;

Œa; Lj Lj ;

Ap D 0;

A3 .L0 / D ¹0º;

A2 .L0 / F t˛

hold. Lemma 12.3.2. The vector space N WD Ap

2

.L/ \ .ker A/

is a nilpotent T -invariant graded subalgebra of L, and Ap of N . In particular, Ap 1 .L/ \ C.N / ¤ ¹0º.

1 .L/

is a nonzero ideal

Proof. It is obviously true, that N is a T -invariant graded subalgebra of L, and that Ap 1 .L/ is an ideal of N . Note that N \ L0 Ap 2 .L0 / D ¹0º and therefore every N \ Li acts nilpotently on N . The Engel–Jacobson theorem shows that N is nilpotent. As Ap 1 .L/ ¤ ¹0º by Lemma 12.3.1, one has Ap 1 .L/ \ C.N / ¤ ¹0º by Engel’s theorem.

184

12


Lemma 12.3.3. The following holds: (1) p D 5; (2) A2 .L0 / D F t˛ ; (3) if y 2 Lk is a T -eigenvector satisfying A4 .y/ 2 C.N / n ¹0º, then we have ŒA3 .y/; A4 .y/ ¤ 0. Proof. (a) Suppose y 2 Lk (for some k) is a T -eigenvector for which Ap 1 .y/ ¤ 0 and ŒAp 1 .y/; N D ¹0º. We intend to show by induction on q that for every such element q Ap 1 Y Ap 2 Y Ap 1 D 0 holds. Note that for arbitrary u 2 L one has Ap 1.u/ 2 N and thus Ap 1 YAp 1.u/ D ŒAp 1 .y/; Ap 1 .u/ 2 ŒAp 1 .y/; N D ¹0º. This settles the case q D 0. By induction hypothesis one gets for q 1 Ap

2

Y

q

Ap

1

.u/ D Ap

2

Y Ap

2

Y

q

1

Ap

1

.u/ 2 Ap

2

.L/ \ .ker A/ D N;

and therefore Ap

1

Y Ap

2

Y

q

Ap

1

.u/ D ŒAp

1

2 ŒAp

1

.y/; Ap

2

Y

q

Ap

1

.u/

.y/; N D ¹0º:

(b) Next we intend to show that for arbitrary r 1 Ai1 Y YAir D 0

if i1 C C ir r.p

2/ C 2:

The claim holds if r D 1; 2 by assumption and (a). Suppose r 3 and assume that there is an index s for which 1 < s < r and is < p 2. We may assume by induction i1 C C is

1

.s

1/.p

2/ C 1;

isC1 C C ir .r

s/.p

2/ C 1:

2/ C 1 < r.p

2/ C 2;

Then i1 C C ir is C .s

1/.p

2/ C 1 C .r

s/.p

a contradiction. P Hence we may assume is p 2 for allp s 1 D 2; : : : ; r 1. The assumption iq r.p 2/ C 2 implies that the factor A occurs at least twice. Then (a) proves the claim (b). (c) In a third step we assume that r D p, i1 C C ip D .p 1/2 and is 2 for some index s. We intend to prove that Ai1 Y YAip D 0

or p D 5 and is D 2 and i1 C C is

1

D 3s

2:

12.3

Suppose such an expression is nonzero. As .p obtain by (b) s ¤ 1; p and i1 C C is

1

.s

2/ C 1;

1/.p

185

Rigid roots

1/2

2 .p

isC1 C C ip .p

1/.p s/.p

2/ C 2, we 2/ C 1:

Then .p 1/2 .p 1/.p 2/ C 2 C is , hence is p 3. As is 2, this is only possible if p D 5, is D 2 and i1 C C is 1 D 3.s 1/ C 1 D 3s 2. (d) After these preliminaries we prove the lemma. According to Lemma 12.3.2 there exists a T -eigenvector y 2 Lk (for some k), for which z WD Ap 1 .y/ ¤ 0 and Œz; N D ¹0º. According to Lemma 12.3.1 there is w 2 Ll (for some l) such that Z p 1 .w/ ¤ 0. Let w be an arbitrary such element. We consider nonzero terms Q WD Ai1 Y YAip .w/ occurring in X Z p 1 .w/ D ˛.i1 ; : : : ; ip /Ai1 Y YAip .w/: .i1 ;:::;ip /

Since Ap 1 .y/ ¤ 0, one has k 6 0 .p/. Determine the index s from the congruence .p s/k C l 0 .p/ (and observe that w 2 L0 if s D p). If s < p, then YAisC1 Y YAip .w/ 2 L0 . The assumption on A implies is 2 in both cases s < p and s D p. We proved in (c) that p D 5 and is D 2. Then Ais Y YAip .w/ D t˛ for some ¤ 0 necessarily holds. This proves (1) and (2).P p (e) The definition of s implies s D l=k. If s D 1, then lD2 il D .p 1/2 2 > .p 1/.p 2/ C 2 and Q D 0 by (b), which is not true. Similarly one gets s ¤ p. As p D 5, only s D 2; 3; 4 are possible. We have Q D Ai1 Y YAis 1 Y.t˛ // D

kAi1 Y YAis 1 .y/

and i1 C C is 1 D 3s 2 by (c). In order to prove (3) we now assume ŒA3 .y/; A4 .y/ D 0 and eventually derive the contradiction Z 4 .w/ D 0. By Equation (12.1.1) A3 YA4 .y/ D ŒA3 .y/; A4 .y/ D 0; 4

3

4

3

./ 3

4

A YA .y/ D ŒA .y/; A .y/ C 4ŒA .y/; A .y/ D 0:

./

The second equation gives A3 .Œy; A3 .y// 2 A3 .L/ \ .ker A/ D N . Note that N contains A3 YA3 .y/ D A3 .Œy; A3 .y// D 3ŒA2 .y/; A4 .y/: Therefore 1 ŒA4 .y/; ŒA2 .y/; A4 .y/ D ŒA4 .y/; A3 .Œy; A3 .y// 2 Œz; N D ¹0º: 3 Expanding this equation by use of Equation (12.1.1) and applying ./ we get 0 D ŒA4 .y/; ŒA2 .y/; A4 .y/ D ŒA4 .y/; A2 YA4 .y/ D A4 YA2 YA4 .y/:

186

12


Finally, ./ shows that A4 YA3 YA3 .y/ D ŒA4 .y/; A3 YA3 .y/ D 3ŒA4 .y/; ŒA2 .y/; A4 .y/ D 0: (f) Suppose s D 4. Then i1 C i2 C i3 D 10 and Ai1 YAi2 YAi3 .y/ is one of A4 YA2 YA4 .y/;

A4 YA3 YA3 .y/;

All these elements vanish. Suppose s D 3. Then i1 C i2 D 9

A3 YA4 YA3 .y/;

A3 YA3 YA4 .y/:

2 D 7 and Ai1 YAi2 .y/ is one of

A4 YA3 .y/ D 0 or A3 YA4 .y/ D 0: As a consequence, s D 2. Recall that y; z 2 Lk . We have now shown that the assumption Z 4 .w/ ¤ 0 implies w 2 L2k . This gives Z 4 .Lik / D ¹0º if i ¤ 2:

./

In particular, Z 5 .L/ D Z 5 .L2k / Z 4 .L3k / D ¹0º. (g) Expand Z 3 .w/ ¤ 0 and observe that Z 3 .w/ 2 L0 , X Z 3 .w/ D ˇ.j1 ; : : : ; j4 /Aj1 YAj2 YAj3 YAj4 .w/; where j1 C C j4 D 12. Let Aj1 YAj2 YAj3 YAj4 .w/ be any nonzero term. Then j1 2 as YAj2 YAj3 YAj4 .w/ 2 L0 . By (b), only j1 D 2 is possible and Z 3 .w/ 2 F t˛ necessarily holds. Consequently, Z 3 .L2k / D F t˛ : Without loss of generality we may assume Z 3 .w/ D t˛ . Equation (12.1.1) gives ad t˛ D ad.Z 3 .w// D Z 3 W

3Z 2 W Z C 3ZW Z 2

W Z3:

Let u 2 Li k . We compute (as Z 5 D 0) ikZ 3 .u/ D Z 3 ı .ad t˛ /.u/ D 3Z 4 W Z 2 .u/ .3 C i/kZ 3 .u/ D .ad t˛ / ı Z 3 .u/ D Z 3 W Z 3 .u/

Z 3 W Z 3 .u/; 3Z 2 W Z 4 .u/:

Then .1

i/kZ 3 .u/ D Z 4 W Z 2 .u/

Z 2 W Z 4 .u/:

Recall that z 2 Lk , w 2 L2k , u 2 Lik . Then W Z 2 .u/ 2 L.4Ci/k . If i D 0; 4, then ./ yields Z 4 .u/ D Z 4 W Z 2 .u/ D 0, whence by the preceding result Z 3 .L4k / D Z 3 .L0 / D ¹0º: If i D 3, then one gets 2Z 3 .u/ 2 Z 4 .L2k / D Œz; F t˛ D F z, i.e., Z 3 .L3k / F z:

12.3

Rigid roots

187

(h) Next for u 2 L0 , 2kZ 2 .u/ D .ad t˛ / ı Z 2 .u/ D Z 3 W Z 2 .u/ 2 Z 3 .L4k / D ¹0º: Then Z 2 .L0 / D ¹0º and kZ.u/ D .ad t˛ / ı Z.u/ D Z 3 W Z.u/ 2 Z 3 .L3k / F z: This gives Œz; L0 F z, whence ŒL0 .1/ ; z D ¹0º. Recall that there is v 2 L0 for which A2 .v/ D t˛ (by (2) of this lemma). Then kz D Œt˛ ; z ŒL0 .1/ ; z D ¹0º, a contradiction. This contradiction shows that Z 4 .w/ D ¹0º and thereby proves (3). Fix according to Lemma 12.3.3 T -eigenvectors b; v 2 L.˛/ for which Œa; v D b;

Œa; b D t˛ :

We observe that t˛ 2 L. Lemma 12.3.4. (1) .ker Al / \ Li D A5 l .Li / for all i ¤ 0 and 1 l 4. (2) N D A4 .L/. (3) A4 .y/ 2 C.N / if and only if A3 .y/ 2 NorL .N /. Proof. (1) Let x 2 .ker A/ \ Li . We construct inductively elements uk 2 Li (1 k 4) for which x D Ak .uk /. Note that t˛ acts as i Id on Li . Put u0 WD x and assume inductively that x D Ak .uk / for some k 3. Then AkC1 .uk / D A.x/ D 0. Observing that A.b/ D t˛ and A2 .b/ D 0 one obtains AkC1 .Œb; uk / D .k C 1/ŒA.b/; Ak .uk / D .k C 1/Œt˛ ; x D .k C 1/ix: Put ukC1 WD

1 Œb; uk . .kC1/i

The preceding step proves (1) for l D 1. Now assume inductively that .ker Al / \ Li D A5 l .Li / for some l < 4. Take x 2 .ker AlC1 / \ Li . Then A.x/ 2 .ker Al / \ Li D A5 l .Li /. Choose y 2 Li for which A.x/ D A5 l .y/. Then x A4 l .y/ 2 .ker A/ \ Li D A4 .Li /, and this gives x 2 A5 .lC1/ .Li /. (2) is a direct consequence of (1) as A3 .L0 / D N \ L0 D ¹0º. (3) Note that we have ŒA3 .y/; N D A3 .Œy; N / A3 .L/ and ŒA4 .y/; N D A.ŒA3 .y/; N /. Therefore ŒA4 .y/; N D ¹0º if and only if ŒA3 .y/; N A3 .L/ \ .ker A/ D N . Set M0 WD F t˛ , Mi WD A3 .Li / \ NorL .N / for i D 1; : : : ; 4 and M WD

4 X i D0

Mi © N:

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12


Lemma 12.3.5. M is a T -invariant graded subalgebra of L having a 1-dimensional CSA F t˛ of toral rank 1. Moreover, rad M D N is abelian. Proof. (a) It is clear that M is a T -invariant and graded subspace of L. In particular, ŒM0 ; M M . (b) In order to prove that M is a subalgebra let i; j ¤ 0 and x D A3 .x 0 / 2 Mi with 0 x 2 Li , y D A3 .y 0 / 2 Mj with y 0 2 Lj . By definition, A.x/; A.y/ 2 A4 .L/ D N , A2 .x/ D A2 .y/ D 0. Therefore A2 .Œx; y/ D 2ŒA.x/; A.y/ D 2A.Œx; A.y// 2 A.N / D ¹0º: If i C j 6 0 .mod .5//, then Lemma 12.3.4(1) implies Œx; y 2 A3 .Li Cj /. Clearly, Œx; y 2 NorL .N / holds. Then Œx; y 2 M . Now suppose i C j 0 .mod .5//. Applying the equation A3 .L0 / D ¹0º one obtains 0 D A3 .Œx 0 ; A3 .y 0 // D ŒA3 .x 0 /; A3 .y 0 / C 3ŒA2 .x 0 /; A4 .y 0 / D Œx; y C 3A2 .Œx 0 ; A4 .y 0 // 2 Œx; y C A2 .L0 / D Œx; y C F t˛ : Therefore Œx; y 2 F t˛ . This completes the proof of M being a subalgebra. Obviously, F t˛ is a 1-dimensional CSA. (c) If there is an index i ¤ 0 for which ŒM Pi ; M i ¤ ¹0º, then t˛ 2 ŒMi ; M0 i . In this case M is not solvable. So assume i ¤0 ŒMi ; M i D ¹0º. Then M WD P 0 i ¤0 Mi is a graded subalgebra of L. As M \ L0 D ¹0º, this algebra is nilpotent. By definition, N is an ideal of M 0 . By Engel’s theorem there is i ¤ 0 and nonzero y 2 Li for which y 2 C.M 0 / \ N C.N /. For this element there is y 0 2 Li with y D A4 .y 0 / (as N D A4 .L/). In particular, A4 .y 0 / 2 C.N / n ¹0º. Lemma 12.3.4(3) shows that A3 .y 0 / 2 M 0 . But then ŒA3 .y 0 /; A4 .y 0 / 2 ŒM 0 ; C.M 0 / D ¹0º, and this contradicts Lemma 12.3.3(3). (d) Theorem P 11.1.3 implies that rad M is abelian. Set I WD l0 .ad T /l .rad M /. Since M is T -invariant, it is easily seen that I is an ideal of M . We proved in (c) that t˛ 62 rad M . Then .rad M / \ M0 D ¹0º. But then I \ M0 D ¹0º as well. This implies that I is nilpotent, whence I rad M . As a consequence, rad M is T -invariant. If rad M ¤ N , then there is a T eigenvector A3 .y/ 2 rad M , A3 .y/ 62 N D A3 .L/ \ .ker A/. But then A4 .y/ ¤ 0 and ŒA3 .y/; A4 .y/ .rad M /.1/ D ¹0º, which contradicts Lemma 12.3.3(3). Lemma 12.3.6. (1) A3 .Li / ¤ A4 .Li / for all i 2 F5 . (2) For every i 2 F5 the space Li is (F a C F b C F t˛ )-irreducible; more exactly there are wi 2 Li and i 2 F such that Œb; wi 2 F wi ;

Œb Œ5

i t˛ ; Li D ¹0º;

Li D ˚4sD0 FAs .wi /:

(3) There is k 2 F5 for which ŒLk ; Li D Li Ck for all i 2 F5 .

12.3

Rigid roots

189

Proof. (1) Suppose there is an index i ¤ 0 for which A3 .Li / D A4 .Li / holds. Then A4 .Li / D A5 .Li / D ¹0º. Lemma 12.3.4(1) gives .ker A/ \ Li D ¹0º. This implies Li D ¹0º. Next observe that there is an index j ¤ 0 for which Lj ¤ ¹0º, because otherwise A3 .L/ D A3 .L0 / D ¹0º which contradicts Lemma 12.3.1. As Li D ¹0º, then every nonzero T -eigenvector z in Lj satisfies .adL z/4 D 0, whence z 2 N4 .L; T / D ;, a contradiction. (2) As N D rad M is abelian (Lemma 12.3.5), Lemma 12.3.4(3) gives M D A3 .L/ C F t˛ . Note that X dim M= rad M D dim M=N D 1 C dim A3 .Li /=A4 .Li / 5: i 2F5

Theorem 11.1.3 then implies that M= rad M Š W .1I 1/ is 5-dimensional. This in turn means dim A3 .Li /=A4 .Li / D 1 for all i ¤ 0: The short exact sequence (cf. Lemma 12.3.4(1)) 0 ! As .Li / ! As gives dim As

1 .L

i /=A

s .L

dim As

1

i/

1

A5

s

.Li / ! A4 .Li / ! 0

D dim A4 .Li /, hence

.Li /=As .Li / D dim A3 .Li /=A4 .Li / D 1

for all s D 1; : : : ; 4. As a consequence, dim Li D 5 for all i ¤ 0. Choose a common eigenvector wi 2 Li for the abelian subalgebra F bCF t˛ . Note that F a C F b C F t˛ is a Heisenberg algebra. Every faithful irreducible module for this algebra is at least 5-dimensional. Consequently, Li D ˚4sD0 FAs .wi / if i ¤ 0, and Li is (F a C F b C F t˛ )-irreducible. Observe that Œb Œ5 ; a D 0. Choose 0i 2 F for which Œb; wi D 0i wi . Then Œib Œ5 .0i /5 t˛ ; Li D ¹0º. Set i WD 1i .0i /5 . This proves (2). (3) Recall that M=N Š W .1I 1/ and N is abelian. The image of the 1-dimensional torus F t˛ is conjugate to F x@ or to F .1 C x/@. Choose in both cases k 2 F5 for which .M=N /k Š F @. Then one has ŒMk ; Ml ¤ ¹0º whenever k ¤ l. Next we show that ŒMk ; Nk ¤ ¹0º. Let V N be an irreducible M -module (which means a W .1I 1/-module). Since N \ L0 D ¹0º, t˛ acts invertibly on N and every element x 2 Mi (i ¤ 0) satisfies .ad x/5 .V / D ¹0º. Then V is a restricted module without 0 weight for F t˛ . Theorem 7.6.10 tells us that V Š O.1I 1/=F . Note that Nk \ V has the same t˛ -weight as F @. Then Nk \ V Š F x 4 or Nk \ V Š F .1 C x/4 . In either case we have @ .Nk \ V / ¤ ¹0º. This gives ŒMk ; Nk ¤ ¹0º. As a result, ŒLk ; Ll ¤ ¹0º for all l 2 F5 . Observe that ŒLk ; Ll is a (F a C F b C F t˛ )-invariant nonzero subspace of LkCl . If k C l ¤ 0, then the (F a C F b C F t˛ )-irreducibility of LkCl shows ŒLk ; Ll D LkCl . Moreover, the simplicity of L P yields L0 D i ¤0 ŒL i ; Li , and the former result then gives L0 D ŒL k ; Lk . This proves (3).

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12


These last results will enable us to prove the theorem we are heading for. Theorem 12.3.7. Suppose the Setting 12.3 is fulfilled. Then N2 .L; T / ¤ ; if and only if there exists a nonzero root ˛ 2 .L; T / for which N .˛/ .L; T / ¤ ;. Proof. (A) Suppose N2 .L; T / ¤ ;. Due to Proposition 12.1.1 there is c 2 N2 .L; T / which satisfies Œad c; Œad c; Der L D ¹0º. In particular, Œad c; Œad c; adL L D ¹0º, and this shows c 2 N2 .L; T /. Choose for which c 2 L . If ¤ 0, then c 2 N . / .L; T /. If D 0, then choose any nonzero root ˛ and obtain c 2 N .˛/ .L; T /. (B) Suppose N .˛/ .L; T / ¤ ;. Assume in a first step that Np 1 .L; T / D ;: Then the general assumption for Lemmas 12.3.1–12.3.6 is satisfied. Fix accordingly T -eigenvectors a; b; v with a 2 N .˛/ .L; T /, b; v 2 L.˛/, and elements wi 2 Li for i ¤ 0 satisfying Œa; v D b; Œa; b D t˛ ; Œb; wi D i wi ; Œt˛ ; wi D iwi ;

Li D ˚4sD1 FAs .wi /:

Recall that p D 5. Take k 2 F5 as in Lemma 12.3.6(3). (a) Suppose a 2 CLŒ5 .T /. Then the T -eigenvectors b; v can be taken from CL .T /. Recall that L is an ideal of LŒ5 , whence Lk is LŒ5 .˛/-invariant. Therefore there is a root for which Lk \L ¤ ¹0º. As this space is .F aCF bCF t˛ /-invariant, it coincides with Lk (Lemma 12.3.6(2)). Lemma 12.3.6(3) then gives LD

p X1

.ad Lk /l .Lk /

lD0

X

Ll :

l2F5

Consequently, .L; T / F5 . But T acts on L as a 2-dimensional torus (since we may regard LŒ5 as a restricted subalgebra of Der L). This contradiction shows that ŒT ; a ¤ ¹0º. We may adjust ˛ ¤ 0 to obtain a 2 L˛ . Then we may take b 2 L ˛ , v 2 L 2˛ . (b) Next we observe that the action of LŒ5 on L induces a restricted representation W LŒ5 .˛/ ! gl.Lk /. For x 2 LŒ5 .˛/ set X 0 WD .x/. Lemma 12.3.6(2) shows that the Heisenberg algebra F a C F b C F t˛ acts irreducibly on Lk , and in particular dim Lk D 5;

.t˛ / D kId;

.a/5 D 0;

.b/5 D kk Id

hold. Therefore we have End Lk D ˚4i;j D0 FA0i B 0j : Moreover, Lemma 12.3.6(3) shows that ker annihilates L. Since the unique minimal ideal L of LŒ5 is a simple algebra, one concludes ker D ¹0º. Then is faithful. (c) We identify LŒp .˛/ via with its image in gl.Lk / D ˚4i;j D0 FA0i B 0j . Put P F ŒA0 D 4sD0 FA0s the set of polynomials in A0 . The equation Œa; b D t˛ yields ŒA0 ; B 0 D kId:

12.3

191

Rigid roots

Therefore the equation Œa; Œa; L.˛/ D F t˛ yields L.˛/ FB 02 C F ŒA0 B 0 C F ŒA0 : Let t0 2 T be toral with ˛.t0 / D 1. The equations Œt0 ; a D a, Œt0 ; b D t0 D

A0 B 0 C Id;

./ b yield

for some 2 F:

This gives T D F t0 C F t˛ D FA0 B 0 C F Id. The equations Œa; v D b and Œt0 ; v D 2v yield 1 V 0 D B 02 C 0 A03 ; 0 2 F: 2 Let an element V 0 C g1 .A0 /B 0 C g2 .A0 / be contained in L.˛/. Then L.˛/ also contains Œ2V 0 ; g1 .A0 /B 0 C g2 .A0 / D ŒB 02 C 20 A03 ; g1 .A0 /B 0 C g2 .A0 / ŒB 02 ; g1 .A0 /B 0 2ŒB 0 ; g1 .A0 /B 02

.mod F ŒA0 B 0 C F ŒA0 /:

By ./ this gives ŒB 0 ; g1 .A0 / 2 F Id, and this is only possible if g1 .A0 / D ˇ1 A0 Cˇ0 Id for some ˇ1 ; ˇ0 2 F . Consequently, L.˛/ FB 02 C FA0 B 0 C FB 0 C F ŒA0 D F V 0 C FA0 B 0 C FB 0 C F ŒA0 : (d) We now determine ŒLk ; L

k .

Recall the definition of wi from Lemma 12.3.6.

Set u WD Œwk ; A4 .w

k /:

Let U L0 denote the minimal vector space containing u and being invariant under ad ad b several times one sees that the vector space U contains P4a, ad b. Applying s .w Œw ; A / D Œwk ; L k , and applying ad a several times one obtains that k sD0 P4 k s U contains sD0 ŒA .wk /; L k D ŒLk ; L k . This gives (cf. Lemma 12.3.6(3)) U D ŒLk ; L

k

D L0 D L.˛/:

On the other hand, as .u/ 2 .L.˛// F V 0 CFA0 B 0 CFB 0 CF ŒA0 , one computes that F .u/ C FB 0 C F ŒA0 is invariant under A0 , B 0 , whence under the identification by L.˛/ D U F .u/ C FB 0 C F ŒA0 : Then L.˛/ D F .u/ C FB 0 C F ŒA0 \ L.˛/, and this implies by dimension reasons L.˛/ D F V 0 C FB 0 C F ŒA0 \ L.˛/:

192

12


Let h.A0 / 2 F ŒA0 \ L.˛/. Then L.˛/ contains Œ2V 0 ; h.A0 / D ŒB 02 ; h.A0 / 2ŒB 0 ; h.A0 /B 0

.mod F ŒA0 /:

This is only possible if ŒB 0 ; h.A0 / 2 F Id, i.e. if h.A0 / 2 FA0 C F Id. Finally, L.˛/ contains Œ2V 0 ; B 0 D ŒB 02 C 20 A03 ; B 0 D 60 A02 : The former result shows 0 D 0, whence V 0 D 12 B 02 . (e) Taking into account again the mapping we have now shown that L.˛/ Š .L.˛// D FB 02 C FB 0 C FA0 C F Id: We conclude first, that L.˛/ is nilpotent. Next, since B 05 2 F Id and A05 D 0, L.˛/ is a restricted subalgebra of LŒ5 . The uniquely determined maximal torus of the nilpotent algebra L.˛/ is 1 1 .F Id/ D F t˛ . Hence L.˛/ is a Cartan subalgebra of L having toral rank 1 in L. Theorem 9.2.11 implies that it acts trigonalizably on L. But this is not true in the present case, and this gives us the final contradiction. As a result, Np 1 .L; T / ¤ ;. (f) It is clear that the Setting 12.2 holds. By Theorem 12.2.6, either N2 .L; T / ¤ ; or L is classical of type A2 , B2 , G2 and T D CL .T / is a 2-dimensional CSA which satisfies the requirements of Theorem 4.1.2. Suppose L is classical. Then it has nonsingular Killing form ([Sel 67, p. 47]) and therefore the 1-section L.˛/ is isomorphic to gl.2/ ([Sel 67, Lemma II.2.7]). Now let a 2 gl.2/ be a nonzero Œp-nilpotent root vector. Then Œa; Œa; gl.2/ 6 T , and this implies that N .˛/ .L; T / D ;. Therefore L cannot be classical. These deliberations lead to the following Definition 12.3.8. Let L and T satisfy the assumptions of the Setting 12.3. A nonzero root ˛ 2 .L; T / is called rigid if N .˛/ .L; T / D ;. The torus T is called rigid if all nonzero roots ˛ 2 .L; T / are rigid. The theorem shows that there exist T -sandwiches if and only if T is non-rigid, i.e., if and only if not all nonzero T -roots are rigid. We will address this problem in the following section.

12.4

Rigid tori

For a first application of the preceding deliberations let L be a simple Lie algebra of absolute toral rank TR.L/ D 2, LŒp be the p-envelope of L in Der L and T LŒp be a torus of maximal toral rank. Since LŒp is centerless, T is a torus of maximal dimension 2. Set H WD CL .T / and HQ WD CLŒp .T /: Clearly, HQ contains T and therefore is a CSA of LŒp , while H in some case even might be ¹0º.

12.4

193

Rigid tori

ForPevery 2 .L; T / and x 2 L one has x Œp 2 HQ . Therefore §1.1 shows that Q H C 2.L;T / L is a restricted subalgebra of LŒp containing L, whence LŒp D HQ C

X

L :

2.L;T /

Consequently, LŒp; D L for every T -root ¤ 0. Therefore the -section in LŒp is X LŒp . / D HQ C Li : i 2Fp

Every 2 .L; T / may be regarded a mapping on H and HQ (see the definition of extended roots in §1.3). Note that L D LŒp WD LŒp and T WD T satisfy the assumptions of the Setting 12.3. Then N .˛/ .L; T / D N .˛/ .L; T / by definition. In the following we will investigate the question when a root S ˛ 2 .L; T / is rigid. In order to do so we will consider root vectors x 2 HQ [ i 2Fp Li˛ satisfying Œx; Œx; L.˛/ T . Such a root vector is contained in N .˛/ .L; T / if and only if x Œp D 0. One can weaken this condition. Namely, ad x is nilpotent if and only if ˇ.x Œp / D .x Œp / D 0 for independent r 1 roots ˇ, . If this is true and x ¤ 0, then there is r > 0 for which x Œp ¤ 0 and r r 1 Œp Œp x D 0. Then y WD x is a root vector satisfying Œy; Œy; L.˛/ D ¹0º and .ad y/p D 0. Therefore in order to show N .˛/ .L; T / ¤ ; it suffices to find a root vector x satisfying Œx; Œx; L.˛/ T with either one of the conditions that ad x is nilpotent

or

ˇ.x Œp / D .x Œp / D 0 for independent roots ˇ; :

Set T \ ker ˛ DW F t˛ with some toral element t˛ . Recall that LŒp .˛/ has absolute toral rank 1 (Theorem 1.3.11(3)). Lemma 12.4.1. Suppose ˛ 2 .L; T / n ¹0º is a rigid root. (1) Let A be a .T C L.˛//-invariant subalgebra of LŒp .˛/ such that A.1/ T \ ker ˛. Then A \ HQ T \ ker ˛ and dim A \ Li˛ 1 for each i 2 Fp . (2) Let G LŒp .˛/ be a subalgebra containing T C L.˛/. Then L=L.˛/ considered as a G-module has a faithful composition factor. If ˛ is solvable, then every composition factor of L=L.˛/ is faithful. S (3) If rad L.˛/ 6 T , then there is w 2 i 2Fp .rad L.˛//i˛ such that .w Œp / ¤ 0 P for all 2 .L; T / n Fp ˛. Every such w acts invertibly on j 2Fp L Cj˛ . In particular, dim L D dim L Cj˛ ;

2 .L; T / n Fp ˛; j 2 Fp :

(4) If rad LŒp .˛/ 6 T , then HQ acts trigonalizably on LŒp .˛/.

194

12


Proof. (1) Let A be a .T C L.˛//-invariant subalgebra of LŒp .˛/ and A.1/ T \ ker ˛. For a 2 A \ HQ denote by .F a/Œp the p-envelope of F a in LŒp . It is straightforward that, for each u 2 .F a/Œp , .ad u/2 .L.˛// T \ ker ˛. As ˛ is rigid we have uŒp ¤ 0, whence .F a/Œp has no Œp-nilpotent element ¤ 0. This means that every a 2 A \ HQ is semisimple, whence A \ HQ T \ ker ˛. Let now x; y 2 A \ Li˛ n ¹0º, i ¤ 0. Then, for each 2 F , ad.x C 2 y/ .L.˛// T \ ker ˛. Since 3 .A/ D ¹0º, .x C y/Œp D p x Œp C y Œp holds ([S-F, Lemma 2.1.2]). Recall that x Œp ; y Œp 2 A \ HQ D T \ ker ˛ D F t˛ and x Œp ¤ 0, y Œp ¤ 0. Therefore there are 1 ; 2 2 F such that x Œp D 1 t˛ , p p Œp Œp y D 2 t˛ . Now it is clear that .0 x C y/ D 0 for 0 D 2 =1 . As ˛ is rigid, y D 0 x. (2) Let M be a composition factor of the G-module L=L.˛/ and W G ! gl.M / denote the corresponding representation. Assume that ker ¤ ¹0º. (i) Suppose ker contains a nilpotent subalgebra I ¤ ¹0º which is stable under T C L.˛/. Then A Decompose PWD C.I / ¤ ¹0º is abelian and stable under T C L.˛/. S A D A \ HQ C i 2Fp Ai˛ into root spaces and let x 2 .A \ HQ / [ i 2Fp Ai˛ , x ¤ 0. Since ˛ is a root and P .ad x/2 L.˛/ D ¹0º, we have ˛.x Œp / D 0. Moreover, .x/ acts trivially on M D ˇ 62Fp ˛ Mˇ . Thus there is a root 2 .L; T / n Fp ˛ satisfying .x Œp / D 0. As ˛ is rigid, a remark at the beginning of the section gives x D 0. Therefore ¹0º is the only nilpotent subalgebra of ker which is stable under T C L.˛/. P (ii) We mention that .ker / \ i ¤0 Li˛ ¤ ¹0º, because otherwise ker HQ would be nilpotent contrary to the preceding step. P P (iii) Suppose ˛ is solvable. Recall from Theorem 11.2.3 that N.˛/ D i ¤0 Li˛ C i ¤0 ŒL i ˛ ; Li˛ is a nilpotent ideal of T C L.˛/. By (ii), .ker / \ N.˛/ is nonzero, and this contradicts (i). We have now proved (2) for solvable roots. (iv) Suppose ˛ is non-solvable. By Theorem 11.1.2(1), Theorem 11.1.1 applies and shows that L.˛/=L.˛/.1/ and rad L.˛/.1/ are nilpotentP and L.˛/.1/ = rad L.˛/.1/ .1/ is simple. Set I WD .ker / \ L.˛/ .ker / \ i¤0 Li˛ ¤ ¹0º. This is a .T C L.˛//-invariant subalgebra of ker . Next observe that I 6 rad L.˛/.1/ , .1/ = rad L.˛/.1/ because otherwise I would be nilpotent. The simplicity of L.˛/ .1/ .1/ .1/ shows that L.˛/ D I C rad L.˛/ . This gives L.˛/ D rad L.˛/.1/ , .1/ in combination with the solvability of rad L.˛/.1/ and the perfectness of L.˛/ .1/ then implies L.˛/ D ¹0º. Now assume that the assertion is not true. Then L.˛/.1/ annihilates all composition factors M of L=L.˛/. Since L.˛/.1/ is perfect then L.˛/.1/ annihilates L=L.˛/. Clearly, L.˛/.1/ annihilates L.˛/=L.˛/.1/ , hence also L=L.˛/.1/ . Consequently, L.˛/.1/ is a nonzero ideal of L, and this contradicts the simplicity of L. (3) Recall that rad L.˛/ is T -invariant (Theorem 11.2.3). By assumption, we have rad L.˛/ 6 T . Then there exists a .T C L.˛//-invariant subalgebra A rad L.˛/ minimal subject to the condition A 6 T \ ker ˛. The minimality of A and the

12.4

Rigid tori

195

solvability of rad L.˛/ imply that A.1/ T \ ker ˛. If A HQ , then the first part of this lemma shows that A T \ ker ˛, a contradiction. Thus there is k 2 Fp such that A \ Lk˛ 6D ¹0º. Let w be a nonzero element of A \ Lk˛ . Then .ad w/2 .L.˛// T \ ker ˛. As ˛ is rigid we get .w Œp / 6D 0 for some 2 .L; T / n Fp ˛. As ˛.w Œp / D 0, we get .w Œp / 6D 0 for all 2 .L; T / n Fp ˛. Apply Proposition 1.3.2(2), (3). (4) If rad LŒp .˛/ 6 T , then there is an ideal A rad LŒp .˛/ of LŒp .˛/ minimal subject to the condition A 6 T \ ker ˛. The minimality of A and the solvability of rad LŒp .˛/ imply that A.1/ T \ ker ˛. If A HQ then the first part of this lemma shows that A T \ ker ˛, a contradiction. Thus there is k 2 Fp such that A \ Lk˛ 6D ¹0º. The first part of this lemma shows that dim A \ Lk˛ D 1. Since .1/ ; A \ L D ¹0º. This implies ˛.H Q .1/ / D 0. HQ acts on A \ Lk˛ , we obtain ŒHQ P k˛ .1/ Q Therefore H acts nilpotently on i ¤0 Li˛ . It clearly acts nilpotently on HQ . Then P it also acts nilpotently on HQ C i ¤0 Li˛ D LŒp .˛/. Recall that LŒp Œ˛ WD LŒp .˛/= rad LŒp .˛/. Proposition 12.4.2. Let ˛ 2 .L; T /n¹0º be a non-solvable rigid root. Then LŒp Œ˛ has a unique minimal ideal S . This ideal is a simple Lie algebra, S 2 ¹sl.2/; W .1I 1/; H.2I 1/.2/ º: If S 2 ¹sl.2/; W .1I 1/º or rad LŒp .˛/ 6 T , then HQ acts trigonalizably on LŒp .˛/. Proof. (1) Suppose rad LŒp .˛/ 6 T . Lemma 12.4.1(4) proves that HQ acts trigonalizably on LŒp .˛/. Theorem 11.1.2 then shows that LŒp Œ˛ 2 ¹sl.2/; W .1I 1/º or H.2I 1/.2/ LŒp Œ˛ H.2I 1/. For this case the claim follows. (2) Suppose rad LŒp .˛/ T . (a) As ˛ ¤ 0, one has L˛ 6 T , hence L˛ 6 rad LŒp .˛/. Therefore not only rad LŒp .˛/ T , but rad LŒp .˛/ T \ ker ˛ holds. Theorem 11.1.1 (with g D LŒp .˛/ and t0 D T ) shows that LŒp Œ˛ has a unique minimal ideal s, and there is a realization s Š s0 ˝ O.mI 1/;

m 0; W LŒp Œ˛ ,! .Der s / ˝ O.mI 1/ Ì Id ˝ W .mI 1/ ; 0

.T / D F t0 ˝ 1; where s0 is a simple Lie algebra, Cs0 .t0 / is a CSA of toral rank 1 in s0 and t0 2 s0Œp is toral. Q p 1 (b) The above shows that, if m ¤ 0, then Cs0 .t0 / ˝ F m ¤ ¹0º is annii D1 x i hilated by .T /. Since LŒp Œ˛ embeds into .Der s0 / ˝ O.mI 1/ Ì Id ˝ W .mI 1/ ,

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every nonzero element v of this space satisfies Œv; Œv; .LŒp Œ˛/

X ai 2p 3

s0 ˝

m Y

xiai D ¹0º:

i D1

Let vQ be an inverse image of v in HQ . We have Œv; Q Œv; Q L.˛/ rad LŒp .˛/ D T \ ker ˛: r

Since ˛ is rigid, vQ is not Œp-nilpotent. Therefore there is a Œp-power vQ Œp 2 T \ Q ker ˛ D F t˛ . But then for some 2 F one has that 0 ¤ vCt Q ˛ 2 H is Œp-nilpotent, contradicting the rigidity of ˛. Consequently, m D 0. (c) We have now proved that the simple algebra s D s0 is the unique minimal ideal of LŒp Œ˛. Theorem 10.6.1 proves the statement about s. (d) Finally suppose that s 2 ¹sl.2/; W .1I 1/º. In both cases all derivations are inner. Then s D LŒp Œ˛ and all root spaces LŒ˛i˛ for i ¤ 0 are 1-dimensional. Note Q .1/ Q .1/ that HQ acts P on these root spaces. As a result, ˛.H / D 0, and H acts nilpotently on HQ C i ¤0 Li˛ D LŒp .˛/. Next we are going to study rigid roots in more detail. Due to Proposition 12.4.2 a rigid root ˛ is classical, Witt, or Hamiltonian if and only if the unique minimal ideal of LŒp Œ˛ has the respective structure. Lemma 12.4.3. Suppose ˛ is rigid and solvable. Then (1) LŒp .˛/ is completely solvable, (2) there is k ¤ 0 such that dim Lk˛ 21 .p 1/, (3) if there is a root 2 .L; T / n Fp ˛ for which dim L < p, then (a) LŒp .˛/ contains an abelian ideal A of codimension 2, (b) A \ HQ T \ ker ˛, dim A \ Li˛ 1 for all i 2 Fp , and there is k ¤ 0 such that Li˛ A for all i 2 Fp n ¹kº and dim Lk˛ =Ak˛ 1, (c) dim Li˛ 2 for all i 2 Fp , (d) HQ D T . Proof. (1) Note that LŒp .˛/ is solvable. Lemma 12.4.1(4) shows that HQ .1/ acts nilpotently on P P LŒp .˛/. Moreover, the solvability of L.˛/ implies that N.˛/ D L C i¤0 i ˛ i ¤0 ŒL i˛ ; Li˛ is a nilpotent ideal of LŒp .˛/ (Lemma 11.2.3). By the P P Engel–Jacobson theorem we now obtain that HQ .1/ C i ¤0 Li˛ C i ¤0 ŒL i˛ ; Li˛ D LŒp .˛/.1/ is nilpotent. This means that LŒp .˛/ is completely solvable. (2) Let A be an ideal of LŒp .˛/ maximal subject to the condition A.1/ T \ker ˛. According to Lemma 12.4.1(1) we have A \ HQ T \ ker ˛;

dim Ai˛ 1

Set I WD ¹x 2 N.˛/ j Œx; A T \ ker ˛º.

8 i 2 Fp :

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Suppose I D A. Then N.˛/=A maps injectively into gl.A=T \ ker ˛/. Moreover, as N.˛/ acts nilpotently, one can simultaneously represent these endomorphisms by strictly upper triangular matrices. Then 1 dim N.˛/=A .dim A=T \ ker ˛/.dim A=T \ ker ˛ 2

1 1/ .p 2

1/.p

2/:

Take k ¤ 0 with dim N.˛/k˛ =Ak˛ minimal. Then (as Li˛ D N.˛/i˛ for all i 2 Fp ) .p

1/.dim Lk˛

1/

X i 2Fp

1 dim Li˛ =Ai˛ dim N.˛/=A .p 2

1/.p

2/;

whence dim Lk˛ 12 p. Thus we have even more dim Lk˛ 12 .p 1/. Now let I ¤ A. Since LŒp .˛/=A acts trigonalizably on A=T \ ker ˛ by (1), there is an ideal I0 I of LŒp .˛/ with dim I0 =A D 1. Write I0 D F x C A. By definition of I , one has I0 .1/ ŒI; A T \ ker ˛. This contradicts the maximality of A. (3) Suppose there is a root 2 .L; T / n Fp ˛ for which dim L < p. Lemma 12.4.1(3) proves dim L Cj˛ D dim L < p for all j 2 Fp . Every HQ -composition factor of L Cj˛ has dimension less than p. On the other hand, its dimension is a p-power as HQ is nilpotent. So every HQ -composition factor is 1-dimensional. As a P consequence, HQ acts trigonalizably on every L Cj˛ . Let W j 2Fp L Cj˛ be a minimal LŒp .˛/-submodule. Lemma 12.4.1(2) shows that W is an irreducible faithful LŒp .˛/-module satisfying dim W < p 2 . Theorems 11.3.1 and 11.3.4 describe all possible Lie algebras M in this situation. Accordingly, only Theorem 11.3.4(1),(2) can occur. In these cases M WD LŒp .˛/ has an abelian ideal A of codimension 2. This proves (a). Next we apply Lemma 12.4.1(1) for A and obtain A \ HQ T \ ker ˛ and dim Ai ˛ 1 if i ¤ 0. Being an ideal A is T -invariant. Then there is an at most 2-dimensional T -invariant complement U , LŒp .˛/ D U ˚ A: Moreover, since dim.A \ T / 1, one has U \ T ¤ ¹0º. Therefore at most one nonzero root sticks out of A. Assertions (b) and (c) are now proved. If dim HQ 2, then HQ D T . So assume dim HQ > 2. As dim A \ HQ 1, this forces U HQ , dim HQ D 3. Write HQ D T ˚ F x, where x is Œp-nilpotent. Observe that now all root spaces Li˛ for i ¤ 0 are containedPin A and therefore are 1-dimensional. One immediately sees that Œx; HQ D ¹0º, Œx; i ¤0 Li˛ D ¹0º. But ˛ is rigid, a contradiction. So this final case does not happen. Lemma 12.4.4. Suppose ˛ is rigid and classical or Witt. Then .1/ (1) rad LŒp .˛/ T \ ker ˛, (2) if rad LŒp .˛/ 6 T \ ker ˛, then rad LŒp .˛/ = T \ ker ˛ is an irreducible L.˛/-module,

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(3) dim Li ˛ \ rad L.˛/ 1 and dim Li˛ 2 for each i 2 Fp , (4) HQ D T , (5) if ˛ is Witt, then ˛ is improper or dim Li˛ D 1 for all i ¤ 0 or there is 2 .L; T / n Fp ˛ for which dim L p. Proof. Due to Proposition 12.4.2 LŒp Œ˛ has a unique minimal ideal S isomorphic to sl.2/ or W .1I 1/ if ˛ is classical or Witt, respectively. Both these algebras coincide with their derivation algebras. So we have in the respective cases LŒp Œ˛ Š sl.2/;

LŒp Œ˛ Š W .1I 1/:

Since these algebras are simple, the canonical embedding L.˛/= L.˛/ \ rad LŒp .˛/ ,! LŒp .˛/= rad LŒp .˛/ D LŒp Œ˛ is a bijection. As a conclusion, LŒp .˛/ D L.˛/ C rad LŒp .˛/;

rad L.˛/ D L.˛/ \ rad LŒp .˛/:

To simplify notation we set r D rad LŒp .˛/. We stated in Lemma 11.2.3(2) that r is a nilpotent ideal. Choose an LŒp .˛/-composition series of r through T \ ker ˛, ¹0º T \ ker ˛ DW I0 ¨ I1 ¨ ¨ In D r: Since all Ij C1 =Ij are LŒp .˛/-irreducible and r acts nilpotently on these spaces, r annihilates these. So they are in fact irreducible as modules for LŒp Œ˛ and L.˛/. In addition, ŒIi ; Ij ŒIi ; r Ii 1 for 1 i j n. By definition, all Ij C1 =Ij are restricted modules. (a) Suppose n 2. (i) Consider first the case that LŒp Œ˛ Š sl.2/. As T acts naturally on LŒp Œ˛, N fN/ is an sl.2/there exist s 2 Fp , h 2 HQ , e 2 Ls˛ and f 2 L s˛ such that .e; N h; triple of LŒp Œ˛. Note that ˛.h/ ¤ 0. Therefore the h-weight space of LŒp .˛/ for the weight 0 is HQ . Next observe that I1 \ HQ T \ ker ˛ by Lemma 12.4.1(1). This N means that 0 is not an h-eigenvalue on I1 =I0 . The Lie multiplication in L induces an LŒp Œ˛-module homomorphism ' W .I2 =I1 / ˝ .I2 =I1 / ! I1 =I0 : Due to Proposition 5.3.6, ' .I2 =I1 / ˝ .I2 =I1 / is generated by its 0-weight space. .1/

However, 0 is not a weight of I1 =I0 . Therefore ' D 0. Then I2 I0 D T \ ker ˛. N By Lemma 12.4.1(1), every h-root space of I2 =I0 is 1-dimensional and 0 is not an N h-root. But I2 =I1 and I1 =I0 both are sl.2/-irreducible. Then it can only be that hN has eigenvalue 1 on both these spaces. Therefore h has eigenvalue 1 on I2 with multiplicity at least 2. The respective root space is then 2-dimensional, a contradiction.

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Rigid tori

(ii) Next consider the case that LŒp Œ˛ Š W .1I 1/. Choose a toral element h 2 T n ker ˛. Then CLŒp Œ˛ .T / D F hN is a 1-dimensional torus. Therefore there is a realization in W .1I 1/ as F hN D F x@ or F hN D F .x C 1/@ (Corollary 7.5.2). Observe that all Ij =Ij 1 are restricted LŒp Œ˛-modules. These modules are described in Theorem 7.6.10. It is immediate that they all are irreducible as modules for the Lie N We can now algebra s0 WD F @ C F x@ C F x 2 @ Š sl.2/. This algebra contains F h. argue word-by-word as before to obtain a contradiction. Consequently, n 1. (b) Suppose n D 1. In this case I1 D r. Then assertion (1) holds by definition. We mentioned above that I1 =I0 is L.˛/-irreducible. This is assertion (2). Lemma 12.4.1(1) shows that dim.Li˛ \r/ 1 for i 2 Fp . We mentioned above that Li˛ \r D Li˛ \ rad L.˛/. Since all root spaces in sl.2/ and W .1I 1/ are 1-dimensional, this proves (3). Similarly Lemma 12.4.1 shows that dim I1 \ HQ 1. As every CSA of sl.2/ and W .1I 1/ is 1-dimensional, this gives dim HQ D 2, and this proves (4) if n D 1. (c) Suppose n D 0, whence r T \ ker ˛. Assertions (1)–(3) are obviously true. Moreover, every CSA of LŒp Œ˛ is 1-dimensional. This gives dim HQ D 2, and this proves (4). (d) In order to prove (5) suppose ˛ is proper Witt and dim Li˛ D 2 for some i 2 Fp . Then ¹0º ¤ ri˛ rad L.˛/, whence rad L.˛/ 6 T . Lemma 12.4.1(3) shows that there P is j 2 Fp and a root vector w 2 .rad L.˛//j˛ D rj˛ which acts invertibly on k2Fp L Ck˛ for all 2 .L; T / n Fp ˛. Let V be an irreducible P LŒp .˛/-submodule of k2Fp L Ck˛ . We want to show that dim V p 2 , which would prove the final statement. If r is abelian, then V is induced by an irreducible subrepresentation V0 of a restricted subalgebra P LŒp .˛/ which contains r (§3.2) V D u.LŒp .˛// ˝u.P / V0 ;

r P;

such that ŒP; r acts nilpotently on V . If P D LŒp .˛/, then w 2 ŒT; w ŒLŒp .˛/; r would act nilpotently on V , which is not true. If dim LŒp .˛/=P 2, then we are done. So assume dim LŒp .˛/=P D 1. Then P equals the inverse image of W .1I 1/.0/ in LŒp .˛/ (as this is the only subalgebra of codimension 1 containing r). In particular, P is solvable. Then V0 has p-power dimension. If dim V0 p, then we are done. So assume that V0 is 1-dimensional. Then ŒP; r annihilates V0 . We now recall our assumption that ˛ is a proper root. In this case T P (Theorem 11.2.5), and therefore ŒP; rad L.˛/ contains ŒT; w D F w. But w acts invertibly on V . Finally, consider the case that r is not abelian. Then n ¤ 0, i.e., n D 1. By previous results, r=T \ ker ˛ is a nonzero at most .p 1/-dimensional irreducible restricted W .1I 1/-module. By Theorem 7.6.10, r=.T \ker ˛/ Š O.1I 1/=F . Observe that T \ ker ˛ D F t˛ . Then dim r D p. The Lie multiplication gives rise to a nonzero pairing r r ! F t˛ . Note that ¹x 2 r j Œx; r D ¹0ºº is a W .1I 1/-submodule of r. So this module has to be T \ ker ˛. As a result, r is the p-dimensional Heisenberg algebra. The representation theory of this algebra gives dim V p .p 1/=2 p 2 .

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Lemma 12.4.5. Suppose ˛ is rigid and Hamiltonian. Then (1) rad LŒp .˛/ D T \ ker ˛, (2) ˛ is improper, (3) p D 5, (4) if H acts non-trigonalizably on L, then dim L 5 for all 2 .L; T / n F5 ˛, (5) if H acts trigonalizably on L, then dim L 52 for all 2 .L; T / n F5 ˛. Proof. Due to Proposition 12.4.2 LŒp Œ˛ has a unique minimal ideal S Š H.2I 1/.2/ . Recall that Der H.2I 1/.2/ D CH.2I 1/ and H.2I 1/ is a restricted subalgebra such that CH.2I 1/=H.2I 1/ is a torus. As every torus of LŒp Œ˛ is at most 1-dimensional, one has H.2I 1/.2/ LŒp Œ˛ H.2I 1/: We have proved in Lemma 11.2.3(2) that rad LŒp .˛/ is nilpotent. Proposition 11.2.1 with g D T C LŒp .˛/.1/ and t D T shows that there is h 2 LŒp .˛/.1/ such that e

T D F hŒp C T \ ker ˛: (1) Suppose rad LŒp .˛/ ¤ T \ker ˛. Let A be an ideal of LŒp .˛/ minimal subject to the condition T \ ker ˛ ¨ A rad LŒp .˛/. Then A.1/ Œrad LŒp .˛/; A ¨ A, whence Œrad LŒp .˛/; A T \ ker ˛. Lemma 12.4.1(1) shows that dim A= T \ ker ˛ p 1, whence dim gl.A=T \ ker ˛/ .p 1/2 < dim H.2I 1/.2/ . As rad LŒp .˛/ annihilates A=T \ ker ˛, we have ŒLŒp .˛/.1/ ; A T \ ker ˛. But then ŒLŒp .˛/.1/ ; A D ¹0º. Therefore A CLŒp .T / D HQ . Then Lemma 12.4.1(1) yields A T \ ker ˛, a contradiction. (2) The image hN of h in LŒp Œ˛ is contained in H.2I 1/.2/ . By Theorem 7.5.8 N D FDH .x1 x2 / there is 2 Autc O.2I 1/ such that ˆ 2 Aut H.2I 1/ and F ˆ .h/ N or F ˆ .h/ D FDH ..1 C x1 /x2 /. In particular, ˆ .LŒ˛/ H.2I 1/ holds. We N and obtain after an adjustment by a nonzero scalar therefore substitute F hN by F ˆ .h/ N N h D DH .x1 x2 / or h D DH ..1 C x1 /x2 /. N DH .x p 2 x p 2 / D 0 and Suppose hN D DH .x1 x2 /. Then Œh; 1 2 p 2 p 2 2 x2 / .LŒ˛/

ad DH .x1

p 2 p 2 2 x2 // .H.2I 1//

.ad DH .x1

D ¹0º:

p 2 p 2 Pick u 2 HQ with uN D DH .x1 x2 /. Substituting u by u0 WD u C t˛ 2 HQ with a suitable 2 F if necessary we get u0Œp D 0. But .ad u0 /2 .L.˛// F t˛ , contradicting the rigidity of ˛. Thus hN D DH ..1 C x1 /x2 /. This means that ˛ is improper (Theorem 11.2.5). p 2 (3) Pick u1 2 HQ with uN 1 D DH ..1 C x1 /p 2 x2 / and observe that DH ..1 C p 2 x1 /p 2 x2 / increases the x2 -degree by p 3. Argue as before to obtain 2.p 3/ 1 p 2, whence p D 5.

12.4

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Rigid tori

(4) Suppose H acts non-trigonalizably on L. Set WD ¹ 2 .L; T / j .H .1/ / ¤ 0º, which is non-empty by assumption. Proposition 1.3.6 (with 0 WD .L; T / n ) shows that contains a root independent of ˛. Since every irreducible H -module has p-power dimension and .i/.H .1/ / ¤ 0 for all i 2 F5 , dim Li p D 5. Next observe that ŒDH ..1 C x1 /x2 /; DH .x24 / D 4DH .x24 / and .ad DH .x24 //2 .H.2I 1// D ¹0º: Choose a root vector w as a preimage of DH .x24 /. Then .ad w/2 .L.˛// T \ ker ˛. The rigidity of ˛ implies w Œ5 ¤ 0. Then w Œ5 2 rad P LŒp .˛/ n ¹0º D F t˛ . As .t˛ / ¤ 0, this means that ad w acts invertibly on j 2F5 LiCj˛ for all i 2 F5 . Then dim Li Cj˛ D dim Li 5. (5) Finally, assume that H .1/ acts nilpotently on L. Normalize hN D DH ..1 C x1 /x2 /. Let " W L.˛/ ! LŒp .˛/= rad LŒp .˛/ H.2I 1/ denote the canonical homomorphism. Note that H.2I 1/.2/ .L.˛// H.2I 1/ D H.2I 1/.2/ ˚ F .1 C x1 /4 @2 ˚ FDH ..1 C x1 /4 x24 / ˚ F x24 @1 : P Set H0 WD 1i 3 FDH ..1 C x1 /i x2i / H.2I 1/.2/ and H WD ".H /. Then H0 H H0 C F .1 C x1 /4 @2 C FDH ..1 C x1 /4 x24 / C F x24 @1 : Suppose a nonzero element E D 0 .1 C x1 /4 @2 C 1 DH ..1 C x1 /4 x24 / C 2 x24 @1 is contained in H . If 0 ¤ 0, then ŒE; DH ..1 C x1 /2 x22 / D 20 DH ..1 C x1 / x2 / 2 H .1/ , whence .1/ H acts non-nilpotently on L.˛/. Thus we have 0 D 0. A direct computation then shows that ad E increases the x2 -degree by at least 3 D .1=2/.p C 1/ and therefore .ad E/2 .LŒ˛/ D ¹0º holds. Choose an inverse image w 2 H of E. There is 2 F such that w Œ5 D 5 t˛ . Then v WD w t˛ has the property that .ad v/2 .L.˛// F t˛ and v Œ5 D 0. Thus the rigidity of ˛ yields v D 0 thereby proving that H D H0 . This gives LŒ˛ D H.2I 1/.2/ . Now observe that L.˛/ is a central extension of H.2I 1/.2/ and according to Equations (11.4.2) and (11.4.1) is given by the associative form .i / .j /

.k/ .l/

.DH .x1 x2 /; DH .x1 x2 // D . 1/i Cj ı.i C k; p

1/ı.j C l; p

1/

and a derivation D 2 H.2I 1/. Any other representative in the residue class D C ad H.2I 1/.2/ describes an equivalent extension. Such an equivalent extension is obtained by substituting preimages u in L.˛/ by one of the form .u/ D u C `.u/t˛ with `.u/ 2 F . Then .H / H C F t˛ is trigonalizable as well. As a result, for every equivalent extension the inverse image of CH.2I1/.2/ .DH ..1 C x1 /x2 // in L.˛/ is a trigonalizable CSA of L. Therefore we may take D D 0 .1 C x1 /4 @2 C 1 DH ..1 C x1 /4 x24 / C 2 x24 @1

for some 0 ; 1 ; 2 2 F:

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12


Then ŒD; DH ..1 C x1 /3 x23 /; DH ..1 C x1 /2 x22 / D Œ0 .1 C x1 /4 @2 ; 3.1 C x1 /2 x23 @2 3.1 C x1 /3 x22 @1 ; DH ..1 C x1 /2 x22 D 30 DH ..1 C x1 /2 x22 /; DH ..1 C x1 /2 x22 / .2/ .2/ .2/ .2/ D 30 24 DH .x1 x2 /; DH .x1 x2 / D 30 : If 0 ¤ 0, then t˛ 2 H .1/ . This contradicts our assumption that H .1/ acts nilpotently on L. Then j

j C3

ŒD; DH ..1Cx1 /i x2 D 1 .i j /DH ..1Cx1 /i C3 x2

/C2 iDH ..1Cx1 /i

1 j C4 x2 /:

Consequently, j ŒD; DH ..1 C x1 /i x2 /; DH ..1 C x1 /k x2l / D 0

if j C l 2:

Set S.l/ WD "

1

® ¯ j span DH ..1 C x1 /i x2 / j i 0; j l C 1 ;

1 l 3;

S.4/ WD T \ ker ˛ \ L.˛/: We obtain a filtration L.˛/ D S.

1/

S.3/ S.4/ ¹0º:

The above shows that the extension splits when restricted to S.0/ . In particular, ŒS.1/ ; S.3/ D ¹0º: P Fix 2 .L; T / n F5 ˛. Let V be a composition factor of the L.˛/-module i 2F5 L Ci ˛ . Let stand for the corresponding representation of L.˛/ on V . Since S.1/ acts nilpotently on L.˛/, there exists an eigenvalue function on S.1/ such that .x/ .x/IdV is nilpotent for any x 2 S.1/ . Also, S is linear on S.3/ since S.3/ is abelian. Let e1 ; f1 ; f2 be the unique elements in i ¤0 Li˛ satisfying ".e1 / D DH ..1 C x1 /2 x2 /; ".f1 / D DH ..1 C x1 /2 x24 /; ".f2 / D DH ..1 C x1 /3 x24 /: Pick an arbitrary e2 2 H with ".e2 / D DH ..1Cx1 /x2 /. The splitting of the extension when restricted to S.0/ gives Œe1 ; f1 D .8 Œe2 ; f1 D 2f1 ;

2/f2 D f2 ;

Œe1 ; f2 D .8

3/DH ..1 C x1 /4 x24 / D 0;

Œe2 ; f2 D f2 :

An easy calculation shows that .ad fi /2 .L.˛// ker " F t˛ for i D 1; 2. As ˛ Œ5 Œ5 is rigid, this forces f1 ; f2 2 F t˛ , whence .f1 / ¤ 0, .f2 / ¤ 0. Then the

12.4

Rigid tori

203

2 2-matrix .Œei ; fj / is lower triangular with nonzero elements on the diagonal. Therefore Proposition 3.2.11 applies with M D S.1/ and I D S.3/ and gives the estimate dim V 52 dim V.1/ , where V.1/ is an S.1/ -submodule of V . Moreover, ŒDH .x23 /; DH ..1 C x1 /3 x22 / D DH ..1 C x1 /2 x24 / and hence f1 2 ŒS.1/ ; S.1/ . In other words, the nilpotent algebra S.1/ acts non-trigonalizably on V.1/ . This yields dim V.1/ 5, whence dim V 53 . Since f2 acts invertibly we obtain dim L dim V D 5 1 dim V 52 : Let us combine all these results. Theorem 12.4.6. Let L be a simple Lie algebra of absolute toral rank 2 and T a 2dimensional torus in the semisimple p-envelope of L. Suppose that T is a rigid torus. Then all root spaces have equal dimension dim L D d

for all 2 .L; T / n ¹0º;

and one of the following is true: (1) d D p D 5, all nonzero roots are improper Hamiltonian, and there is a root

2 .L; T / for which .H .1/ / ¤ 0; (2) d D 2, HQ D T , j.L; T /j D p 2 , all nonzero roots are improper Witt, and rad L. / is abelian for all nonzero roots ; (3) d D 1, HQ D T and either (a) L is classical, or (b) L Š H.2I 1I ˆ.//.1/ , or (c) there are non-solvable roots; every non-solvable root ˛ is improper Witt and the central extension F t˛ ,! LŒp .˛/ W .1I 1/ splits. Proof. (1a) Suppose there is a Hamiltonian root ˛. Then Lemma 12.4.5 applies and shows that p D 5 and that every root space for every root independent of ˛ is at least 5-dimensional. Lemmas 12.4.3 and 12.4.4 show that no root can be solvable, classical or Witt. This means that all roots are Hamiltonian, and then Lemma 12.4.5(2) implies that all roots are improper. Proposition 12.4.2 shows that every L (for ¤ 0) is isomorphic as a vector space to a full root space of LŒp Œ˛.1/ Š H.2I 1/.2/ with respect to the torus FDH ..1 C x1 /x2 /. Then the dimension of every root space for a nonzero root is 5. Finally, Lemma 12.4.5(5) then implies that H .1/ acts nonnilpotently on L. This means that there is root 2 .L; T / for which .H .1/ / ¤ 0. (1b) We next assume that there are no Hamiltonian roots. Then all 1-sections L.˛/ contain a space Lk˛ of dimension less than p (Lemmas 12.4.3 and 12.4.4). Applying these lemmas again one obtains that all root spaces have dimension 2. Lemmas 12.4.3 and 12.4.4 show that HQ D T . (2) Let us assume that there is a root ˛ for which dim L˛ D 2. Then rad L.˛/ 6 H (as ˛ is solvable, classical or Witt). According to Lemma 12.4.1(3) there is k 2 Fp and u 2 .rad L.˛//k˛ such that .uŒp / ¤ 0 for all 2 .L; T / n Fp ˛.

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12


(a) Suppose there is a root ˇ independent of ˛ such that rad L.ˇ/ 6 H . According to Lemma 12.4.1(3) there is l 2 Fp and v 2 .rad L.ˇ//lˇ such that .v Œp / ¤ 0 for all

2 .L; T / n Fp ˇ. Take any 2 .Fp ˛ C Fp ˇ/ n ¹0º, write P D i˛ C jˇ and assume by symmetry j ¤ 0. Then ad u acts invertibly on every i 2Fp Li˛Cjˇ . In particular, dim L D dim L˛Cjˇ . We argue similarly with ad v and obtain dim L˛Cjˇ D dim L˛ D 2. Lemma 12.4.3(3b) shows that no root is solvable. Suppose there is a classical root

. Since p > 3, there is i 2 Fp for which Li rad L. /. Lemma 12.4.4(3) then yields dim Li D 1, a contradiction. As a consequence, all roots are Witt. Lemma 12.4.4(5) proves that all roots are improper. The L.˛/-module rad L.˛/=T \ ker ˛ is .p 1/-dimensional. Let V be a composition factor. Since rad L.˛/ is nilpotent, it annihilates V . Therefore V is an irreducible W .1I 1/-module of dimension p 1. This dimension estimate forces that it is a restricted module. Theorem 7.6.10 then shows that dim V D p 1, i.e., V D rad L.˛/=T \ ker ˛. As a consequence, rad L.˛/ is abelian or there is a nondegenerate bilinear pairing rad L.˛/ rad L.˛/ ! T \ ker ˛ D F t˛ : The first case is the claim. In the second case rad L.˛/ is the p-dimensional HeisenP berg algebra. The representation theory of this algebra shows that i 2Fp Li˛Cˇ has dimension p .p 1/=2 > 2p, a contradiction. (b) Suppose that every root independent of ˛ has the property that rad L. / H . We want to derive a contradiction. The present assumption yields that no such is solvable. Recall that .uŒp / 6D 0. Suppose there is ˇ 2 .L; T / n Fp ˛ which is a classical root. Then it is classical reductive (recall that rad L.ˇ/ HQ D T ). More exactly, L.ˇ/.1/ Š sl.2/. The only multiples of ˇ which P are roots, are ˙ˇ; 0. Let iˇ C j˛, i 2 Fp , be any root. Since u acts invertibly on l2Fp Liˇ Cl˛ , one has Liˇ ¤ ¹0º. This gives i D ˙1. As a first consequence, no such root is Witt. Then all roots independent of ˛ are classical. Let .e; h; f / be a T sl.2/-triple for L.ˇ/.1/ . Observe that U WD Lˇ C˛ C L˛ C L

ˇ C˛

is an L.ˇ/.1/ -module of dimension 4. Computing the trace of adU h gives 4˛.h/ D 0, whence ˛.ŒL ˇ ; Lˇ / D 0. Since all roots 2 .L; T / n Fp ˛ are classical, this reasoning gives ˛.ŒL ; L / D 0 for all 2 .L; T / n Fp ˛. Proposition 1.3.6 yields H ker ˛. Then L.˛/ is a CSA of toral rank 1 in L (Proposition 11.2.6). Setting X Li WD Liˇ Ck˛ ; i D ˙1; 0; k2Fp

one defines a Z-grading of L. Corollary 4.1.5 shows that L is classical. Then L Š sl.2/, a contradiction.

12.4

Rigid tori

205

As a result, all 2 .L; T / n Fp ˛ are Witt and all respective root spaces L are 1-dimensional. Suppose ˛.H / D 0. Then L.˛/ is a trigonalizable CSA of toral rank 1 in L (Proposition 11.2.6). Consequently, L.˛/.1/ is nilpotent and, moreover, acts nilpotently on L. If L.˛/.1/ ¤ ¹0º, then C.L.˛/.1/ / DW A is a nonzero abelian ideal of .T C L.˛// which acts nilpotently on L. In this case ˛ cannot be rigid. Thus L.˛/.1/ D ¹0º. Set A WD L.˛/ in Lemma 12.4.1(1) and obtain that L˛ is 1dimensional, a contradiction. So we have ˛.H / ¤ 0. Then there is ˇ 2 .L; T / n Fp ˛ with ˛.ŒL ˇ ; Lˇ / ¤ 0. Choose x˙1 2 L˙ˇ such that ˛.Œx 1 ; x1 / ¤ 0 and set h WD Œx 1 ; x1 2 T . If ˇ.h/ D 0, then F x 1 CP F h C F x1 is the 3-dimensional Heisenberg algebra. This algebra acts on W WD j 2Fp L˛Cjˇ , and h acts on this module invertibly. The representation theory of the Heisenberg algebra implies dim L˛ D dim L˛Cˇ D 1, a contradiction. This gives ˇ.h/ ¤ 0. Adjusting the elements in question we may Œp assume that .x1 ; h; x 1 / is an sl.2/-triple, whence ˇ.h/ D 2. If ˛.x1 / ¤ 0, then Œp dim L˛ D dim L˛Cˇ D 1, a contradiction. Hence ˛.x1 / D 0, whence x1 acts nilpotently on W . The representation theory of sl.2/ shows (cf. §5.3) that the hweights come in pairs, i and i occur with the same multiplicity on W . Since ˛.h/ occurs twice, so does ˛.h/. This means that 2 D dim L˛ D dim L˛ ˛.h/ˇ . This final contradiction completes the case that there is a 2-dimensional root space. (3) We now consider the case that dim L D 1 for all ¤ 0. Lemmas 12.4.3 and 12.4.4 show that HQ D T . Let us treat first several distinguished cases. (a) Claim: For every classical root ˛ the 1-section LŒp .˛/ is reductive. Namely:POtherwise there is a classical root ˛ such that rad LŒp .˛/ 6 T \ ker ˛. Let W j 2Fp Lˇ Cj˛ be an irreducible LŒp .˛/-submodule with representation W LŒp .˛/ ! gl.W /. Since W is T -invariant, it contains a root vector. Since root spaces are 1-dimensional, Lˇ Ci˛ W for some i . Lemma 12.4.1(3) shows that there is w 2 .rad L.˛//k˛ for some P k 2 Fp which acts invertibly on W , whence Lˇ Cj˛ ¤ ¹0º for all j and W D j 2Fp Lˇ Cj˛ . In particular, ˇ C Fp ˛ consists of roots. Note that .ker / \ T D ¹0º and all root vectors u 2 [i 2Fp .ker / \ Li˛ satisfy .uŒp / D 0 for all 2 .L; T /. By the Engel–Jacobson theorem ker therefore is an ideal of LŒp .˛/ acting nilpotently on L. If ker ¤ ¹0º, then C.ker / ¤ ¹0º, but this contradicts the rigidity of ˛. Therefore LŒp .˛/ acts faithfully on W . P Next we observe that dim W D j 2Fp dim Lˇ Cj˛ D p. Theorems 11.3.1 and 11.3.4 show that LŒp .˛/=C.LŒp .˛// Š sl.2/, or LŒp .˛/ Š H1 Ìsl.2/ or LŒp .˛/ Š O.1I 1/ Ì sl.2/. In the first case LŒp .˛/ is reductive, in the last case there is a root space Lk˛ , k 2 Fp , of dimension 2, a contradiction. Hence we have LŒp .˛/ Š H1 Ì sl.2/. Choose elements (not necessarily root vectors) x; y 2 rad LŒp .˛/ such that x Œp D y Œp D 0;

Œx; y D t˛ :

206

12


P Then F x C F y C F t˛ D rad LŒp .˛/ Š H1 . Set Lj WD i 2Fp Li˛Cjˇ , j 2 Fp , which is a restricted .rad LŒp .˛//-module of dimension p. Find wj 2 Lj such that Œx; wj D 0; Œt˛ ; wj D jˇ.t˛ /wj ¤ 0; j 2 Fp : p 1

The representation theory of the Heisenberg algebra shows Lj D ˚kD0 F .ad y/k .wj /. Put uj WD Œw j ; .ad y/p 1 .wj / 2 L.˛/; j 2 Fp : Since Œx; .ad y/k .wj / D k.ad y/k ŒL

j ;.ad y/

k 1

X

ŒL

1 .Œt

˛ ; wj /

D kjˇ.t˛ /.ad y/k

1 .w /, j

one gets

.wj / j ; .ad y/

l

.wj / C

lk

X

Œx; ŒL

j ; .ad y/

l

.wj /;

k 1:

lk

As .ad y/p .wj / D 0, one has ŒL

j ; .ad y/

p 1

.wj / D

p X1

.ad y/k Œw

j ; .ad y/

p 1

.wj / :

kD0

Therefore ŒL

j ; Lj

F uj C Œx; ŒL

j ; Lj

C Œy; ŒL

j ; Lj

holds. Consequently, ŒL j ; Lj F uj Crad LŒp .˛/. Note that ŒL j ; Lj is an ideal of LŒp .˛/ and LŒp .˛/= rad LŒp .˛/ Š sl.2/. Then P it can only be that ŒL j ; Lj rad LŒp .˛/. The simplicity of L yields L.˛/ D j 2Fp ŒL j ; Lj rad LŒp .˛/, a contradiction. (b) Claim: Let ˛ be a Witt root. Then the central extension F t˛ ,! LŒp .˛/ W .1I 1/ splits and ˛ is improper. Namely: Suppose the extension does not split. The nonsplit extensions have been described in Proposition 7.6.9. According to that proposition one obtains a basis .eQ 1 ; eQ0 ; : : : ; eQp 2 ; z/ such that

ŒeQp

ŒeQi ; eQj D .j

i/eQi Cj ! j C1 z; j ; eQj D 3

if

1i Cj p

ŒeQi ; eQj D 0

otherwise:

2;

Moreover, the image of eQi in LŒp .˛/=T \ ker ˛ Š W .1I 1/ is x i C1 @. We have F z D F t˛ . Then F eQp 2 C F eQ2 C F t˛ is a Heisenberg algebra. Choose x; y; wj ; Lj ; uj as in (a) and conclude analogously ŒL

j ; Lj

F uj C Œx; LŒp .˛/ C Œy; LŒp .˛/ X X F uj C F ŒeQ2 ; eQi C F ŒeQp i 1

i 1

2 ; eQi

F uj C

X i 1

F eQi C F t˛ :

12.4

207

Rigid tori

Since the former is an ideal and the latter has codimension 1 in LŒp .˛/, it can only P be that ŒL j ; Lj F t˛ . But then L.˛/ D j 2Fp ŒL j ; Lj F t˛ , a contradiction. Hence the extension splits. Suppose ˛ is a proper root. In this case x p 1 @ is a root vector satisfying Œx p 1 @; Œx p 1 @; L.˛/ F t˛ . Moreover, Theorem 11.3.2 shows P that x p 1 @ acts nilpotently on every L.˛/-module i 2Fp Li˛C for 62 Fp ˛. But then x p 1 @ acts nilpotently on L, and this implies (by the introductory remark) that N .˛/ .L; T / 6D ;. (c) Suppose ˛, ˇ are independent roots, ˇ is Witt, and ˛.ŒLˇ ; L ˇ / ¤ 0. Let i˛ C jˇ be a root and i 2 Fp . Claim: i ˛ is a root. P Namely: Consider the L.ˇ/-module W WD k2Fp Li˛Ckˇ ¤ ¹0º. Note that dim W p, and if W is p-dimensional, then Li˛Ckˇ ¤ ¹0º for all k 2 Fp . So assume that dim W < p. The representation theory of the Witt algebra (Š L.ˇ/.1/ by (b)) shows that W Š O.1I 1/=F (Theorem 7.6.10). Recall that ˛.ŒLˇ ; L ˇ / ¤ 0 by assumption and therefore ŒLˇ ; L ˇ D F t ¤ ¹0º for some t 2 T . One has t 2 T n ker ˇ because the central extension splits, i.e., ˇ.t/ ¤ 0. Then F t represents a torus in W .1I 1/ and as such it is conjugate to F x@ or F .1 C x/@. Set t 0 WD x@ or t 0 WD .1 C x/@. In both cases t 0 has p 1 different eigenvalues on O.1I 1/=F , these eigenvalues are contained in Fp , and only 0 is missing. This in turn means that all those i ˛ C kˇ are roots for which i˛.t 0 / C kˇ.t 0 / ¤ 0. In particular, i˛ is a root. We now turn to the general proof. (d) Consider the case that all nonzero roots are solvable. Arguing as in (2a) we show that all nonzero i˛ C jˇ (˛, ˇ independent and i; j 2 Fp ) are roots. For any nonzero root ˛ set B˛ WD ¹x 2 L˛ j Œx; L ˛ D ¹0ºº: For ˇ 62 Fp ˛ one obtains ŒŒB˛ ; Lˇ ; L

˛ ˇ

ŒB˛ ; L

˛

C ŒLˇ ; L

ˇ

D ŒLˇ ; L

ˇ

H T:

Since all roots are solvable, this gives T \ ker.˛ C ˇ/ \ ker ˇ D ¹0º: P Then ŒB˛ ; Lˇ B˛Cˇ . Next observe that Li˛ D 62Fp ˛ ŒLi˛ ; L , and therefore for i ¤ 1 X X ŒB˛ ; Li ˛ ŒŒB˛ ; Li˛ ; L C ŒŒB˛ ; L ; Li˛ B.iC1/˛ : ŒŒB˛ ; Lˇ ; L

62Fp ˛

P

˛ ˇ

62Fp ˛

Consequently, B WD ˛¤0 B˛ is an ideal of L. Lemma 12.4.3(3b) yields, for every nonzero root ˛, the existence of an abelian ideal A.˛/ of L.˛/ which contains L r˛ C Lr˛ for some r 2 Fp . Then Br˛ D Lr˛ ¤ ¹0º. The simplicity of L gives L D B. Therefore H D ¹0º, L.˛/ is a CSA of

208

12


toral rank 1 in L for every ˛ ¤ 0 (Proposition 11.2.6), and dim L p 2 1. Applying Theorem 9.2.11 one gets that L is one of sl.2/, W .1I 1/, H.2I 1/.2/ , H.2I 1I ˆ.//.1/ . Since the algebras sl.2/, W .1I 1/, H.2I 1/.2/ have absolute toral rank 1, only L Š H.2I 1I ˆ. //.1/ is possible. From now on we may assume that there are non-solvable roots. (e) Consider the case that all nonzero roots are classical. By (a), every LŒp . / is reductive. If for given independent roots ˛, ˇ not all of ˛ C kˇ, k 2 Fp , are roots, then Theorem 4.1.2 applies and yields that L is classical. So suppose there are independent roots ˛, ˇ such that all ˛ C kˇ, k 2 Fp are roots. As ˛ C kˇ is classical reductive, the only multiples which are roots are ˙.˛ C kˇ/, 0. Consequently, X X LD L˛Ckˇ ˚ L.ˇ/ ˚ L ˛Ckˇ : k2Fp

k2Fp

P Set Li WD k2Fp Li˛Ckˇ . Then L D L1 ˚ L0 ˚ L 1 , and this defines a Z-grading. Corollary 4.1.5 shows that L is classical. (f) Consider the case that there is at least one classical root ˛. Suppose there is a root ˇ which is solvable. Then rad L.ˇ/ 6 H . Let i˛ C jˇ, iP2 Fp be a root. Recall that rad L.ˇ/ contains a root vector acting invertibly on k2Fp Li ˛Ckˇ . Then i˛ is a root. Suppose there is a Witt root ˇ for which ˛.ŒLˇ ; L ˇ / ¤ 0 and i˛ C jˇ is a root, then (c) implies that i˛ is a root. Since ˛ is classical reductive, in both these cases we obtain i D ˙1. As in (e) we conclude that L is classical. As a result, we now may assume in this case that every root independent of ˛ is classical reductive, or it is Witt and ˛.ŒL ; L / D 0 for such . Since P H D 62Fp ˛ ŒL ; L ¤ ¹0º and ˛.H / ¤ 0, there is a classical reductive root ˇ independent of ˛. According to (e) we may assume that there is a Witt root ı (otherwise L is classical). Adjust ı so that ı.ŒLı ; L ı / ¤ 0. Note that this setting implies ˇ.ŒLı ; L ı / ¤ 0, because otherwise all i˛ C jˇ would annihilate ŒLı ; L ı , and then so would ı. We now substitute ˛ by ˇ and obtain by the former reasoning that L is classical. (g) We end up with the case that there are non-solvable roots and every nonsolvable root is Witt. By (b), this is Case (3c) of the theorem. Corollary 12.4.7. Let L be a simple Lie algebra of absolute toral rank 2. One of the following occurs: (1) L is classical; (2) L Š H.2I 1I ˆ.//.1/ ; (3) there is a 2-dimensional torus T in the minimal p-envelope of L for which N2 .L; T / ¤ ;.

12.4

209

Rigid tori

Proof. Suppose the corollary is not true. Due to Theorem 12.3.7 and Definition 12.3.8 we may assume that every 2-dimensional torus T in the minimal p-envelope of L is rigid. Let T be such. According to Theorem 12.4.6 in all cases there is a root ˛ which is improper and Hamiltonian or Witt. Choose a semisimple preimage h 2 LŒp .˛/ of FDH .x1 x2 / if ˛ is Hamiltonian and of x@ if ˛ is Witt. Put T 0 WD F h C T \ ker ˛. This is a 2-dimensional torus and contains T \ ker ˛. Therefore CL .T \ ker ˛/ is a 1-section with respect to T 0 of proper Hamiltonian or Witt type. But then T 0 cannot be rigid. Anticipating the classification result of this volume one can prove (but not use in the following) Corollary 12.4.8. Let L be a simple Lie algebra of absolute toral rank 2 and T a 2-dimensional torus in the semisimple p-envelope of L. Suppose that T is a rigid torus. Then one of the following is true: (1) L is classical and T is any 2-dimensional torus; (2) L Š H.2I 1I ˆ.//.1/ and T is any 2-dimensional torus; (3) L Š W .2I 1/ and T is conjugate to F .1 C x1 /@1 ˚ F .1 C x2 /@2 ; Œp

(4) L Š W .1I 2/ and T is conjugate to F u0 ˚ F u0 , where u0 is as in Theorem 7.6.3; (5) L Š H.2I 1I ˆ.1//; (6) L Š M.1; 1/ and T is conjugate to F .1 C x1 /@1 ˚ F .1 C x2 /@2 . Proof. If in Theorem 12.4.6 d D 5, then H is a non-trigonalizable CSA. In the next section we will show that Case 6 of this corollary occurs. If d D 2, then dim L 2 ¹2p 2 ; 2p 2 1º. The only simple Lie algebra of Cartan or Melikian type of absolute toral rank 2 with this dimension is W .2I 1/ (Theorem 10.6.3). Due to Corollary 7.5.2 T is conjugate to one of F x1 @1 ˚ F x2 @2 , F .1 C p 1 x1 /@1 ˚ F x2 @2 , F .1 C x1 /@1 ˚ F .1 C x2 /@2 . In the first and the second case x2 @2 is a T -sandwich. So these tori do not occur here. Finally, in the last case every root vector is of the form r.1 C x1 /i C1 .1 C x2 /j @1 C s.1 C x1 /i .1 C x2 /j C1 @2 . No such vector is a sandwich, and hence Theorem 12.3.7 shows that this torus is rigid. Now suppose d D 1. Then dim L p 2 C 1. If L is classical, no T -sandwiches occur for any 2-dimensional torus (Lemma 10.6.2). The only simple Lie algebras of Cartan or Melikian type of absolute toral rank 2 with this dimension are W .1I 2/, H.2I 1I ˆ.//.1/ , H.2I 1I ˆ.1// (Theorem 10.6.3). In case L Š W .1I 2/ the 2-dimensional tori are described in Theorems 7.6.3 and 7.6.5. If the torus is ruled by Theorem 7.6.3, then no root vector is Œp-nilpotent. This torus is rigid. The torus ruled by Theorem 7.6.5 has a proper Witt root ˛ by part (3) of that theorem. That torus is not rigid (Theorem 12.4.6(3c)). In case L Š H.2I 1I ˆ.//.1/ the 2-dimensional tori are described in Theorem 10.3.2. No root vector is Œp-nilpotent. Thus every 2-dimensional torus is rigid.

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12


We have not determined all 2-dimensional tori for H.2I 1I ˆ.1//. Therefore we refrain from describing all rigid tori in this case. However, the torus T WD F t ˚F t Œp , p 1 where t D DH;1 .1 C x2 C x1 x2 / is as in Proposition 10.4.11, has the property that T D F .@1 C O.0// ˚ F .@2 C O.0//. Then no T -root vector is contained in H.2I 1I ˆ.1//.0/ , and therefore every T -root vector u satisfies .ad u/p 1 ¤ 0. Again, no sandwiches exist, and therefore T is rigid.

12.5

Trigonalizability

Another important application of §12.2 and §12.3 aims to the problem of trigonalizability of nilpotent subalgebras. We assume in the following Setting 12.5: (1) Let g be a finite dimensional simple Lie algebra and gŒp denote its semisimple p-envelope. (2) Let t gŒp be a not necessarily maximal torus. (3) Set h WD Cg .t/ and assume that h is nilpotent. (4) Assume that the maximal torus of hŒp gŒp is contained in t. (5) Assume that h acts non-trigonalizably on g. Recall that for any restricted Lie algebra M the maximal ideal consisting of Œpnilpotent elements is denoted by radp M . Assumption (5) of Setting 12.5 implies that h 6 t C radp hŒp . Then .t C hŒp /=.t C radp hŒp / is a nonzero restricted hŒp -module. Lemma 12.5.1. There exists b 2 hŒp such that (1) b Œp 2 radp hŒp , (2) Œb; hŒp t C radp hŒp , Œb; Œb; hŒp radp hŒp , (3) Œb; h 6 radp hŒp , Œb; Œh; h radp hŒp . Proof. As hŒp is nilpotent and t centralizes hŒp , there exists b 2 t C hŒp , b 62 t C radp hŒp , such that Œb; hŒp t C radp hŒp . Then Œb; Œb; hŒp radp hŒp . Decompose b D bs C bn into the sum of the semisimple part bs and the Œp-nilpotent part bn . Note that according to Setting 12.5(4) bs 2 t. Then b bs has the same properties. Thus we may assume that b D bn is Œp-nilpotent. As Œb Œp ; hŒp .ad b/p 1 .t C radp hŒp / radp hŒp ; F b Œp C radp hŒp is a Œp-nilpotent ideal of hŒp . The maximality of radp hŒp gives b Œp 2 radp hŒp .

12.5

211

Trigonalizability

If Œb; h radp hŒp , then Œb; hŒp

X

.ad h/n .b/ radp hŒp :

n>0

In this case F b Cradp hŒp is a Œp-nilpotent ideal of hŒp , and therefore b 2 radp hŒp . This contradicts the choice of b. Finally, observe that and Œb; Œh; h ŒŒb; h; h Œt C radp hŒp ; h radp hŒp :

Note that g0 WD F b C g gŒp is a Lie algebra, suppose ˛ 2 .g; t/ is such that g0 .˛/ is non-solvable. By Proposition 11.2.1, there is h 2 h \ g.˛/.1/ , such that e t D F hŒp C t \ ker ˛ and the nilpotent subalgebra F b C h is a CSA of g0 .˛/ of toral rank 1 . Similarly, h is a CSA of g.˛/ of toral rank 1. Lemma 12.5.2. For any root ˛ 2 .g;t/ one has ŒF bCt; rad g.˛/.1/ rad g.˛/.1/ . Proof. There is nothing to prove if g.˛/ is solvable. Therefore we may assume g0 .˛/.1/ D g.˛/.1/ ¤ ¹0º. Let r rad g.˛/.1/ denote the maximal solvable F binvariant ideal of g.˛/.1/ . Due to the preceding remark rad g.˛/.1/ is t-invariant. In order to show that it is also b-invariant, apply Theorem 11.1.1 for F b C g.˛/.1/ (instead of g) and F b C h \ g.˛/.1/ (instead of h). Due to that theorem there is a realization g.˛/.1/ =r Š s0 ˝ O.mI 1/; t ! F t0 ˝ 1;

t0 2 s0Œp Der s0 ;

0 0 where s0 is simple, Cs0 .t0 / is a CSA of s , and b is mapped into .Der s /˝O.mI 1/ Ì Id ˝ W .mI 1/ . Suppose m ¤ 0. Recall that g.˛/.1/ =r Š s0 ˝ O.mI 1/ has no proper b-invariant solvable ideals. Therefore 2 .b/ 62 W .mI 1/.0/ . But then Œb; Œb; Cs0 .t0 / ˝ O.mI 1/ contains non-nilpotent elements. Lifting this information back to g.˛/.1/ one obtains that Œb; Œb; h contains elements which are not Œp-nilpotent. This contradicts Lemma 12.5.1(2). As a consequence, m D 0 and rad g.˛/.1/ D r. Then rad g.˛/.1/ is b-invariant. Every 2 .g; t/ may be regarded as a .t C hŒp /-root. As such it is not necessarily a linear form. However, when restricted to Œb; hŒp t C radp hŒp it is linear. Although ker is a well-defined set, it is not necessarily a linear space. Lemma 12.5.3. .Œb; Œg ; g

/

D 0 holds for all 2 .g; t/.

Proof. Suppose the claim is not true. Let be such that .Œb; Œg ; g is q 2 Œb; Œg ; g h for which .q/ ¤ 0.

/

¤ 0. There

212

12


(a) Since gi D Œq; gi for all i 2 Fp , we observe that g ; g g./.1/ and q 2 g./.1/ . Theorem 11.1.1 shows that M WD g./.1/ = rad g./.1/ is a simple e Lie algebra contained in ¹sl.2/; W .1I n/; H.2I nI ˆ/.2/ º. The semisimple part q Œp of e q is contained in t and is not contained in ker . Then t D F q Œp Cker . This shows that h \ g./.1/ is selfnormalizing. As a consequence, the image H of h \ g./.1/ in M is a CSA of toral rank 1 in M . In particular, H .1/ acts nilpotently on M . Lemma 12.5.2 shows that F b C t acts on M . We denote the images of b and t in e Der M by bN and Nt, respectively, and observe that Nt D F .adM q/p . Adjust q such that e

tN WD .adM q/p is toral: N ŒM ; M / D 0, then q acts nilpotently on M , and this implies Nt.M / D If .Œb; N ŒM ; M / ¤ 0. But then M would be nilpotent, a contradiction. Therefore .Œb; 0. Since H acts trigonalizably on M , this gives N H / ¤ 0 .Œb;

and

bN 62 H:

N N (b) Choose a maximal b-invariant subalgebra M.0/ of M containing H and a bN invariant subspace M. 1/ M.0/ such that M. 1/ =M.0/ is .M.0/ C F b/-irreducible. Look at M CF bN as a filtered Lie algebra with standard filtration (see Definition 3.5.1) N and let defined by M.0/ C F bN and M. 1/ C F b, N G WD ˚i 2Z gri .M C F b/ denote the associated graded algebra. Note that G satisfies (g1)–(g3) of Notation 3.5.2 (by Proposition 3.5.3). Recall that H M.0/ and Nt HŒp Der M (the p-envelope of H in the semisimple p-envelope of M ). Therefore the filtration is Nt-invariant. N H .M C F b/ N .1/ , then .Œb; N H / D 0 which contradicts a former (c) If Œb; N H 6 .M C F b/ N .1/ and bN … .M C F b/ N .1/ hold. statement in (a). Consequently, Œb; N Let N be the homomorphic image of HPC F b in G0 . As 0 is not an HP -weight of M=M.0/ , 0 is not an N -weight of G WD k0 I .l/ . Consequently, X

g C Œg

; g

.1/ D I .1/ D t C hŒp C g.˛; ˇ/ :

20 n¹0º

P As 20 n¹0º g 6 r.˛; ˇ/, we obtain that L is not solvable and L ¤ ¹0º is the uniquely determined minimal ideal of L. Lemma 12.5.6. (1) N2 .L; T / D ;. (2) H \L is a CSA of L and T is a 2-dimensional torus contained in the p-envelope of H \ L in Der L. (3) L is a simple Lie algebra. Proof. (1) Suppose N2 .L; T / ¤ ;. Note that CL .T / H is nilpotent. According to Corollary 12.1.3 there is 2 .L; T /, ¤ 0, and c 2 L n ¹0º for which .ad c/2 .L/ D ¹0º. Let x 2 L be an arbitrary element. Since .adŒc; x/3 D 0 by Proposition 12.1.1(3), 1 exp.adŒc; x/ D Id C adŒc; x C .adŒc; x/2 2

12.5

219

Trigonalizability

is an automorphism of L. Consider 1 exp.adŒc; x/..b// D .b/ C ŒŒc; x; .b/ C ŒŒc; x; ŒŒc; x; .b/: 2 Since exp.adŒc; x/ is an automorphism and b is Œp-nilpotent, the left-hand side exp.adŒc; x/..b// of this equation acts nilpotently on L. Next we conclude from Lemma 12.5.1(2) that ŒŒc; x; .b/ is contained in .tCradp hŒp /. Write accordingly ŒŒc; x; .b/ D t C n where t 2 .t/ and n 2 .radp hŒp /. Finally, we obtain by the same reasoning that ŒŒc; x; ŒŒc; x; .b/ .radp hŒp /. Since .b/ acts nilpotently on L and Œt; F .b/ C .radp hŒp / D ¹0º, we obtain exp.adŒc; x/..b// D t C n0 ;

Œt; n0 D 0;

ad n0 is nilpotent:

As a consequence, t D 0 and hence ŒŒc; x; .b/ 2 .radp hŒp / for all x 2 L . Choose an inverse image cQ 2 g.˛; ˇ/ of c. The former reasoning shows that ŒŒc; Q g.˛; ˇ/

; b

radp hŒp C .ker / \ h \20 ker :

But then cQ r by definition, and therefore c D .c/ Q D 0. This contradiction proves Claim (1). (2) The construction yields that T D .t/ is at most 2-dimensional. Even more, the space Œb; Œg˛ ; g ˛ C Œb; Œgˇ ; g ˇ h \ g.˛; ˇ/.1/ distinguishes the roots of 0 , and therefore T is a 2-dimensional torus contained in the p-envelope of H \ L. (3) Since L is L-simple (Lemma 12.5.5), there is a realization L Š S ˝ O.mI 1/; L .Der S/ ˝ O.mI 1/ Ì Id ˝ W .mI 1/ ; where S is a simple Lie algebra. Since T LŒp , one has T .Der S/ ˝ O.mI 1/, and then Theorem 3.6.1 shows that even more one may take a realization of L such that T .Der S / ˝ F holds. Note thatQ CL .T / ¤ ¹0º. If m ¤ 0, then there is p 1 a T -eigenvector of the form v D v0 ˝ m , and this eigenvector satisfies i D1 xi Œv; Œv; L D ¹0º. But there are no T -sandwiches, and therefore m D 0. This means that L D S is simple. Next we verify the Setting 12.3 for L and T : It is clear from Lemmas 12.5.5 and 12.5.6 that L is semisimple with unique minimal ideal L, which is a simple Lie algebra in its own right. It is also clear that H D CL .T / and T is a 2-dimensional torus contained in the p-envelope of H \ L. In addition, since t C hŒp is closed under Œp and is nilpotent with maximal torus t and the homomorphism in Lemma 12.5.5 when restricted to t C hŒp is a p-homomorphism, H is restricted and nilpotent with unique maximal torus T . As T LŒp , L D H C L holds. Theorem 12.3.7 now says that N ./ .L; T / D ; for all 2 0 n ¹0º.

220

12


Theorem 12.5.7. Suppose the assumptions of the Setting 12.5 hold. Let ˛; ˇ, r.˛; ˇ/, and , L, H , T be as in Lemmas 12.5.4 and 12.5.5. The following is true: (1) p D 5; (2) L.1/ is simple; (3) T is a 2-dimensional torus contained in L.1/ ; (4) for every 2 .L; T / n ¹0º one has (a) H.2I 1/.2/ L. /=.T \ker / H.2I 1/.2/ CF D for some D 2 H.2I 1/, (b) is Hamiltonian improper, (c) H \ L. /.1/ acts non-trigonalizably on L.1/ , (d) C.L.1/ . // D T \ ker . Proof. The proof of this theorem has several steps. Let LŒp denote the semisimple p-envelope of L. We identify L Š ad L LŒp Der L. Recall that H is closed under the mapping Œp and T H L holds. Choose any nonzero 2 .L; T / and a toral element t 2 T with ker D F t . Set L0 WD L. / and L0 WD L0 =F t . We have already established that N . / .L; T / D ;. (a) As T L0 , one has that C.L0 / D ¹x 2 H j Œx; L0 D ¹0ºº is a restricted subalgebra of LŒp . Then C.L0 / contains the semisimple and nilpotent components of every its element. If C.L0 / does contain a nonzero nilpotent element, then it contains a nonzero element a for which aŒp D 0. But such an element is contained in N . / .L; T / D ;. Therefore C.L0 / D F t holds. (b) Suppose that L0 contains an ideal A satisfying F t ¨ A;

A.1/ F t :

P Then A D A0 ˚ i 2Fp Ai , where Ai D A \ L0;i for i ¤ 0 and A0 D A \ H . For any x 2 A we have .ad x/2 .L0 / F t . Suppose A0 ¤ F t . As A0 Œp C.L0 /, this yields A0 Œp D F t . Let a0 2 Œp A0 n F t . Then a0 D a1 C t where a1 ¤ 0 and a1 D 0. But the element a1 is contained in N . / .L; T / D ;. Therefore A0 D F t . Then Ak ¤ ¹0º for some k 2 Fp . Choose a nonzero element a 2 gk which is mapped onto a nonzero element .a/ 2 Ak . One has Œ.a/; L k˛ A0 D F t , and this in turn gives Œb; Œa; g k r0 : As r0 \20 ker (Lemma 12.5.5), we obtain a 2 rk by definition, whence .a/ D 0. Consequently, rad L0 D F t D C.L0 / and L0 D L=C.L0 / is a semisimple Lie algebra. (c) Identify L0 with the subalgebra ad L0 Der L0 . Clearly, H WD H =F t is a restricted subalgebra in Der L0 and L0 has toral rank 1 with respect to H . Proposition

12.5

Trigonalizability

221

11.2.1 gives that H is a CSA of toral rank 1 in L0 . Then Theorem 11.1.1 shows that L0 contains a unique minimal ideal I , and there is a realization I Š S ˝ O.rI 1/ L0 .Der S/ ˝ O.rI 1/ Ì Id ˝ W .rI 1/ ; where S is a simple Lie algebra, adS T D F t ˝ 1

for some toral element t 2 Der S;

CS .t / is a CSA of S of toral rank 1, and t is contained in the p-envelope of CS .t/. that r ¤ 0. Let c be a preimage of a nonzero element from CS .t/ ˝ Qr Suppose p 1 x under the canonical epimorphism L0 ! L0 . Then c 2 H , .ad c/2 .L0 / i D1 i F t and c Œp D t for some 2 F (because c Œp 2 C.L0 //. The element c 1=p t is contained in N . / .L; T /, a contradiction. Consequently, L0 D S C CL0 .t/ D S C H : Recall that CS .t / D S \ H is a CSA of toral rank 1 in S . Theorem 9.2.11 shows S 2 ¹sl.2/; W .1I n/; H.2I nI ˆ/.2/ º: If S Š sl.2/ or S Š W .1I 1/, then Der S D ad S and L0 D S holds. But then any Cartan subalgebra of L0 is one-dimensional, and therefore H is 2-dimensional abelian, a contradiction. (d) The image of T in Der S is a 1-dimensional torus F t with a toral element t . Suppose there is c 2 S \ H satisfying Œc; Œc; S D ¹0º. Observe that .ad c/3 .L0 / .ad c/2 .S / D ¹0º. If Œc; Œc; L0 ¤ ¹0º, then there is d 2 H such that c1 WD Œc; Œc; d ¤ 0. Now apply Equation (12.1.4) for X D ad c and y D d . One obtains .ad c1 /2 .L0 / D .ad c/2 .ad d /2 .ad c/2 .L0 / .ad c/2 .S/ D ¹0º: Therefore we may assume in any case Œc; Œc; L0 D ¹0º. Let c 0 2 H be a preimage. Then c 0 Œp 2 C.L0 / D F t , and there is an element c 0 t (for suitable 2 F ) contained in N . / .L; T /. But there are no such elements, and this shows that S has no sandwiches contained in H . (e) Let S D S. 1/ S.0/ S.1/ S.s/ ¹0º be the natural filtration of the Lie algebra S of one of the mentioned Cartan types. Suppose S.4/ contains a t-eigenvector. Let m 4 be the largest index for which S.m/ contains such an eigenvector u. The subspace Œu; Œu; S S.2m 1/ S.mC1/ is invariant under t. By choice of m, .ad u/2 .S/ D ¹0º. By (d), u … H . Then there is k 2 Fp such that u 2 Sk . Note that Œu; S k Œu; S \ H S.m 1/ \ H . Suppose Œu; S k D ¹0º. Choose an inverse image u0 2 Lk of u. The former observation shows that Œu0 ; L k F t . Let u 2 gk be an inverse image of u0 in g. The above shows

222

12


that Œb; Œu ; g k r0 . Then Lemma 12.5.5(2) gives u 2 r , and this implies the contradiction u D 0. Consequently, there is a nonzero element v 2 Œu; S k S.m 1/ \ H . By (d), Œv; Œv; S ¤ ¹0º. As Œv; Œv; S S.2m 3/ S.mC1/ this contradicts the choice of m. One can do more. Suppose S.3/ \ H ¤ ¹0º. Let m 3 be the largest index for which S.m/ \ H ¤ ¹0º and u 2 S.m/ \ H be nonzero. The subspace Œu; Œu; S S.2m 1/ S.4/ is invariant under t. By (d), .ad u/2 .S/ ¤ ¹0º. This contradicts the former result. Consequently, S.3/ \ H D ¹0º. Note that, since t Œp D t, .ad t/p 1 Id .S / CS .t/ H . n (f) Suppose that S D W .1I n/ and n 2. Recall that x .p 1/ @ 2 W .1I n/.pn 2/ . Pn 1 i Theorem 7.1.2 shows that Der W .1I n/ D iD1 F @p C W .1I n/. This means that tD

n 1 X

i

i @p C d

i D0

for some i 2 F and d 2 W .1I n/.0/ . If 0 D D n 1 D 0, then t D d 2 W .1I n/.0/ , and W .1I n/.pn 2/ contains a t -eigenvector. So this case is impossible k by (e). Then .ad t/p 1 D k .ad @p /p 1 C for some 0 k n 1 and k ¤ 0. This is of the form k .ad @/r C with r p n p n 1 . Then ¹0º ¤ ..ad t/p

1

Id/.W .1I n/.pn

2/ /

W .1I n/.pn

1

2/

\ H:

As p n 1 2 3, this case is impossible by (e). (g) Let S D H.2I nI ˆ/.2/ with n1 n2 (see Theorems 7.1.2 and 7.1.3). It follows from Theorems 7.1.2 and 7.1.3 that tD

nX 1 1

pi

i @1 C

i D0

nX 2 1

pi

i @2 C d;

i D0

where the derivation d preserves S.0/ . If all i and i vanish, then t D d preserves the highest term in the filtration of S . This contradicts (e). The compatibility property (6.2.2) yields .p n1 2/ .p n2 1/ x2 /

DH .x1 whence S.pn1 Cpn2

5/

2 H.2I n/.2/ .pn1 Cpn2

5/

gr S;

¤ ¹0º and

¹0º ¤ .ad t/p

1

Id .S.pn1 Cpn2

5/ /

S.m/ \ H

with m .p n1 C p n2

5/

.p

1/p n2

1

p n1 C p n2

1

5:

If .n1 ; n2 / ¤ .1; 1/, then m p 5. This contradicts (e). Similarly, if p 7, then m 3 again contradicting (e). So we have n1 D n2 D 1 and p D 5.

12.5

223

Trigonalizability

Next we go into the details of the classification of Hamiltonian algebras in 2 generators (see §6.3 and in particular Theorem 6.3.10), and recall that now n1 D n2 D 1 and p D 5 hold. If ˆ D ˆ./ or ˆ D ˆ.l/, then according to the mentioned theorem S.7/ D ¹0º, S.6/ D FDH .x14 x24 / ¤ ¹0º hold. Then w WD DH .x14 x24 / 2 S.6/ ; 0 ¤ u WD ..ad t/4

Id/.w/ 2 S.2/ \ H ;

Œw; Œu; S S.7/ D ¹0º; .ad t /i .Œu; S/ D Œu; .ad t/i .S/ Œu; S S.1/ ; 4

Œu; Œu; S D Œ.ad t/ .w/

0 i 5;

w; Œu; S

3 X .ad t/ .Œw; Œu; S/ C Œ.ad t/i .w/; Œu; .ad t/4 i .S/ 4

i D0

¹0º C

3 X

ŒS.6

i/ ; ŒS.2/ ; S

S.4/ :

i D0

According to (d), Œu; Œu; S ¤ ¹0º holds. Again this contradicts (e), and this shows that S Š H.2I 1/.2/ . (h) Theorem 7.5.8 shows that an arbitrary toral element of H.2I 1/.2/ is conjugate under Aut H.2I 1/.2/ either to some element of F5 DH .x1 x2 / or to DH ..1 C x1 /x2 /. The centralizer of DH .x1 x2 / contains the sandwich DH .x13 x23 / 2 S.4/ \ H . By (d), t cannot be conjugate to DH .x1 x2 /. Then we may assume that t D DH ..1 C x1 /x2 /. In particular, is an improper Hamiltonian root (Definition 11.2.2). Note that Der H.2I 1/.2/ Š H.2I 1/ ˚ F ..1 C x1 /@1 C x2 @2 / (Theorem 7.2.1). Moreover, H.2I 1/ is a restricted ideal of Der H.2I 1/.2/ and .1 C x1 /@1 C x2 @2 is a toral element. If H contains an element u D .1 C x1 /@1 C x2 @2 C y, where y 2 H.2I 1/, then (as t D .1 C x1 /@1 x2 @2 2 H ) one has that adS T is 2-dimensional, a contradiction. This contradiction shows that L0 H.2I 1/ D H.2I 1/.2/ ˚ FDH ..1 C x1 /4 x24 / ˚ F .1 C x1 /4 @2 ˚ F x24 @1 : Suppose L0 contains a nonzero element v WD aDH ..1 C x1 /4 x24 / C bF x24 @1 2 H . Obviously, ad v increases the x2 -degree of any monomial by at least 3 D .p C 1/=2. Therefore .ad v/2 .S/ D ¹0º. Since no such elements exist by (d), we have v D 0. Therefore L0 D H.2I 1/.2/ ˚ F D, where D D0

or

D D .1 C x1 /4 @2 C a1 DH ..1 C x1 /4 x24 / C a2 x24 @1 ;

(i) Note that X i 2F5

Li L.1/ D L:

ai 2 F:

224

12


Moreover, we have shown in (h) that dim H = H \L. /.1/ 1. Then dim H = H \ L 1, and this shows that L D L.1/ . Recall that L is simple (Lemma 12.5.6). (j) Recall that the image .b/ of b in L0 satisfies .Œ.b/; Œ.b/; H / D 0 (Lemma 12.5.1(2)). Suppose .b/ … H.2I 1/.2/ . Then .b/ D a0 .1 C x1 /4 @2 C a1 DH ..1 C x1 /4 x24 / C a2 x24 @1 C d with a0 ¤ 0 (by (h)) and d 2 S \ H . Note that S \ H is abelian. But then .b/; Œ.b/; DH ..1 C x1 /3 x23 / D .b/; Œa0 .1 C x1 /4 @2 C a1 4.1 C x1 /3 x24 @2 4.1 C x1 /4 x23 @1 C a2 x24 @1 ; 3.1 C x1 /2 x23 @2 3.1 C x1 /3 x22 @1 D .b/; a0 9.1 C x1 /x22 @2 6.1 C x1 /2 x2 @1 C 12.1 C x1 /x22 @2 D a0 a0 .1 C x1 /4 @2 C a1 4.1 C x1 /3 x24 @2 4.1 C x1 /4 x23 @1 C a2 x24 @1 ; .1 C x1 /x22 @2 .1 C x1 /2 x2 @1 D a02 x2 @2 .1 C x1 /@1 C 4a0 a1 x24 @1 C 3x24 @1 C 2x24 @1 4x24 @1 D a02 DH ...1 C x1 /x2 / is not Œp-nilpotent. Consequently, a0 D 0 and this gives .b/ 2 S \ H . Then .b/ 2 L0 .1/ C F t . .1/ (k) Since S \ H is abelian, we have L0 .1/ \ H F t . On the other hand, .1/ if Œ.b/; L0 \ H D ¹0º, then in particular Œb; Œg ; g r0 . Lemma 12.5.5(2) .1/ gives g r , whence 62 .L; T /. This contradiction shows L0 .1/ \ H D .1/ F t . Consequently, L0 \ H acts non-trigonalizably. Moreover, for independent roots ˛; ˇ one obtains T D T \ ker ˛ C T \ ker ˇ L.1/ . Next we will characterize the Lie algebras and tori described in Theorem 12.5.7. Theorem 12.5.8. Suppose p D 5, L is a simple Lie algebra, and T is a 2-dimensional torus of LŒp which is contained in L. Assume that for every 2 .L; T / n ¹0º the following holds: (a) H.2I 1/.2/ L. /=.T \ ker / H.2I 1/.2/ C F D for some D 2 H.2I 1/; (b) is Hamiltonian improper; (c) CL. /.1/ .T / acts non-trigonalizably on L; (d) C.L. // D T \ ker . Then L Š M.1; 1/ under an isomorphism which maps T onto F .1 C x1 /@1 ˚ F .1 C x2 /@2 .

12.5

Trigonalizability

225

Proof. The proof of this theorem is done in steps by a subtle computation in terms of a suitable basis and their multiplication coefficients. Indeed, we will prove that L and T are uniquely determined up to isomorphisms and then detect these by showing that the algebra M.1; 1/ and the torus F .1 C x1 /@1 ˚ F .1 C x2 /@2 do have the required properties. (I) We fix independent roots ˛; ˇ 2 .L; T /. Choose a toral element t˛ 2 T \ ker ˛ such that ˇ.t˛ / D 1. Put X Li WD Liˇ Cj˛ ; i 2 F5 : j 2F5

Then adLi t˛ D i IdLi ;

i 2 F5 :

Since H.2I 1/=H.2I 1/.2/ is p-nilpotent, the homomorphism W L.˛/ ! L.˛/=F t˛ H.2I 1/.2/ C F D maps T onto a 1-dimensional torus of H.2I 1/.2/ . We will adjust by an automorphism of H.2I 1/.2/ . This adjustment will also modify D. Since ˛ is an improper root, Theorem 7.5.8 shows that we may assume .T / D P FDH ..1 C x1 /x2 /. Since CL.˛/.1/ .T / acts non-trigonalizably but .CL.˛/.1/ .T // D 3i D1 FDH ..1 C x1 /i x2i / is abelian, one has F t˛ D ker L.˛/.1/ . Then T L.˛/.1/ . Let the central extension 0 ! F t˛ ! L.˛/.1/ ! H.2I 1/.2/ ! 0 .5/ .4/ .4/ .5/ be given by the derivation D WD DH 0 x1 C 1 x1 x2 C 2 x2 and the bilinear P form (see Equations (11.4.2) and (11.4.1)). As .CL.˛/.1/ .T // D 3i D1 FDH ..1C x1 /i x2i /, we have that CL.˛/.1/ .T / D T C F u2 C F u3 , where .ui / D DH ..1 C x1 /i x2i / for i D 2; 3. By assumption, CL.˛/.1/ .T / acts non-trigonalizably. This gives .ŒD; .u2 /; .u3 // ¤ 0 or .ŒD; .u3 /; .u2 // ¤ 0. As .4/ .4/ .5/ ŒDH 1 x1 x2 C 2 x2 ; DH ..1 C x1 /i x2i / D 0 for i D 2; 3, we conclude 0 ¤ 0. Lemma 11.4.6 gives rise to an automorphism .5/ of O.2I 1/ such that ı D ı 1 D DH .0 x1 / and ı DH ..1 C x1 /x2 / ı 1 D .5/ DH ..1 C x1 /x2 /. By Lemma 11.4.4 we may assume that D D DH .0 x1 /. Then .2/ .3/ ŒD; DH .x2 /; DH .x2 / ¤ 0 and therefore the central extensions of this type are described in Equation (11.4.3): there is a basis of L.˛/.1/ j

.zI x1i x2 j .0; 0/ < .i; j / < .4; 4// such that j

j

.x1i x2 / D DH .x1i x2 /;

.0; 0/ < .i; j / < .4; 4/;

.1/

226

12


and (setting z D qt˛ with q 2 F ) the multiplication is given by 8 ˆ 0:

.1/

By Equation (2), L0;.0/ is a subalgebra. Moreover, Equation (2) shows that the central .1/

extension splits when restricted to L0;.1/ : .1/

L0;.1/ D F x23 C F x24 C

X iCj 3; i ¤0

j

F x1i x2

228

12

Sandwich elements and rigid tori .1/

.1/

.1/

is closed under multiplication and ŒL0;.i/ ; L0;.j / L0;.i Cj / holds whenever i; j > P .1/ j 0. Similarly, L0;.2/ D F x24 C i Cj 4; i ¤0 F x1i x2 and therefore .1/

.1/

j

X

ŒL0;.0/ ; L0;.2/ D F Œx22 ; x24 C F Œx23 ; x24 C

.1/

F Œx1i x2 ; x1k x2l L0;.2/

i Cj 2; kCl4 i Ck¤0

.1/

.1/

.1/

holds. Then ŒL0;.0/ ; L0;.m/ L0;.m/ holds for all m 2. To derive a contradiction assume that .x1r x2s / ¤ 0 for some .r; s/ with r C s 3 or r D 2; s D 0, and choose such a pair for which r C s is maximal. Set m WD .1/ r C s 2 0. The assumption implies that x1r x2s 2 L0;.m/ acts invertibly on L1 . Suppose r D 4, whence s 3 and m 2. Set e1 WD x12 ;

f1 WD x13 x2sC1 ;

e2 WD x1 x2 ;

f2 WD x14 x2s :

Then Œe1 ; f1 D 2.s C 1/x14 x2s ¤ 0;

Œe1 ; f2 D 0;

Œe2 ; f2 D .s

.1/

.1/

4/x14 x2s ¤ 0:

Apply Proposition 3.2.11 with G WD L0 .1/ ;

M WD L0;.1/ ;

I WD L0;.m/ M .1/

and V a composition factor of the L0 .1/ -module L1 . One has f1 ; f2 2 L0;.m/ , hence .1/

Œei ; fj ; Œei ; Œej ; fk 2 L0;.m/ M .1/

.1/

and ŒM; I L0;.mC1/ . Set R L0;.m/ the Lie algebra generated by

P

i;j F Œei ; fj . .1/ As m 2, we have R.1/ By choice of m every x1k x2l 2 L0;.mC1/ acts .1/ nilpotently on L1 . The Engel–Jacobson theorem shows that the subalgebra L0;.mC1/ itself acts nilpotently on L1 . Then ŒM; I C R.1/ C F Œe1 ; f2 acts nilpotently on L1 . Proposition 3.2.11 now shows that 25 D dim L1 dim V 52 dim V0 , where V0 is an irreducible M -module. This gives dim V0 D 1. But Œx12 x2 ; x13 x2s D .2s 3/x14 x2s ¤ 0 is contained in M .1/ and acts invertibly on L1 . .1/ L0;.mC1/ .

One argues analogously interchanging x1 ; x2 if s D 4. Consequently, r; s 3. Next suppose r C s 3, choose such a pair with maximal r. Note that m 1. Set e1 WD x1 ;

f1 WD x1r x2sC1 ;

e2 WD x2 ;

f2 WD x1rC1 x2s :

12.5

229

Trigonalizability

By Equation (2) and the choice of r one has Œe1 ; f1 D .s C 1/x1r x2s ¤ 0;

.Œe1 ; f2 / D 0;

.r C 1/x1r x2s ¤ 0:

Œe2 ; f2 D

Apply Proposition 3.2.11 with G WD L0 .1/ ;

.1/

M WD L0;.0/ ;

.1/

.1/

I WD L0;.mC1/ L0;.2/ :

Then I is an ideal of M by an earlier remark and .1/

f1 ; f2 2 I;

Œei ; fj 2 L0;.m/ ;

.1/

Œei ; Œej ; fk 2 L0;.0/ D M; .1/

.1/

and ŒM; I C R.1/ C F Œe1 ; f2 L0;.mC1/ C F Œe1 ; f2 . As before, L0;.mC1/ and .1/

F Œe1 ; f2 act nilpotently on V , whence the subalgebra L0;.mC1/ CF Œe1 ; f2 acts nilpo.1/

tently. The proposition shows that L1 contains a 1-dimensional L0;.0/ -submodule. However, Œx22 ; x23 D 2t˛ acts invertibly on L1 . .1/ It remains to consider the case r D 2; s D 0. Under the present assumption L0;.1/ acts nilpotently on L1 . Put in Proposition 3.2.11 G WD L0 .1/ ;

.1/

M WD L0;.0/ ;

.1/

I WD L0;.2/ ;

e1 WD x2 ;

f1 WD x24

.1/

to obtain that L1 contains an at most 5-dimensional L0;.0/ -submodule. Then set .1/

.1/

G WD L0;.0/ ;

M WD L0;.1/ C F x12 C F x1 x2 C F t˛ ;

e1 WD x22 ;

f1 WD x23

.1/

I WD L0;.1/ ;

and observe that also in this case I is an ideal of M . We obtain that L1 contains a 1-dimensional M -submodule. However, Œx1 x2 ; x12 D 2x12 acts invertibly on L1 . Œ5

In order to obtain x2 D 0 and .x22 /Œ5 D 0 we have to apply a suitable automorphism. Note that ad.1 C x1 /3 x23

3

x1k x2l 2 Œ.1 C x1 /3 x23 ; O.2I 1/x24 C F t˛ O.2I 1/x26 D ¹0º

3 for all k; l, whence adL0 .1/ .1 C x1 /3 x23 D 0. One similarly proves adL0 .1/ .1 C 3 x1 /4 x24 D 0. As Œ.1 C x1 /3 x23 ; .1 C x1 /4 x24 D 0, this shows that for arbitrary 3 ; 4 2 F the mapping WD exp adL0 .1/ 3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24

230

12


is an automorphism of L0 .1/ satisfying .t˛ / D t˛ ;

.1 C x1 /x2

.1/

.1 C x1 /x2 2 ker D F t˛ : .1/

.1/

Then .T / D T . Moreover, .L0;.1/ / D L0;.1/ and .x12 / 2 x12 C L0;.1/ . Therefore .1/ F x12 C L0;.1/ still acts nilpotently on L1 . Next .x22 / D x22 C Œ3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; x22 1 C 3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; 2 Œ3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; x22 D x22 C 3 .1 C x1 /2 x24 C 33 t˛ 1 C Œ3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; 3 .1 C x1 /2 x24 C 33 t˛ 2 2 D x2 C 3 .1 C x1 /2 x24 C 33 t˛ ; .x2 / D x2 C Œ3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; x2 1 C 3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; 2 Œ3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; x2 D x2 C 33 .1 C x1 /2 x23 C 44 .1 C x1 /3 x24 4 t˛ 1 C Œ3 .1 C x1 /3 x23 C 4 .1 C x1 /4 x24 ; 2 33 .1 C x1 /2 x23 C 44 .1 C x1 /3 x24 D x2 C 33 .1 C

x1 /2 x23

C 44 .1 C

x1 /3 x24

4 t˛

4 t˛ : j

Next we have to compute the Œ5-powers of these elements. Recall that .x1i x2 /Œ5 D 0 if i C j > 2 by Claim 2. Also, Œx22 ; .1 C x1 /2 x24 D Œx22 ; t˛ D Œ.1 C x1 /2 x24 ; t˛ D 0; Œ.1 C x1 /2 x23 ; .1 C x1 /3 x24 D Œ.1 C x1 /2 x23 ; t˛ D Œ.1 C x1 /3 x24 ; t˛ D 0: Therefore .x22 /Œ5 D .x22 C 3 .1 C x1 /2 x24 C 33 t˛ /Œ5 D .x22 /Œ5 C 335 t˛ D ..x22 / C 33 /5 t˛ and 33 .1 C x1 /2 x23 C 44 .1 C x1 /3 x24

4 t˛

Œ5

D

45 t˛ :

12.5

231

Trigonalizability

Put in Lemma 10.5.6 a WD x2 , b WD 33 .1 C x1 /2 x23 C 44 .1 C x1 /3 x24 observe that Œb; .ad a/i .b/ 2 Œ33 .1 C x1 /2 x23 C 44 .1 C x1 /3 x24

4 t˛ and

4 t˛ ; O.2I 1/x23 C F t˛ D ¹0º

for all i 0. This gives s1 D s2 D s3 D 0 and Œ5

.x2 /Œ5 D x2

45 t˛ C 44 .ad x2 /4 .x13 x24 / D ..x2 /5

45 C 4 /t˛ :

There is a choice of 3 ; 4 such that .x2 /Œ5 D .x22 /Œ5 D 0. Substituting by ı we may assume that

1

j

Equations (1)–(3) and .x1i x2 /Œ5 D 0 if .i; j / ¤ .1; 1/; .1; 0/ hold. (III) Next we determine the action of L0 .1/ on each Lr for r ¤ 0. Set m WD F x12 C

X

j

F x1i x2 :

i Cj 3

This is a subalgebra which according to Claim 2 acts nilpotently on Lr . Then F x1 x2 C m acts trigonalizably on Lr , and therefore there is a common eigenvector erC 2 Lr , j

Œx1i x2 ; erC D 0 for i C j 3 or i D 2; j D 0;

Œx1 x2 ; erC D r erC ; r 2 F: (4)

P Note that 4i D1 F x2i C F t˛ is isomorphic to the Heisenberg algebra H2 . Since t˛ acts invertibly on Lr , every H2 -composition factor of Lr is 25-dimensional. Then Claim 1 shows that Lr is induced by any 1-dimensional subrepresentation of the restricted subalgebra F x23 C F x24 C F t˛ . The definition of erC and (II) then show that Lr D ˚4i;j D0 F .ad x2 /i .ad x22 /j .erC /;

.ad x2 /5 D .ad x22 /5 D 0:

(5)

Set er WD .ad x2 /4 .ad x22 /4 .erC /; whence Œx2 ; er D Œx22 ; er D 0

(6)

by (5). Arguing as above one gets Lr D ˚4i;j D0 F .ad x23 /i .ad x24 /j .er /;

.ad x23 /5 D .ad x24 /5 D 0:

(7)

It is a property of induced modules that the only subspace of Lr annihilated by F x23 C F x24 is F erC and the only subspace of Lr annihilated by F x2 C F x22 is F er .

232

12


Since Œx24 ; Œx1 ; erC D Œx1 ; Œx24 ; erC

4Œx23 ; erC D 0;

we obtain by (5) Œx1 ; erC 2

4 X

F .ad x22 /i .erC /:

i D0

Using the fact that Œx1 ; erC is an eigenvector of the operator ad x1 x2 with eigenvalue 1 C r , we conclude that Œx1 ; erC D r .ad x22 /2 .erC / for some r 2 F . As Œx23 ; Œx1 ; erC D 3Œx22 ; erC , one has (see Lemma 2.1.5(1) and Equation (2)) 3.ad x22 /.erC / D Œx23 ; Œx1 ; erC D r .ad x23 /.ad x22 /2 .erC / D r .ad x22 /2 .ad x23 / 2.ad x22 / adŒx22 ; x23 .erC / D

2r .ad x22 /2.ad t˛ /.erC / D rr .ad x22 /.erC /:

Then r D 2=r, whence 2 Œx1 ; erC D .ad x22 /2 .erC /: r

(8)

By (6), Lemma 2.1.5(1) and (8), Œx1 ; er D .ad x1 /.ad x2 /4 .ad x22 /4 .erC / D .ad x2 /4 .ad x1 /.ad x22 /4 .erC / D .ad x2 /4 .ad x22 /4 .ad x1 / 4.ad x22 /3 . 2/.ad x2 / .erC / 2 D .ad x2 /4 .ad x22 /6 .erC / C 8.ad x2 /5 .ad x22 /3 .erC / D 0: r We note Œx1 ; er D 0: By Lemma 2.1.5(1), r r 0 D .ad x1 /.ad x12 /.erC / D .ad x12 /.ad x1 /.erC / 2 2 D .ad x12 /.ad x22 /2 .erC / D .ad x22 /2 .ad x12 / 2.ad x22 / adŒx22 ; x12 C ad.Œx22 ; Œx22 ; x12 / .erC / D 2.ad x22 /.ad x1 x2 / 2.ad x22 / .erC / D

2.r C 1/.ad x22 /.erC /:

(9)

12.5

233

Trigonalizability

This yields Œx1 x2 ; erC D

erC ;

Œx1 x2 ; er D Œx1 x2 ; .ad x2 /4 .ad x22 /4 .erC / D er :

(10)

As a result, the equations (4), (5), (8), (10) determine the multiplication of L0 .1/ on every Lr for r ¤ 0. Note that erC is only determined up to a nonzero scalar factor. (IV) Next we determine the multiplication Lr L r ! L0 and the complete action of L0 on every Lr . Equations (5) and (7) give ŒLr ; L

r

4 X

D

.ad x2 /i .ad x22 /j .ŒerC ; L

r /

i;j D0 4 X

D

(11) .ad x2 /i .ad x22 /j .ad x23 /k .ad x24 /l .ŒerC ; e

r /;

r ¤ 0:

i;j;k;lD0

We have to determine ŒerC ; e r 2 L0 . By assumption (a) of the present theorem we find D 2 L0 such that L0 D L0 .1/ C F D, where .D/ D 1 x14 @2 C 2 DH .x14 x24 / C 3 x24 @1 for suitable 1 ; 2 ; 3 2 F . Here D D 0 might be possible. Using (10) one gets ŒerC ; e r 2 CL0 .x1 x2 /, i.e., ŒerC ; e

r

.r/

.r/

.r/

.r/

.r/

D 0 D C 1 x1 x2 C 2 x12 x22 C 3 x13 x23 C 4 t˛ :

Note that Œ.D/; DH .x1 / D Œ1 x14 @2 D

2 x13 x24 @2 C 2 x14 x23 @1 C 3 x24 @1 ; @2

2 x13 x23 @2

32 x14 x22 @1 C 3 x23 @1

D 2 DH .x14 x23 / C 3 DH .x24 /; and similarly Œ.D/; DH .x22 / D 21 DH .x14 x2 /. By (1), this means ŒD; x1 D 2 x14 x23 C 3 x24 C 1 t˛ ; ŒD; x22 D 21 x14 x2 C 2 t˛ ;

1 ; 2 2 F:

Hence .ad x1 /.ŒerC ; e

r / .r/

.r/

.r/

.r/

.r/

D Œx1 ; 0 D C 1 x1 x2 C 2 x12 x22 C 3 x13 x23 C 4 t˛ D

.r/

.r/

.r/

.r/

0 .2 x14 x23 C 3 x24 C 1 t˛ / C 1 x1 C 22 x12 x2 C 33 x13 x22

234

12


and .ad x1 /.ŒerC ; e

r /

D ŒŒx1 ; erC ; e

r

2 2 D Œ.ad x22 /2 .erC /; e r / D .ad x22 /2 .ŒerC ; e r / r r 2 .r/ .r/ .r/ .r/ .r/ D .ad x22 /2 0 D C 1 x1 x2 C 2 x12 x22 C 3 x13 x23 C 4 t˛ r 2 .r/ 2 .r/ Œx2 ; 21 x14 x2 2 t˛ C 1 Œx22 ; 2x22 D r 0 .r/ .r/ 42 Œx22 ; x1 x23 3 Œx22 ; x12 x24 D

2 .r/ .r/ 0 1 x13 x22 C 32 x24 : r

It follows that .r/

.r/

.r/

.r/

0 2 D 0 1 D 1 D 2 D 0; .r/

.r/

.r/

1 .r/ .r/ 0 3 D 2 ; r

.r/

2 .r/ .r/ 33 D 0 1 : r

.r/

.r/

.r/

If 0 1 D 0, then 1 D 2 D 3 D 0 and 0 1 D 0 2 D 0 3 D 0. .r/ This gives 0 D 2 F t˛ , ŒerC ; e r 2 F t˛ and ŒLr ; L r F t˛ by (11). However, .r/ in this case Claim 1(2) shows that L0 D F t˛ , a contradiction. Thus 0 ¤ 0, 1 ¤ 0, 2 D 3 D 0. Adjusting D by a nonzero scalar we may assume 1 D 1, i.e., D is a preimage of x14 @. In particular, .ad .D//2 .H.2I 1/.2/ / D ¹0º, whence .ad D/5 .L0 / D ¹0º holds. As L1 is L0 .1/ -irreducible, there is 2 F such that .adL1 D/5 5 adL1 t˛ D 0, and then Claim 1(2) gives .ad D/5 D 5 ad t˛ . We now take another preimage of x14 @2 , namely D t˛ and obtain .ad D/5 D 0. We note .D/ D x14 @2 ; .r/

.ad D/5 D 0:

(12)

.r/

Set .r/ WD r0 ¤ 0 and .r/ WD 4 =.r/ . The above gives ŒerC ; e

r

1 D .r/ . D r

1 3 3 x x C .r/ t˛ /; r2 1 2

.r/ ; .r/ 2 F:

Next we observe that Œ.ad x2 /i .ad x22 /j .erC /; .ad x2 /k .ad x22 /l .e Cr / . 1/kCl Œ.ad x2 /i Ck .ad x22 /j Cl .erC /; e Cr

.mod L0 .1/ /:

This gives 1 .r/ D ŒerC ; e r

r

D ŒerC ; .ad x2 /4 .ad x22 /4 .e Cr /

Œ.ad x2 /4 .ad x22 /4 .erC /; e Cr D Œer ; e Cr D

Œe Cr ; er

.

r/

1 D r

12.5

235

Trigonalizability

modulo L0 .1/ , and so .r/ D . r/ holds. Substitute e1C by ..1/ / 1 e1C and e2C by ..2/ / 1 e2C . Then we get .r/ D 1 for all r D 1; : : : ; 4. As a consequence, ŒerC ; e

r

1 3 3 .x x / C .r/ t˛ : r2 1 2

1 D D r

(13)

Next we determine products of D with arbitrary elements of L. Note that j

j

Œ.D/; DH .x1i x2 / D Œx14 @2 ; ix1i 1 x2 @2 j 1

D ijx1i C3 x2

@2 j 1

D jDH .x1i C4 x2

j 1

jx1i x2

@1 j 2

1/x1i C4 x2

j.j

@1

j 1

jx1i C3 x2

@2

/;

whence j

j 1

ŒD; x1i x2 D jx1i C4 x2 j

j

C .x1i x2 /t˛ j

for some .x1i x2 / 2 F . As ŒD; x1i x2 is an eigenvector with respect to ad x1 x2 of j eigenvalue j i , one has .x1i x2 / D 0 if i ¤ j . If i D j ¤ 0, then 1 i; j < 4 and we obtain j

1 j C1 ŒD; Œx1 ; x1i x2 j C1 1 1 j C1 j C1 D Œx1 ; ŒD; x1i x2 C ŒŒD; x1 ; x1i x2 j C1 j C1 1 j 1 j 1 j C1 Œ.x1 /t˛ ; x1i x2 D jx1i C4 x2 : D jx1i C4 x2 C j C1

ŒD; x1i x2 D

As a result, j

j 1

ŒD; x1i x2 D jx1i C4 x2

;

.0; 0/ < .i; j / < .4; 4/:

(14)

In particular, ŒD; erC is annihilated by F x23 C F x24 . Then (III) shows that ŒD; erC 2 F erC . Since D acts nilpotently on L, we get ŒD; erC D 0:

(15)

We have now determined the action of L0 on Lr and the multiplication Lr L r ! L0 for all r 2 F5 up to the numbers .r/ . (V) Arguing as in (11) the multiplication Lr Ls ! LrCs for r; s; r C s ¤ 0 is given by ŒerC ; es . Using (10) one gets ŒerC ; es 2 CLrCs .x1 x2 /, whence ŒerC ; es D 0 .ad x22 /3 C 1 .ad x2 / C 2 .ad x2 /2 .ad x22 /2 C C 3 .ad x2 /3 .ad x22 /4 C 4 .ad x2 /4 .ad x22 / .erCs /

236

12


for some 0 ; : : : ; 4 2 F . Observe that Œx1i x24 ; x22 D 0 and Œx1i x24 ; e C t D 0 for all i D 0; : : : ; 4 and t ¤ 0. Lemma 2.1.5(1) then gives C / Œx13 x24 ; ŒerC ; es D 4 ad .ad x2 /4 .x13 x24 / .ad x22 /.erCs C D 4 .3Š/. 1/3 .adŒx2 ; x24 /.ad x22 /.erCs / C 4 .ad t˛ /.ad x22 /.erCs /D

D

C 4 .r C s/.ad x22 /.erCs /;

Œx13 x24 ; ŒerC ; es D ŒerC ; Œx13 x24 ; es D ŒerC ; .ad x13 x24 /.ad x2 /4 .ad x22 /4 .esC / D ŒerC ; .ad t˛ /.ad x22 /4 .esC / D

sŒerC ; .ad x22 /4 .esC /:

As k/t˛k .ad x24 /4

Œ.ad x2 /; : : : ; Œ.ad x2 /; .ad x24 /4 : : : D 4 .5 „ ƒ‚ …

k

k times

for 1 k 4, Lemma 2.1.5(1) yields C .ad x24 /4 .ŒerC ; es / D 4 .4Š/.ad t˛ /4 .ad x22 /.erCs /

D

C .r C s/4 4 .ad x22 /.erCs /D

C 4 .ad x22 /.erCs /;

.ad x24 /4 .ŒerC ; es / D ŒerC ; .ad x24 /4 .es / D ŒerC ; .ad x24 /4 .ad x2 /4 .ad x22 /4 .esC / D .4Š/ŒerC ; .ad t˛ /4 .ad x22 /4 .esC / D

s 4 ŒerC ; .ad x22 /4 .esC / D

ŒerC ; .ad x22 /4 .esC /:

Combining these four equations one gets C ŒerC ; .ad x22 /4 .esC / D 4 .ad x22 /.erCs /D

s Œe C ; .ad x22 /4 .esC /; r Cs r

and this gives ŒerC ; .ad x22 /4 .esC / D 0, hence 4 D 0. Using (8), (9) and (6) one gets 2 Œx1 ; ŒerC ; es D .ad x22 /2 .ŒerC ; es /: r From this we deduce the equations (observe 4 D 0) 2 Œx1 ; ŒerC ; es D .ad x22 /2 .ŒerC ; es / r 2 D .ad x22 /2 0 .ad x22 /3 C 1 .ad x2 / C 2 .ad x2 /2 .ad x22 /2 r C C 3 .ad x2 /3 .ad x22 /4 .erCs / 2 C D 1 .ad x2 /.ad x22 /2 C 2 .ad x2 /2 .ad x22 /4 .erCs / r

12.5

237

Trigonalizability

and (by (8)) Œx1 ; ŒerC ; es D .ad x1 / 0 .ad x22 /3 C 1 .ad x2 / C 2 .ad x2 /2 .ad x22 /2 C C 3 .ad x2 /3 .ad x22 /4 .erCs / D 0 .ad x22 /3 C 1 .ad x2 / C 2 .ad x2 /2 .ad x22 /2 C C 3 .ad x2 /3 .ad x22 /4 .ad x1 /.erCs / C 60 .ad x2 /.ad x22 /2 C 42 .ad x2 /3 .ad x22 / C C 83 .ad x2 /4 .ad x22 /3 .erCs / 2 C D 1 .ad x2 /.ad x22 /2 C 2 .ad x2 /2 .ad x22 /4 .erCs / r Cs C 0 .ad x2 /.ad x22 /2 2 .ad x2 /3 .ad x22 / C C 33 .ad x2 /4 .ad x22 /3 .erCs /: We obtain

Set .r;s/ WD

2 2 1 D 1 C 0 ; 2 D 3 D 0: r r Cs r rCs 1 .

Then 0 D 21

s s D 2 2 .r;s/ r.r C s/ r

and ŒerC ; es D .r;s/

r C s 2s C .ad x2 / C 2 .ad x22 /3 .erCs /; r r

r; s; r C s ¤ 0:

(16)

(VI) It remains to determine the coefficients .r/ and .r;s/ . Since Œer ; e r is annihilated by ad x1 (by (9)) and ad x2 (by (6)), we have Œer ; e r 2 F x1 C F x2 C F t˛ . Moreover, it is an .ad x1 x2 /-eigenvector with eigenvalue 2 (by (10)). This gives Œer ; e r D 0. Similarly, if r C s ¤ 0, then, as Œer ; es is annihilated by ad x22 and ad x2 , we have Œer ; es 2 F erCs by (III). Considering .ad x1 x2 /-eigenvalues one obtains Œer ; es D 0. We note Œer ; es D 0 8r; s ¤ 0: Observe that ŒD; x22 2 m by (14), whence ŒD; .ad x22 /i .esC / D 0 by (15). Next, by (14), .ad x2 /i .D/ D .ad x2 /i 1 .x14 / D .i 1/Šx15 i

238

12


for i D 1; : : : ; 4, and this gives ad .ad x2 /i .D/ .ad x22 /4 .esC / ! 4 X b 4 D .i 1/Š . 1/ .ad x22 /4 b

b

ad .ad x22 /b .x15 i / .esC /:

bD0

If i D 1, the above is contained in equals 4 X

. 1/

b

bD0

D

! 4 .ad x22 /4 b

b

P

l

.ad x22 /l .ad m/.esC / D ¹0º. If i D 2, the above

ad .ad x22 /b .x13 / .esC /

Œ.ad x22 /4 .x13 /; esC D

. 1/3 .3Š/23 ŒŒx22 ; x23 ; esC D 8Œ2t˛ ; esC D sesC :

If i D 3, the above equals 2

4 X

. 1/b

bD0

! 4 .ad x22 /4 b

b

ad .ad x22 /b .x12 / .esC /

D . 2/. 4/.ad x22 /3 . 4/.ad x1 x2 /.esC / C . 2/.ad x22 /2 8.ad x22 /.esC / D .ad x22 /3 .esC /: If i D 4, the above equals 4 X

. 1/b

bD0

D

! 4 .ad x22 /4 b

b

ad .ad x22 /b .x1 / .esC /

2 .ad x22 /6 .esC / C 4.ad x22 /3 . 2/.ad x2 /.esC / D 2.ad x2 /.ad x22 /3 .esC /: ƒ‚ … s ƒ‚ … „ „ bD0

bD1

Then ŒD; es D ŒD; .ad x2 /4 .ad x22 /4 .esC / ! 4 X i 4 D . 1/ .ad x2 /4 i ad .ad x2 /i .D/ .ad x22 /4 .esC / i i D0

D s.ad x2 /2 .esC / C .ad x2 /.ad x22 /3 .esC / C 2.ad x2 /.ad x22 /3 .esC /: We note ŒD; es D 3.ad x2 /.ad x22 /3 C s.ad x2 /2 .esC /:

12.5

Trigonalizability

239

Further, ad .ad x2 /i .x13 x23 / .ad x22 /4 .esC / D 0 for i ¤ 3 (by (2) and (4)). Using the equation .ad x23 /.ad x22 /4 .esC / D 4 3.ad x22 /3 .ad t˛ /.esC / D 2s.ad x22 /3 .esC / one deduces ! 3 3 3 4 Œx1 x2 ; es D . 1/ .ad x2 / ad .ad x2 /3 .x13 x23 / .ad x22 /4 .esC / 3 D .ad x2 /. 1/3 3Š.ad x23 /.ad x22 /4 .esC / D

2s.ad x2 /.ad x22 /3 .esC /:

Applying (13) and the former equations yields ŒŒerC ; es ; e

r

1 3 3 1 DŒ D .x x / C .r/ t˛ ; es r r2 1 2 3 s 2s D .ad x2 /.ad x22 /3 C .ad x2 /2 C 2 .ad x2 /.ad x22 /3 r r r C s .r/ .ad x2 /4 .ad x22 /4 .esC / D ŒŒerC ; e

D

r ; es

s .ad x2 /.ad x22 /3 C .ad x2 /2 r C s .r/ .ad x2 /4 .ad x22 /4 .esC /:

2.s

r/

r2

On the other hand, by (6) and (16), ŒŒerC ; es ; e

r

r C s 2s C .ad x2 / C 2 .ad x22 /3 .erCs /; e r r r r C s 2s C D .r;s/ .ad x2 / C 2 .ad x22 /3 .ŒerCs ; e r / r r r C s 2s D .r;s/ .ad x2 / C 2 .ad x22 /3 r r s 2r 2 3 .ad x2 / C .ad x / .esC / .rCs; r/ 2 r Cs .r C s/2 s 2 .ad x2 /.ad x22 /3 D .r;s/ .rCs; r/ .ad x2 /2 r .r C s/ 2s 2 C 2 .ad x2 /.ad x22 /3 .esC / r .r C s/ s 2.s r/ 2 3 D .r;s/ .rCs; r/ .ad x2 /2 C .ad x /.ad x / .esC /: 2 2 r r2

D Œ.r;s/

We conclude .r/ D 0;

.r;s/ .rCs;

r/

D 1;

r; s; r C s ¤ 0:

.17/

240

12


Choose 1 ; 2 2 F such that 21 2 1 D .1;1/ and 1 22 D .2;2/ and substitute eQ1C WD 1 1 e1C ;

eQ2C WD 2 1 e2C ;

eQ3C WD 2 e3C ;

eQ2 D 2 1 e2 ;

eQ3 D 2 e3 ;

eQ4C WD 1 e4C :

Observe that by (6) eQ1 D 1 1 e1 ;

eQ4 D 1 e4 :

Therefore (4)–(15) still hold. In addition, by (16) ŒeQ1C ; eQ1 D 1 2 2 .1;1/ 2.ad x2 / C 2.ad x22 /3 .eQ2C / D 2.ad x2 / C 2.ad x22 /3 .eQ2C /; ŒeQ2C ; eQ2 D 2 2 1 1 .2;2/ 2.ad x2 / C .ad x22 /3 .eQ4C / D 2.ad x2 / C .ad x22 /3 .eQ4C /: Then we may assume .1;1/ D .2;2/ D 1. Denote by the linear operator acting on the space F5 F5 by the rule .u; v/ D .u C v; u/;

.u; v/ 2 F5 F5 :

The set D ¹.u; v/ 2 F5 F5 j u; v; u C v ¤ 0º is -stable and consists of 12 elements. It is easy to check that 6 D 1 and D ¹ i .1; 1/ j i D 0; : : : ; 5º [ ¹ i .2; 2/ j i D 0; : : : ; 5º: i

i

By (17), .r;s/ D 1=.r;s/ , and this gives .1;1/ D .2;2/ D 1 for all i . This means .r;s/ D 1 for all r; s; r C s ¤ 0. (VII) As a result of these deliberations we get that there is at most 1 isomorphism class of the pair .L; T / satisfying the assumptions of the theorem. We have to detect this isomorphism type. Recall that M.1; 1/ is a simple restricted Lie algebra of absolute toral rank 2, and T WD F .1 C x1 /@1 C F .1 C x2 /@2 is a torus of maximal dimension. Moreover, H WD CM.1;1/ .T / is a non-trigonalizable CSA (Volume 1, p. 209). Clearly, the assumptions of the Setting 12.5 hold for g D M.1; 1/, t D T and h D H . Since M.1; 1/ is restricted, one has t C hŒp D H . Let ˛, ˇ, r.˛; ˇ/ and L, H , T be as in Lemmas 12.5.4 and 12.5.5. Since T is 2-dimensional, the roots ˛ and ˇ span the full root lattice, whence g.˛; ˇ/ D M.1; 1/. Moreover, r.˛; ˇ/ D ¹0º as r.˛; ˇ/ is a proper ideal and t C hŒp C g.˛; ˇ/ D M.1; 1/ is simple. We obtain L D L D M.1; 1/, H D H , T D T . Theorem 12.5.7 states that the assumptions of the present theorem are satisfied by M.1; 1/ and T . This completes the proof of the theorem. The following corollaries are immediate consequences.

12.5

Trigonalizability

241

Corollary 12.5.9. Suppose the assumptions of the Setting 12.5 hold. Let ˛, ˇ, r.˛; ˇ/ be as in Lemmas 12.5.4 and 12.5.5. Then g.˛; ˇ/=r.˛; ˇ/ Š M.1; 1/ under an isomorphism which maps t=t \ ker ˛ \ ker ˇ onto F .1 C x1 /@1 ˚ F .1 C x/2 @2 . Proof. Theorem 12.5.7 shows that the assumptions of Theorem 12.5.8 hold for L WD L.1/ and T WD T . Then L.1/ Š M.1; 1/ and dim L.1/ D 125. Now observe that dim L 125 by Theorem 12.5.7(4a). Then L itself is simple. But g.˛; ˇ/=r.˛; ˇ/ is an ideal of L. This gives g.˛; ˇ/=r.˛; ˇ/ Š L.1/ D L. Corollary 12.5.10. Let L be a simple Lie algebra of absolute toral rank 2. If the semisimple p-envelope LŒp of L contains a maximal torus T such that CL .T / acts non-trigonalizably on L, then L Š M.1; 1/ under an isomorphism, which maps T onto F .1 C x1 /@1 ˚ F .1 C x/2 @2 . In addition, N2 .L; T / D ;. Proof. Put in the Setting 12.5 g D L, t WD T . The assumptions of this setting hold as T is a maximal torus. Let ˛; ˇ and r.˛; ˇ/ be as in Lemmas 12.5.4 and 12.5.5. Since dim T D 2, these roots span the full root lattice, whence g.˛; ˇ/ D L. This algebra is simple, and this gives r.˛; ˇ/ D ¹0º. Consequently, Corollary 12.5.9 yields L Š g.˛; ˇ/=r.˛; ˇ/ Š M.1; 1/ under an isomorphism which map T D t onto F .1 C x1 /@1 ˚ F .1 C x/2 @2 . In addition, Lemma 12.5.6(1) shows that N2 .L; T / D ;. We mentioned already that in fact T WD F .1 C x1 /@1 ˚ F .1 C x/2 @2 is a torus of maximal dimension in L WD M.1; 1/ and that it is non-trigonalizable. Lemma 12.5.6 shows that N2 .L; T / D ;, and then Theorem 12.3.7 yields that T is a rigid torus. This is the proof for statement (6) of Corollary 12.4.8.

Chapter 13

Towards graded algebras

In this chapter we are going to construct filtrations for simple Lie algebras of absolute toral rank 2, which will allow to do investigations in the associated graded algebra rather than in the original Lie algebra. As a general assumption of this chapter L always denotes a simple Lie algebra of absolute toral rank 2 and T is a 2-dimensional torus in the semisimple p-envelope LŒp of L.

13.1

The pentagon

Throughout this chapter we impose the general assumption that L denotes a finite dimensional simple Lie algebra of absolute toral rank 2 and T stands for a 2-dimensional torus in the semisimple p-envelope LŒp in Der L for which CL .T / acts trigonalizably on L. Set H WD CL .T /, HŒp the p-envelope of H in LŒp and HQ WD CLŒp .T /. We r r treat .L; T / as a set of functions on HQ by setting ˛.h/ D ˛.hŒp /p (cf. Volume 1, §1.3). Recall that radp HŒp is the maximal ideal of HŒp consisting of Œp-nilpotent elements. Set nil H WD H \ .radp HŒp /: As H .1/ acts nilpotently on L, each 2 .L; T / vanishes on H .1/ and so may be viewed as a linear function on H (Proposition 1.3.2). Note that radp HŒp and nil H are ideals of HŒp and H , respectively, which both contain .HŒp /.1/ D H .1/ . As a consequence, HŒp D T ˚ radp HŒp . Set H˛ WD H \ ker ˛. Then nil H D H˛ \ Hˇ for independent roots ˛, ˇ 2 .L; T /. Given Fp -independent ˛; ˇ 2 .L; T / put K˛ WD ¹x 2 L˛ j Œx; L

˛

H˛ º;

.RK/˛ WD ¹x 2 K˛ j Œx; K

˛

nil H º;

M˛.ˇ / WD ¹x 2 L˛ j Œx; L

˛

Hˇ º;

R˛ WD ¹x 2 L˛ j Œx; L

˛

nil H º:

13.1

243

The pentagon .ˇ /

Obviously, all subspaces K˛ , .RK/˛ , M˛ , R˛ are T -invariant. The following pentagon illustrates the inclusions between the subspaces defined above: L˛ @ @

K˛

(13.1.1)

.ˇ / M˛

.RK/˛ @ @ @

R˛

Lemma 13.1.1. Let ˛; ˇ 2 .L; T / be Fp -independent. Then (1) L˛ =K˛ Š .L

˛ =K ˛ / ,

(2) the subspace .RK/˛ =R˛ can be embedded into .L .ˇ /

(3) dim L˛ =M˛

˛ =K ˛ / ,

2 dim L˛ =K˛ C dim K˛ =.RK/˛ .

Proof. If H˛ D H , then K˛ D L˛ and K ˛ D L Otherwise we consider the nondegenerate pairing .L˛ =K˛ / .L

˛ =K ˛ /

˛.

In this case, (1) is obvious.

! H=H˛ Š F

to prove (1). In order to prove (2) we observe that, if nil H D H˛ then K˛ D R˛ D .RK/˛ . In this case, (2) is obvious. Otherwise nil H D H˛ \Hˇ and H˛ =nil H Š F . The definitions show that there is an injective linear mapping .RK/˛ =R˛ ,! Hom .L

˛ =K ˛ ; H˛ =nil H /

Š .L

˛ =K ˛ / :

Assertion (3) is obtained by the pentagon and a combination of (1) and (2).

Next we refer to Corollary 11.2.4 and Definition 11.2.2. The semisimple quotient of a 1-section is denoted by LŒ˛ D L.˛/= rad L.˛/. A nonzero root ˛ 2 .L; T / is called solvable if LŒ˛ D ¹0º. Then L.˛/.1/ is nilpotent. Otherwise LŒ˛ 2 ¹sl.2/; W .1I 1/º or H.2I 1/.2/ LŒ˛ H.2I 1/; and ˛ is classical, Witt, or Hamiltonian in the respective cases. Accordingly, the 1-section L.˛/ is said to be solvable, classical, Witt, or Hamiltonian. Suppose L.˛/ is not solvable. Let TN Š T =T \ ker ˛ denote the 1-dimensional image of T in Der LŒ˛. Then CLŒ˛ .TN / acts non-nilpotently (Proposition 1.3.7). Moreover, in the present cases listed above LŒ˛ is a restricted Lie algebra. Therefore

244

13


we have LŒ˛ \ TN ¤ ¹0º. Then TN LŒ˛ is a 1-dimensional maximal torus. Since T has maximal toral rank, Theorem 1.3.11(3) shows that TR.L.˛// 1. Therefore TN is a torus of maximal dimension in LŒ˛. In case that ˛ is Hamiltonian, the quotient H.2I 1/=H.2I 1/.2/ is p-nilpotent. Therefore TN H.2I 1/.2/ . Theorem 11.2.5 describes proper and improper roots in the present setting. According to that theorem the following holds. Solvable and classical roots are always proper. A Witt root ˛ is proper if and only if TN LŒ˛ Š W .1I 1/ is conjugate to F x@ under an automorphism of W .1I 1/, and ˛ is improper if and only if TN is conjugate to F .1 C x/@ under an automorphism of W .1I 1/. A Hamiltonian root ˛ is proper if and only if TN LŒ˛ is conjugate to FDH .x1 x2 / under an automorphism of H.2I 1/.2/ , and ˛ is improper if and only if TN is conjugate to FDH ..1 C x1 /x2 / under an automorphism of H.2I 1/.2/ . Lemma 13.1.2. Let ˛ 2 .L; T / n ¹0º. One of the following occurs: (1) ˛ is solvable and Ki˛ D Li˛ for all i 2 Fp ; (2) ˛ is classical and there is i0 2 Fp such that, for i 2 Fp , dim Li˛ =Ki˛ D 1 if i D ˙i0 and Li˛ D Ki˛ if i ¤ ˙i0 ; (3) ˛ is proper Witt and there is i0 2 Fp such that, for i 2 Fp , dim Li˛ =Ki˛ D 1 if i D ˙i0 and Li˛ D Ki˛ if i ¤ ˙i0 ; (4) ˛ is improper Witt and dim Li˛ =Ki˛ D 1 for all i 2 Fp ; (5) ˛ is proper Hamiltonian and there is i0 2 Fp such that 8 ˆ 3 or 2k i > 3, respectively, these summands are contained in H.2I 1/.2/ , whence they are contained in Ki˛ . Also, DH .x13 / 2 K 3˛ , DH .x23 / 2 K3˛ . This shows that the only root vectors which may stick out of Ki ˛ are represented by those linear combinations of DH .x1 /; DH .x2 /; DH .x12 /; DH .x22 /; DH .x12 x2 /; DH .x1 x22 / lying in LŒ˛i ˛ . Since we have DH .x1 /; DH .x12 x2 / 2 L L˛ , DH .x12 / 2 L 2˛ , DH .x22 / 2 L2˛ and

˛,

DH .x2 /; DH .x1 x22 / 2

ŒDH .x1 /; DH .x1 x22 / D 2DH .x1 x2 /; ŒDH .x2 /; DH .x12 x2 / D

2DH .x1 x2 /;

ŒDH .x12 /; DH .x22 / D 4DH .x1 x2 /; assertion (5) follows. If TN D FDH ..1 C x1 /x2 /, then ˛ is an improper root and one has (adjusting ˛ DH ..1 C x1 /x2 / D 1) for i ¤ 0 LŒ˛i˛ D

p X1

FDH ..1 C x1 /k i x2k /:

kD0

It is clear that, whenever k 3, these summands are contained in Ki˛ . As ŒDH ..1 C x1 /k i x2k /; DH ..1 C x1 /2

kCi 2 k x2 /

D

for k D 0; 1; 2 and i 2 Fp , this shows that LŒ˛i˛ x1 /k i x2k /. This is (6).

2iDH ..1 C x1 /x2 / ¤ 0 P D Ki˛ ˚ 2kD0 FDH ..1 C

Next we set K.˛/ WD H ˚

X

Ki˛ ;

i2Fp

M .˛/ WD K.˛/ ˚

X

M .˛/ ;

…Fp ˛

R WD nil H C

X

R :

¤0

If necessary we indicate the dependency on T or L, K˛ D K˛ .T / D K˛ .L; T /, R D R.T / D R.L; T /, etc. The following is an almost obvious result.

246

13


Lemma 13.1.3. Let ˛ 2 .L; T / n ¹0º. (1) K.˛/ is a completely solvable subalgebra of L. (2) R is a nilpotent subalgebra of L. .˛/

.˛/

(3) M .˛/ is a subalgebra of L. In particular, the inclusions ŒKi˛ ; Mˇ Mˇ Ci˛ .˛/

.˛/

and ŒH; Mˇ Mˇ

hold. .˛/

(4) If H ¤ H˛ , then there is ˇ 2 .L; T / n Fp ˛ for which Lˇ ¤ Mˇ . In particular, L ¤ M .˛/ holds in this case. Proof. Remember that roots are linear on H . (1) Take x 2 Ki˛ , y 2 Kj˛ for i; j 2 Fp and z 2 L .i Cj /˛ . If i D j D 0, then x; y 2 H and Œx; y 2 H .1/ H˛ . Next suppose i ¤ 0, j D 0. Then ŒŒx; y; z D ŒŒx; z; y C Œx; Œy; z 2 ŒH; H C ŒKi˛ ; L

i˛

H˛ :

This shows that Œx; y 2 Ki˛ . One argues similarly to show that Œx; y 2 K.i Cj /˛ if i; j ¤ 0 and i C jP¤ 0. Clearly, Œx; y 2 H˛ if i; j ¤ 0 and i C j D 0. As a result, K.˛/.1/ H˛ C i 2Fp Ki˛ K.˛/. Let x 2 [i 2Fp Ki˛ . The semisimple part xs of x is contained in T (as T is a maximal torus) and ˛.xs / D 0 holds (Proposition 1.3.2(1)). By definition of H˛ the same is true for x 2 H˛ . Since [i 2Fp Ki˛ [ H˛ is a Lie set, the Engel–Jacobson theorem shows that K.˛/.1/ is nilpotent. Then K.˛/ is completely solvable. (2) One proves similarly that R is a nilpotent Lie algebra. .˛/ (3) Take x 2 Ki˛ for i 2 Fp , y 2 Mˇ for ˇ … Fp ˛ and z 2 L i˛ ˇ . Then ŒŒx; y; z D ŒŒx; z; y C Œx; Œy; z 2 ŒL

.˛/ ˇ ; Mˇ

C ŒKi˛ ; L

i˛

H˛ ;

.˛/

and this shows Œx; y 2 Mˇ Ci˛ . One argues similarly to show that Œx; y 2 M .˛/ if .˛/

.˛/

x 2 M , y 2 Mˇ

for ˇ; … Fp ˛. Then

.M .˛/ /.1/ K.˛/.1/ C

X ˇ …Fp ˛

Therefore M .˛/ is a subalgebra of L. (4) By Proposition 1.3.6, X LD L C

2.L;T /nFp ˛ .˛/

Suppose L D M

.˛/

X

ŒK.˛/; Mˇ C

.˛/

ŒMˇ ; M .˛/ M .˛/ :

ˇ; …Fp ˛

X

ŒL ; Lı :

;ı2.L;T /nFp ˛

for each 2 .L; T / n Fp ˛. Then, as M .˛/ is a subalgebra,

13.2

247

An upper bound

L D M .˛/ . But then L D L.1/ D .M .˛/ /.1/ D H˛ C

X i 2Fp

X

Ki˛ C

M .˛/ :

2.L;T /nFp ˛

In particular, H D H˛ , which contradicts the present assumption.

Remark 13.1.4. The proof of Lemma 13.1.2 gives a complete description of K.˛/ in the various cases. Obviously, rad L.˛/ K.˛/ and one computes K.˛/ in LŒ˛. If ˛ is solvable, then clearly K.˛/ D L.˛/ holds. If ˛ is classical or improper Witt, then K.˛/ D H C rad L.˛/. If ˛ is proper Witt, then after P the normalization T =T \ ker ˛ D F x@ one has K.˛/= rad L.˛/ D F x@ C i 3 F x i @ D HN C W .1I 1/.2/ . If ˛ is proper Hamiltonian, then after the normalization T=T \ker ˛ D FDH .x1 x2 / one has X j K.˛/= rad L.˛/ D HN C FDH .x13 / C FDH .x23 / C FDH .x1i x2 / i Cj 4

D HN C FDH .x13 / C FDH .x23 / C H.2I 1/.2/ .2/ HN C H.2I 1/.2/ .1/ and L.˛/=K.˛/ is represented by FDH .x1 /CFDH .x12 x2 / ˚ FDH .x2 /CFDH .x1 x22 / ˚FDH .x12 /˚FDH .x22 /: If ˛ is improper Hamiltonian, then T =T \ ker ˛ D FDH ..1 C x1 /x2 / after a suitable normalization, XX j FDH ..1 C x1 /i x2 / HN C H.2I 1/.2/ .1/ ; K.˛/= rad L.˛/ D HN C i2Fp j 3

and L.˛/=K.˛/

i˛

is represented by

FDH ..1 C x1 / i / C FDH ..1 C x1 /1 i x2 / C FDH ..1 C x1 /2 i x22 /;

13.2

i 2 Fp :

An upper bound for n˛

Lemmas 13.1.1 and 13.1.2 give estimates dim L˛ =K˛ 3 and dim.RK/˛ =R˛ dim L

˛ =K ˛

3:

In this section we derive a weak estimate for the missing number dim K˛ =.RK/˛ . In fact, eventually it will be proved that K˛ D .RK/˛ holds for all ˛ ¤ 0. But that result needs a rather long sequence of arguments and can only be achieved as late as in Theorem 14.1.1.

248

13


For ˛ 2 .L; T / n ¹0º we set n˛ WD dim K˛ =.RK/˛ ; X n.˛/ WD ni˛ ; i 2Fp

(13.2.1)

n.T / WD max ¹n.˛/ j ˛ 2 .L; T / n ¹0ºº; n.L/ WD max ¹n.T 0 / j T 0 is a 2-dimensional torus of LŒp and CL .T 0 /.1/ acts nilpotently on Lº:

Due to Corollary 12.5.10 the exposed set is nonempty and therefore n.L/ is well defined. Lemma 13.2.1. Let M be a T -invariant subalgebra of K.˛/ and W ¤ ¹0º be an M -module such that (a) M \ nil H acts nilpotently on W , (b) M \ H˛ D F h ˚ .M \ nil H /, where h acts invertibly on W . Set .RM /i ˛ WD ¹x 2 Mi˛ j Œx; M

i˛

nil H º. There is m 2 N for which

dim Mi˛ =.RM /i˛ 2m 8i 2 Fp

and

p m dim W:

Proof. We may assume that W is irreducible as an M -module. According to [S-F, Proposition 5.8.3] there exists a subalgebra Q of M such that (i) dim W p dim M=Q , (ii) W contains a 1-dimensional Q-submodule F u. Define bilinear mappings fi W Mi˛ M

i˛

! H˛ =nil H Š F

by Œx; y fi .x; y/h

.mod nil H /:

Note that according to (ii) xuD0

8x 2 Q.1/ :

Put Qi WD Mi˛ \ Q (observe that Q is not necessarily T -invariant, while Qi is so because it consists of T -eigenvectors). Then ŒQi ; Q i Q.1/ \ H˛ has 0 as a common eigenvalue on W . Note that .M \ H /.1/ M \ nil H acts nilpotently on W . Therefore an element rh C h0 (with r 2 F , h0 2 nil H ) acts invertibly on W if and only if r ¤ 0. Since

13.2

249

An upper bound

every ˇ element of ŒQi ; Q i has eigenvalue 0, we obtain ŒQi ; Q i nil H , i.e., fi ˇQ Q D 0. Therefore fi induces a bilinear form i

i

gi W .Qi =Q \ .RM /i˛ / .M

i˛ =Q i /

!F

given by gi .x C Q \ .RM /i˛ ; y C Q i / WD fi .x; y/: The definition of .RM /i˛ shows that this bilinear form is nondegenerate in the first argument, whence dim Qi =Q \ .RM /i˛ dim M

i˛ =Q i :

As a result, dim Mi ˛ =.RM /i˛ D dim Mi˛ =Qi C dim Qi =Q \ .RM /i˛ dim.RM /i˛ =Q \ .RM /i˛ dim Mi˛ =Qi C dim M

i˛ =Q i

2 dim M=Q:

Set m WD dim M=Q.

Theorem 13.2.2. Let ˛ 2 .L; T / be a nonzero root. The following is true: (1) if n˛ ¤ 0, then H ¤ H˛ , HŒp contains T , and there is ˇ 2 .L; T / n Fp ˛ for .˛/ which Lˇ ¤ Mˇ holds; (2) n˛ 3; if n˛ D 3, then ni˛ 2 for i 62 ¹ 1; 0; 1º; (3) dim L˛ =R˛ 9; P .˛/ (4) if n.˛/ > 2, then the K.˛/-module i 2Fp Lˇ Ci˛ =Mˇ Ci˛ is either ¹0º or irreducible of dimension p 2 (for arbitrary ˇ 2 .L; T / n Fp ˛); (5) dim L=M .˛/ < 2p 3 . Proof. (1) Suppose n˛ ¤ 0. If H D H˛ , then the Engel–Jacobson theorem implies that L.˛/ is nilpotent. Proposition 11.2.6 tells us that L.˛/ is a trigonalizable CSA of L. In particular, ŒK˛ ; K ˛ acts nilpotently on L. Then K˛ D .RK/˛ , which contradicts our assumption n˛ ¤ 0. Therefore H ¤ H˛ and there is h1 2 H for which ˛.h1 / ¤ 0. Since n˛ ¤ 0, there are x˙ 2 K˙˛ such that h2 WD ŒxC ; x 62 nil H . On the other hand, the definition of K.˛/ gives ŒxC ; x 2 H˛ . Therefore the restricted subalgebra of HŒp generated by F h1 C F h2 contains a 2-dimensional torus T 0 . The maximality of T gives T 0 T , and a dimension argument yields T 0 D T . .˛/ Lemma 13.1.3(4) shows that there is ˇ 2 .L; T / n Fp ˛ for which Lˇ ¤ Mˇ . (2) (a) Set r WD max¹n j 2 .L; T / n ¹0ºº. Choose a nonzero root such that n D r. We may assume r > 0, whence by (1) there is ˇ 62 Fp such that

250

13


P ./ ./ Lˇ =Mˇ ¤ ¹0º. Put W WD i 2Fp Lˇ Ci =Mˇ Ci ¤ ¹0º, which according to Lemma 13.1.3(3) is a K./-module. Using the pentagon (13.1.1), Lemmas 13.1.1 and 13.1.2 one readily observes that ./

dim Lˇ Ci =Mˇ Ci dim Lˇ Ci =Rˇ Ci 2 dim Lˇ Ci =Kˇ Ci C nˇ Ci 6 C r: Set in Lemma 13.2.1 M WD K./ and observe that .RM / D .RK/ . We proved in (1) that nil H ¤ H . Set H D F h ˚ nil H . Then .h/ D 0 and ˇ.h/ ¤ 0, whence h acts invertibly on W . According to that lemma there is m such that r D n D dim K =.RK/ 2m and p m dim W D

X

./

dim Lˇ Ci =Mˇ Ci p.6 C r/ p.6 C 2m/;

i 2Fp

whence pm

1

6 C 2m:

This inequality forces m 2. Consequently, one obtains for every nonzero root ˛ n˛ r 2m 4: (b) Now let ˛ be a nonzero root for which n.˛/ ¤ 0, take ˇ 2 .L; T / n Fp ˛ .˛/ such that Lˇ =Mˇ ¤ ¹0º and let W ¤ ¹0º denote a composition factor of the P .˛/ nonzero K.˛/-module i 2Fp Lˇ Ci˛ =Mˇ Ci˛ . Put W K.˛/ ! gl.W / the respective representation. As the action of K.˛/ on W is induced by the adjoint action of LŒp on L, it can be uniquely extended to a restricted representation W K.˛/Œp ! gl.W /: As H K.˛/, (1) implies that T K.˛/Œp . In particular, X X K.˛/.1/ D H .1/ C Ki˛ C ŒKi˛ ; K i 2Fp

i˛

i2Fp

and .T / acts as torus on W . Suppose W D W carries only S one T -weight. Then .Ki ˛ / D ¹0º whenever i 2 Fp and therefore every h 2 i 2Fp ŒKi˛ ; K i˛ anniS hilates W . But as n.˛/ ¤ 0, there is h 2 i 2Fp ŒKi˛ ; K i˛ satisfying ˛.h/ D 0, P ˇ.h/ ¤ 0. Such h acts invertibly on i 2Fp Lˇ Ci˛ , hence invertibly on W . This contradiction shows that W carries 2 independent weights ˇ C j1 ˛ and ˇ C j2 ˛ with j1 ¤ j2 . Consequently, an element h0 2 H acts nilpotently on W if and only if h0 2 nil H .

13.2

251

An upper bound

(c) Suppose dim W < p 2 . As M WD K.˛/ is solvable, Theorem 11.3.1 yields that Theorem 11.3.4 applies. In Case 1 of that theorem K.˛/.2/ annihilates W and therefore ŒKi ˛ ; K i˛ nil H for all i 2 Fp . Then n.˛/ D 0, a contradiction. In the second case .K.˛//.1/ D A ˚ F x holds, where A is an abelian ideal of .K.˛//. Moreover, since T is contained in HŒp , the complement of A in .K.˛//.1/ can be chosen T -invariant. That means that x can be chosen as a T -root vector, x 2 .Kk˛ / for some k 2 Fp . Then ni˛ D 0 for all i ¤ ˙k and n k˛ D 1 (if x ¤ 0 and k ¤ 0). As nk˛ D n k˛ , this gives n.˛/ 2. Cases 3 and 4 do not occur here because K.˛/ is solvable. As a consequence, n.˛/ 2 holds if dim W 2. By the above, every composition factor W ¤ ¹0º of P .˛/ the nonzero K.˛/-module i 2Fp Lˇ Ci˛ =Mˇ Ci˛ has dimension p 2 . On the other hand, applying Lemmas 13.1.1, 13.1.2 and part (a) one obtains X X X .˛/ dim W dim Lˇ Ci˛ =Mˇ Ci˛ 2 dim Lˇ Ci˛ =Kˇ Ci˛ C nˇ Ci˛ i 2Fp

i 2Fp

i 2Fp

p.6 C 4/ D 10p 2p 2 : Since K.˛/ is solvable, dim W D p s is a p-power. Then only s D 2 is possible. Lemma 13.1.3(1) states that .K.˛// is completely solvable. It cannot be abelian, as s ¤ 0. Consequently, .K.˛//=C .K.˛// has a 1-dimensional ideal I . Let A WD ¹x 2 .K.˛// j x C C .K.˛// 2 I º be the full preimage. Clearly, A is an abelian non-central ideal of .K.˛/Œp /. There is an eigenvalue function W A ! F , such that x .x/IdW is a nilpotent endomorphism for every x 2 A. By Theorem 3.3.4 W is induced by an irreducible A?; -submodule W0 of W , where A?; D ¹x 2 .K.˛/Œp / j .Œx; A/ D 0º. Suppose A?; D .K.˛/Œp /. Then the ideal Œ.K.˛//; A consists of nilpotent transformations. The irreducibility of W gives Œ.K.˛//; A D ¹0º, whence A C .K.˛// , a contradiction. Consequently, dim W0 < dim W D p 2 . As T K.˛/Œp , one has C .K.˛// CK.˛/ .T / D .H /; A D C .K.˛// ˚ F.u/; where uP2 Kk˛ for some k 2 Fp . The Engel–Jacobson theorem shows that K 0 .˛/ WD H˛ C i 2Fp Ki ˛ K.˛/.1/ is a nilpotent subalgebra. Then it annihilates the 1 dimensional space A=C .K.˛// . This means that there is a bilinear mapping .K 0 .˛// F.u/ ! C .K.˛// :

252

13


Looking at weights one obtains Œ.Ki˛ /; F.u/ D ¹0º if i ¤

k:

Since H .1/ nil H acts nilpotently on W and .u/ lies non-centrally, it therefore must be that k 2 Fp . Set K ?;u WD ¹x 2 K k˛ and

k˛

j .Œx; u/ D 0º

M WD H˛ C K ?;u C k˛

X

Ki˛ :

i ¤0; k

By construction one has M D ¹x 2 K 0 .˛/ j Œ.x/; A D ¹0ºº: Then M is a T -invariant subalgebra of K.˛/ which is contained in A?; . Applying Lemma 13.2.1 for M and W0 gives dim Mi˛ =.RM /i˛ 2

8i 2 Fp :

Let i ¤ ˙k. Then M˙i˛ D K˙i˛ and hence .RM /˙i˛ D .RK/˙i˛ . In this case one obtains ni˛ D dim Ki˛ =.RK/i˛ D dim Mi˛ =.RM /i˛ 2: Consider the case i D k. Observe that Kk˛ D Mk˛ . As W is an irreducible K.˛/module, one has dim C .K.˛// 1. Then dim K whence dim K

k˛ =M k˛

?;u k˛ =K k˛

1;

1. Therefore dim.RM /k˛ =.RK/k˛ 1 and

nk˛ D dim Kk˛ =.RK/k˛ 1 C dim Mk˛ =.RM /k˛ 3: By definition one also has n k˛ D nk˛ 3. This completes the proof of (2). (3) Using (2) this assertion follows immediately from the inequality dim L˛ =R˛ 2 dim L˛ =K˛ C n˛ 6 C 3 D 9: (4) It has been shown in (2)(c) that under the present assumption every comP .˛/ position factor of a nonzero K.˛/-module i 2Fp Lˇ Ci˛ =Mˇ Ci˛ has dimension p 2 . Using the pentagon (13.1.1) and part (3) of this theorem we conclude that .˛/ dim Lˇ Ci ˛ =Mˇ Ci˛ dim Lˇ Ci˛ =Rˇ Ci˛ 9, and this gives X i 2Fp

.˛/

dim Lˇ Ci˛ =Mˇ Ci˛ 9p < 2p 2 :

13.3

253

Filtrations

Therefore there is no room for a second composition factor. This means that the P .˛/ K.˛/-module i 2Fp Lˇ Ci˛ =Mˇ Ci˛ has dimension p 2 . (5) Since dim L =R 9 for every nonzero root , we obtain the estimate dim L=M .˛/ D

X X i 2Fp j 2Fp

X

.˛/

dim Li.ˇ Cj˛/ =Mi.ˇ Cj˛/ C

dim L =R 9.p 2

X

dim Lj˛ =Kj˛

j 2Fp

1/ < 2p 3 :

¤0

13.3

Filtrations

An important technique, which in §12.2 and §12.5 has already proved its usefulness, is to construct an appropriate filtration of L and to investigate the associated graded Lie algebra. That construction will require a subtle choice of the maximal subalgebra. Notation 13.3: Let T .L/ denote the set of tori T 0 LŒp satisfying (1) dim T 0 D 2, (2) TR.CL .T 0 /; L/ D 2, (3) N2 .T 0 / 6D ;. We rephrase assumption (3) as T 0 being non-rigid (Theorem 12.3.7 and Definition 12.3.8). Corollary 12.5.10 shows that every such torus has the property that CL .T 0 /.1/ acts nilpotently on L. So every such torus satisfies the general assumptions of this chapter. As a matter of fact, if L is classical or isomorphic to H.2I 1I ˆ.//.1/ , then any 2-dimensional torus is rigid (Corollary 12.4.8). This makes clear that in general T .L/ might be empty. Assumption (2) implies that the p-envelope CL .T 0 / Œp of CL .T 0 / in LŒp contains a 2-dimensional torus T 00 . Since T 0 is a torus of maximal dimension, T 00 D T 0 holds. This means that the p-envelope of CL .T 0 / contains T 0 . Lemma 13.1.3(4) shows that L ¤ M .˛/ .T 0 / if T 0 2 T .L/. Therefore the following definition makes sense. Definition 13.3.1. A triple .T; ˛; L.0/ / is called admissible if T 2 T .L/, ˛ 2 .L; T / is a nonzero T -root, and L.0/ is a maximal subalgebra of L containing M .˛/ .T /. Lemma 13.3.2. The following is true: (1) N2 .L; T / R.T /; (2) ŒN2 .L; T /; L R.T /.

254

13


Proof. Suppose N2 .L; T / ¤ ;. Then H is trigonalizable by Corollary 12.5.10. (1) As H is trigonalizable, nil H coincides with the set of all ad-nilpotent elements of H . Let c 2 N2 .L; T / \ L , x 2 L . If D 0, then c 2 R.T / by the above reasoning. So assume ¤ 0. By Proposition 12.1.1(3), .adŒc; x/3 D 0: Then Œc; L nil H , proving (1). (2) Let c 2 N2 .L; T /\L . We already mentioned that Œc; L nil H R.T /. Therefore let ı 2 .L; T / n ¹ º, x 2 Lı and y 2 L . Cı/ . We need to prove that ŒŒc; x; y 2 nil H (but observe that we allow D 0). Suppose the contrary. Set WD ¹ 2 .L; T / j .ŒŒc; x; y/ 6D 0º: By assumption, 6D ;. Since L is simple, we have X X LD L C ŒL ; L 2

;2

(Proposition 1.3.6). Since c … C.L/ D ¹0º, there is ˇ 2 such that Œc; Lˇ ¤ ¹0º. As ŒŒc; x; y; c D .ad c/.ad y/.ad c/.x/ D 0 by Equation (12.1.6) and c 2 L , we have ŒŒc; x; y D 0, hence .ˇ C / ŒŒc; x; y ¤ 0. On the other hand, we 6 will show that adŒŒc; x; y .Œc; Lˇ / D 0, which contradicts the preceding statement. Write U WD ad u for an element u 2 L. It follows from Equations (12.1.14), (12.1.16), (12.1.13) and (12.1.10) that 6 4 adŒŒc; x; y .Œc; Lˇ / D 4.CXY /6 C.Lˇ / D .C Y 2 CX 2 CXY /.C Y 2 CX 2 CXY /C.Lˇ / D C Y 2 .CX 2 CXY C Y 2 /CX 2 CXY C.Lˇ / D C Y 2 .C Y 2 CXY CX 2 /CX 2 CXY C.Lˇ / D 0: Hence Œc; Lı R.T / for each ı 2 .L; T /.

From now on let .T; ˛; L.0/ / be an admissible triple. Choose a minimal L.0/ invariant subspace L. 1/ of L containing L.0/ . Then one defines the standard filtration of L associated to the pair .L.0/ ; L. 1/ / by setting ® ¯ L.i C1/ WD x 2 L.i/ j Œx; L. 1/ L.i/ ; i 0; L. i 1/ WD L. i/ ; L. 1/ C L. i/ ; i 1 (Definition 3.5.1). Since L.0/ is maximal in L, this filtration is exhaustive, and since L is simple, it is separating, i.e., there are s1 ; s2 0 with L D L.

s1 /

L.s2 C1/ D ¹0º:

Next consider the associated graded algebra 2 G WD ˚siD

s1 gri

L;

Gi WD gri L D L.i/ =L.iC1/ :

13.3

Filtrations

255

By Proposition 3.5.3(1) G satisfies (g1)–(g3) of Notation 3.5.2. Let M.G/ denote P the sum of all ideals of G contained in j 1 Gj . Note that, if L.1/ ¤ ¹0º, then G 1 \ M.G/ D ¹0º (by (g1) and (g3) of Notation 3.5.2). By Proposition 3.5.3(3),(4), M.G/ is a graded ideal of G, the graded Lie algebra GN WD G=M.G/ D ˚i GN i ;

GN i D Gi =.Gi \ M.G//

inherits the above mentioned properties (g1)–(g3), and, if L.1/ is not an ideal of L (which means if L.1/ ¤ ¹0º), GN satisfies (g4) of Notation 3.5.2. Lemma 13.3.3. The graded Lie algebra GN satisfies (g1)–(g4) and has a unique minN This ideal is a G-simple N imal ideal A.G/. graded ideal, N D ˚ A.G/ N \ GN i ; A.G/ N \ GN 1 ¤ ¹0º holds. and A.G/ Proof. Lemma 13.3.2 shows that ; ¤ N2 .L; T / R.T / M .˛/ .T / L.0/ and, similarly, ŒN2 .L; T /; L L.0/ . Then N2 .L; T / L.1/ . This shows L.1/ ¤ ¹0º, hence GN satisfies (g1)–(g4). This also implies GN 1 ¤ ¹0º. Theorem 3.5.6 shows N DW A D ˚ Ai , which is that GN is semisimple and has a unique minimal ideal A.G/ P N N In addition, it follows from Theorem 3.5.6(3) that a graded ideal of G. i 1 Gi N A.G/. Let us assume A1 D ¹0º. As ŒGN 1 ; GN 2 A1 , property (g3) implies GN 2 D ¹0º. We are in the degenerate case AC D ¹0º which is described in Theorem 3.5.8. According to that theorem there is an isomorphism W A Š S ˝ O.mI n/; where S is simple and O.mI n/ carries a grading such that all xi are homogeneous of degree 0 or 1, and .Ai / D S ˝ O.mI n/i : If m D 0, then A D S would be homogeneous of degree 0. However, A 1 D GN 1 ¤ ¹0º holds. As a consequence, m ¤ 0. Moreover, according to Theorem 3.5.8(3) GN 1 ˚ GN 0 is mapped into .Der S/ ˝ O.mI n/ Ì Id ˝ WP .mI n/ and the image of N N G1 ˚ G0 in W .mI n/ is a transitive subalgebra (recall that i0 (we mentioned that the image of G0 in W .mI 1/ is a transitive subalgebra), this gives N D S and hence GN 0 D A.G/ N 0 D S0 . Consequently, G0 is the m D 0. Then A.G/ p-envelope of S0 in Der S. On the other hand, ˆ.T / is a 2-dimensional torus in G0 . Then TR.S / D 2. Lemma 13.3.6. Suppose TR.S/ D 1. The following holds: (1) S Š H.2I 1/.2/ ; (2) S0 2 ¹sl.2/; W .1I 1/º;

260

13


(3) Der0 S D adS S0 ˚ F ı, where ı is the degree derivation; (4) GN 2 D ¹0º; (5) m ¤ 0. Proof. (1) As TR.S/ D 1, one has S 2 ¹sl.2/; W .1I 1/; H.2I 1/.2/ º (Theorem 10.6.1). Note that S

2

˝ O.mI 1/ D .GN 2 / D Œ.GN 1 /; .GN 1 / D ŒS

1

˝ O.mI 1/; S

1

˝ O.mI 1/ D ŒS 1 ; S 1 ˝ O.mI 1/;

whence S 2 D ŒS 1 ; S 1 . Suppose S Š sl.2/. Then it is clear that S 2 D ¹0º, which means GN 2 D ¹0º. All homogeneous spaces of S are 1-dimensional. Choose h 2 S0 which satisfies Œh; x D x for x 2 S 1 . Then Œh˝1; y D y for all y 2 S 1 ˝O.mI 1/ D .GN 1 /. Lemma 13.3.5 shows that this is not possible. Suppose S Š W .1I 1/. The gradings of the Witt algebra are determined in Theorem 7.4.1. They are given by a degree a on a suitable choice of generator x. Since S 1 ¤ ¹0º, we have a 2 ¹˙1º. If a D 1, then S 1 D F @, S 2 D ¹0º and .x@/ ˝ 1 acts on .GN 1 / D S 1 ˝ O.mI 1/ as Id. Lemma 13.3.5 shows that this is not possible. If a D 1, then S 1 D F x .2/ @. But then F x .3/ @ D S 2 D ŒS 1 ; S 1 D ¹0º, a contradiction. (2) The gradings of S Š H.2I 1/.2/ are given in Theorem 11.4.1. According to that theorem, either S0 2 ¹sl.2/; W .1I 1/º or S0 is solvable. In the latter case we apply the theorem again for M D Der H.2I 1/.2/ and obtain that Der0 S acts trigonalizably (Theorem 11.4.1(4)). On the other hand, Theorem 3.5.7(4) shows that S 1 is (Der0 S )-irreducible. Then S 1 is 1-dimensional. As in (1) it follows that S 2 D ¹0º, while Theorem 11.4.1(4) states that there are homogeneous subspaces of at least 2 different negative degrees. So this grading does not arise from a standard filtration. (3) The gradings under consideration are given as in Cases (2) or (3) in Theorem 11.4.1: there are indeterminates u1 ; u2 homogeneous of degree a1 ; a2 , respectively, and (as we mentioned in the text preceding that theorem) these generators satisfy the equation .i/ .j /

.u/

.i/ .j /

.i 1/ .j / u2 @u2

DH .u1 u2 / D DH .u1 u2 / D u1

.i/ .j 1/

u1 u2

@u1 I .i/ .j /

moreover, either a1 D 0, a2 ¤ 0 or a1 D a2 ¤ 0. An element DH .u1 u2 / is homogeneous of degree .i/ .j / deg DH .u1 u2 / D .i 1/a1 C .j 1/a2 : ./

13.3

261

Filtrations

The theorem (for M D Der H.2I 1/.2/ ) shows Der0 S Š S0 ˚ F u2 @u2 : In the first case the degree derivation ı is given by a2 u2 @u2 , and in the second case it is given by a1 .u1 @u1 C u2 @u2 / D 2a1 u2 @u2 a1 DH .u1 u2 /. In both cases ı 62 S0 , and this gives the result. (4) We now look at the grading of GN in more detail. The case S0 Š sl.2/ occurs when a1 D a2 ¤ 0, and the case S0 Š W .1I 1/ occurs when a1 D 0, a2 ¤ 0. Suppose a2 > 0. Then one has (in both cases) S 2 D ¹0º, whence GN 2 D ¹0º. So assume that a2 < 0 (this means a2 D 1 as S 1 ¤ ¹0º). Then Der2 S D ¹0º, and (see ./) Der1 S Š S1 D F @u1 C F @u2 Der1 S D

p X1 i D0

.i/ F u1 @u2

Š

p X

if a1 D a2 D .i/

FDH .u1 /

if a1 D 0; a2 D

1; 1:

i D1

This gives GN 2 D ¹0º in both cases and .GN 1 / D S1 ˝ O.mI 1/ and .GN 1 /

p X

.i /

FDH .u1 / ˝ O.mI 1/; resp.

i D1

Recall that there are sandwich elements c 2 N2 .L; T /, and by construction the image N cN of such an element in GN is contained in GN 1 (as GN 2 D ¹0º). Moreover, Œc; N G GN 0 C GN 1 (as Œc; L M .˛/ .T / L.0/ ). Write .c/ N D

X

db ˝ x .b/

b

Pp .i/ with db 2 S1 D F @u1 C F @u2 and db 2 i D1 FDH .u1 /, respectively. Then Œdb ; S 2 D ¹0º for all b, because Œˆ.c/; GN 2 .GN 0 C GN 1 / \ GN 1 D ¹0º. Since c ¤ 0, there is b for which db is nonzero. .2/ .2/ If a1 D a2 D 1, then DH .u1 u2 / 2 S 2 . In this case we obtain the contradiction .2/ .2/

.2/

.2/

0 D Œ@u1 C @u2 ; DH .u1 u2 / D DH .u1 u2 / C DH .u1 u2 / for some nonzero .; / 2 F 2 . .3/ If a1 D 0, a2 D 1, then DH .u2 / 2 S 2 . In this case we obtain the contradiction 0D

p hX i D1

p i X .i/ .3/ .i i DH .u1 /; DH .u2 / D i DH .u1

1/ .2/ u2 /

i D1

for some nontrivial choice of the i 2 F . This contradiction proves (4).

262

13


Q / Š H.2I 1/.2/ ˚ F u2 @u2 (cf. Theorem 11.4.1). In (5) If m D 0, then S C ˆ.T this case G0 contains the elements u1 @u1 C u2 @u2 and u2 @u2 . These elements define the degree derivation of S for the respective gradings of S under consideration. As in former cases this contradicts Lemma 13.3.5. From now on we discuss the case m ¤ 0. Lemmas 13.3.4(2) and 13.3.6(5) show that this assumption is equivalent with the condition TR.S/ D 1. Proposition 13.3.7. Suppose m ¤ 0. Let T 0 L.0/ be a 2-dimensional torus such N Š S ˝ O.mI 1/ can that p .CL .T 0 // acts nilpotently on L. The isomorphism A.G/ be chosen such that Q 0 .T 0 / D F .h0 ˝ 1/ ˚ F . 0 ı ˝ 1 C Id ˝ t00 /; ˆ

(13.3.3)

where h0 is a toral element in S0 , 0 2 Fp , ı is the degree derivation of S for the grading under consideration, and t00 is a nonzero toral element in W .mI 1/. More exactly, one can obtain 0 D 0 and t00 D .1 C x1 /@1

or

t00 2

m X

F xi @i ; t00 ¤ 0:

i D1

Q 0 .T 0 /-roots ˇ 0 ; 0 satisfying There are ˆ ˇ 0 .h0 ˝ 1/ D 1;

ˇ 0 . 0 ı ˝ 1 C Id ˝ t00 / D 0;

0 .h0 ˝ 1/ D 0;

0 . 0 ı ˝ 1 C Id ˝ t00 / D 1:

(13.3.4)

If S0 Š sl.2/, then there is i0 2 Fp such that Q 0 .T 0 // D .2i0 ˇ 0 C Fp 0 / [ . 2i0 ˇ 0 C Fp 0 / [ Fp 0 ; .S0 ˝ O.mI 1/; ˆ .S

1

Q 0 .T 0 // D .i0 ˇ 0 C Fp 0 / [ . i0 ˇ 0 C Fp 0 /: ˝ O.mI 1/; ˆ

If S0 Š W .1I 1/, then S

1

Š O.1I 1//F as an S0 -module and

Q 0 .T 0 // D Fp ˇ 0 C Fp 0 ; .S0 ˝ O.mI 1/; ˆ .S

(13.3.5)

1

Q 0 .T 0 // D Fp ˇ 0 C Fp 0 : ˝ O.mI 1/; ˆ

(13.3.6)

Q 0 .T 0 /. Put in Corollary 10.6.6 g WD G, N Proof. (a) We are going to prove the claim on ˆ 0 N sQ WD A.G/ and t WD ˆ.T /. One obtains a realization Q 0 .T 0 / D F .h0 ˝ 1/ ˚ F .d 0 ˝ 1 C Id ˝ t00 /; ˆ where h0 P 2 S is toral, t00 2 W .mI 1/ is toral and nonzero and d 0 D 0, t00 D .1 C x1 /@1 Q 0 .T 0 / is homogeneous of degree 0, one has h0 2 S0 and or t00 2 F xi @i . As ˆ 0 d 2 Der0 S. This is the claim with 0 D 0 if d 0 D 0, t00 D .1 C x1 /@1 . Otherwise Q 0 .T 0 / D F .h0 ˝ 1/ ˚ F .d 0 ˝ 1 C Id ˝ t00 / ˆ

13.3

Filtrations

263

Pm 0 0 where d 0 2 Der0 S , t00 2 i D1 F xi @i and d ˝ 1 C Id ˝ t0 is toral. Then both 0 0 0 0 d , t0 are toral. If F h C F d Der0 S is a 1-dimensional torus, then we may take d 0 D 0, whence 0 D 0. Suppose that F h0 C F d 0 Der0 S is a 2-dimensional torus. According to Lemma 13.3.6(3) Der0 S D S0 ˚ F ı holds, where ı is the degree derivation. As a consequence, F h0 C F d 0 D F h0 C F ı, which gives Q 0 .T 0 / D F .h0 ˝ 1/ ˚ F . 0 ı ˝ 1 C Id ˝ t00 /; ˆ where 0 2 Fp because d 0 is toral. (b) Since S Š H.2I 1/.2/ , all elements of Fp occur as eigenvalues for adS h0 . In particular, there is a homogeneous u 2 Sk for some k satisfying Œh0 ; u D u. Suppose 0 D 0 and t00 D .1 C x1 /@1 . Then w1 WD u ˝ 1 and w2 WD h0 ˝ .1 C x1 / Q 0 .T 0 /-eigenvectors satisfying Œh0 ˝ 1; w1 D w1 , Œh0 ˝ 1; w2 D 0 and ŒId ˝ .1 C are ˆ x1 /@1 ; w1 D 0, ŒId ˝ .1 C x1 /@1 ; w2 D w2 . Denote the respective roots by ˇ 0 and 0 . P 0 Suppose t00 D m are 0 and all i D1 i xi @i ¤ 0. Since t0 ¤ 0 is toral, not all i P these are contained in Fp . Choose b D .b1 ; : : : ; bm / such that 0 ı.u/ C i bi D 0 0 ı ˝ 1 C Id ˝ t 0 ; w D 0. and set w1 WD u ˝ x b . Then Œh0 ˝ 1; wP D w and Œ 1 1 1 0 Next choose c D .c1 ; : : : ; cm / such that i ci D 1 and set w2 WD h0 ˝ x c . Then Œh0 ˝ 1; w2 D 0 and Œ 0 ı ˝ 1 C Id ˝ t00 ; w2 D w2 . Denote the respective roots by ˇ 0 and 0 . (c) Suppose S0 Š sl.2/. The grading of S is described in Theorem 11.4.1(3). Under the present assumptions it is given by homogeneous generators u1 , u2 of degree a1 , a2 . Since S 2 D ¹0º and S 1 ¤ ¹0º, one has a1 D a2 D 1. Then S 1 is spanned by @u1 , @u2 . Due to Theorem 7.5.8 there is an automorphism ˆ of S which maps F h0 onto F .u2 @u2 u1 @u1 /. Since the present associated P filtration of S is the natural one, ˆ respects this filtration. Therefore ˆ D l0 ˆ;l decomposes into the sum of homogeneous linear mappings of degree 0. Then it is immediate that ˆ;0 also is an automorphism of S and maps F h0 onto F .u2 @u2 u1 @u1 /. So we may adjust h0 D i0 .u2 @u2 u1 @u1 /, i0 2 Fp . Since t00 ¤ 0, it is clear by Theorem 13.3.7 and Equation (13.3.4) that Equation (13.3.5) holds. Q 0 .T 0 // is a direct (d) Suppose S0 Š W .1I 1/. The assertion on .S0 ˝ O.mI 1/; ˆ consequence of the first parts of this theorem. The grading of S is described in Theorem 11.4.1(2). Under the present assumptions it is given by homogeneous generators u1 , u2 of degree a1 , a2 where a1 D 0, a2 D 1. Part (d) of that theorem also shows that S 1 is a .p 1/-dimensional nontrivial W .1I 1/-module, whence S 1 Š O.1I 1//F as an S0 -module. Therefore all i 2 Fp occur as eigenvalues for ad h0 . Since t00 ¤ 0, it is clear by the first parts of this theorem that Equation (13.3.6) holds. Q (depending on For the following we fix a restricted Lie algebra homomorphism ˆ T ) such that Q / D F .h ˝ 1/ ˚ F .ı ˝ 1 C Id ˝ t0 /; ˆ.T

264

13


P where h 2 S0 is toral, D 0 and t0 D .1 C x1 /@1 or t0 2 m i D1 F xi @i is nonzero Q / set and toral. We also fix roots ˇ, as in Equation (13.3.4). For any root 2 ˆ.T according to Equation (13.3.3) N WD .h ˝ 1/ and

Q WD .ı ˝ 1 C Id ˝ t0 /;

; N Q 2 Fp ;

and O.mI 1/` WD ¹f 2 O.mI 1/ j t0 .f / D `f º;

S` WD ¹g 2 S j Œh; g D `gº;

W .mI 1/` WD ¹D 2 W .mI 1/ j Œt0 ; D D `Dº for all ` 2 Fp . Then .GN i; / D

X

Si;N ˝ O.mI 1/

iC Q

if N ¤ 0; (13.3.7)

i

.GN 0; / CDer0 S .h/ ˝ O.mI 1/Q C Id ˝ W .mI 1/Q

if N D 0:

Lemma 13.3.8. Suppose m ¤ 0. The following holds: (1) G

3

D ¹0º and G3 ¤ ¹0º;

(2) if T 0 .Der L/.0/ is a 2-dimensional torus, then N2 .L; T 0 / ¤ ;; (3) G

2

G. /; if S0 Š sl.2/, then G

2

D ¹0º;

(4) if n.˛/ ¤ 0 or m D 2, then Fp ˛ D Fp ; (5) dim L D p m for all 2 Fp ˇ C Fp and dim Li < 4p m for all i 2 Fp ; (6) dim GN

1;

D pm

1

for all 2 .GN 1 ; ˆ.T //.

N Proof. (1) Note that GN D G=M.G/ acts on M.G/= 2 .M.G//. Suppose that A.G/ N acts nontrivially on a composition factor W of the G-module M.G/= 2 .M.G//. Theorem 3.3.10 shows that there is an S -module U and a simultaneous realization N ˚ W Š .S ˚ U / ˝ O.mI 1/; A.G/ GN C G0 ! Der0 .S ˚ U / ˝ O.mI 1/ Ì Id ˝ W .mI 1/ ; N on W is given by the action of S ˝ O.mI 1/ on W Š such that the action of A.G/ U ˝O.mI 1/. Moreover, Theorem 3.6.1 implies that we may take this realization such that Equation (13.3.3) holds. Let V denote a composition factor of the S -module U and W S ! gl.V / the representation. Recall that S Š H.2I 1/.2/ is restricted. Then there is a Œp-character S W S ! F such that .g/p .g Œp / D .g/p IdV for all g 2 S . Let g 2 i ¤0 Si .

13.3

Filtrations

265

Then g Œp 2 Sip . Choose inverse images gQ 2 Gi and y 2 Gip of g ˝ 1 and g Œp ˝ 1. As i ¤ 0, .adG g/ Q p .adG y/ is a nilpotent endomorphism, whence adM.G/= 2 .M.G// .gQ C M.G//

p

adM.G/= 2 .M.G// .y C M.G//

is so. Consequently, .ad g ˝ 1/p ad g ŒpS˝ 1 acts nilpotently on U ˝ O.mI 1/. This shows that .g/ D 0 whenever g 2 i ¤0 Si . Next take g 2 S0 . Let again g; Q y 2 G0 be elements which are mapped onto g ˝ 1 and g Œp ˝ 1, respectively. The endomorphism .ad.g˝1//p ad.g Œp ˝1/ annihilates S ˝O.mI 1/. Since S ˝O.mI 1/ N then .ad N .g ˝1//p ad N .g Œp ˝1/ is a nilpotent transformation. is an ideal in .G/, G G But .adGN .g ˝ 1//p adGN .g Œp ˝ 1/ is homogeneous of degree 0 and Gi Š GN i for all i 1. Then .ad Q p adGi y is a nilpotent transformation for all i 1. PGi g/ Now G 1 generates i 0 can occur. Then the theorem gives dim S 1;N D 1 for all 2 .GN 1 ; ˆ.T //. The deliberations in (5) give dim G 1; D dim GN 1; D p m 1 . (4) Write ˛ D iˇ C j . We have to show under the present assumptions that N is an ideal of G, N this i D 0. P Assume the contrary. Then ˛.h ˝ 1/ ¤ 0. As A.G/ N N implies j 2Fp Gj˛ A.G/.

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Suppose n.˛/ ¤ 0. Since K.˛/ L.0/ by construction, there is an element N \ ˆ.H / which satisfies ˛.d / D 0 but is not Œp-nilpotent. Then .d / 2 d 2 A0 .G/ CS0 .h/ ˝ O.mI 1/ D F h ˝ O.mI 1/ (see Lemma 13.3.6(2)). Since .d / is not Œpnilpotent, .d / 62 F h ˝ O.mI 1/.1/ . This gives the contradiction ˛.h ˝ 1/ D 0. Suppose m D 2. Proposition 13.3.7 yields the existence of k 2 Fp for which kˇ C Fp .GN 1 ; ˆ.T //. Observe that every kˇ C l has multiplicity p m 1 D p on GN 1 by (6). But then ki 1 ˛ D kˇ C ki 1 j has the property dim Lki 1 ˛ =Kki 1 ˛ dim GN 1;kˇ Cki 1 j D p > 3 by choice of L.0/ . This contradicts Lemma 13.1.2. L.

It is a remarkable fact that .T; ˛; L.0/ / determines (independent of the choice of N 1/ ) if A.G/ is simple or not.

N is non-simple if and only if the following holds: Proposition 13.3.9. A.G/ (a) rad L.0/ acts nilpotently on L; (b) L.

1/

is the only minimal L.0/ -invariant subspace of L containing L.0/ properly;

(c) TR.ŒL.

1/ ; rad L.0/ /

D 1.

N is not simple. Lemma 13.3.6(2) shows that A.G/ N 0 Š S0 ˝ Proof. (1) Suppose A.G/ N 0 is the unique minimal ideal O.mI 1/ is ˆ.L.0/ /-simple. As ˆ.L.0/ / D G0 , A.G/ of GN 0 Š L.0/ =L.1/ . Then GN 0 is semisimple and therefore L.0/ = rad L.0/ Š GN 0 Š L.0/ =L.1/ holds. In particular, rad L.0/ D L.1/ acts nilpotently on L. Let V ¤ L. 1/ be a minimal L.0/ -invariant subspace of L containing L.0/ properly. Lemma 13.3.8(3) in combination with the definition of in Equation (13.3.4) N 0 annihilates G 2 . Set I WD ˆ 1 .A.G/ N 0 / L.0/ , which is the gives that A.G/ unique ideal of L.0/ minimal subject to the condition that it contains rad L.0/ . As G 3 D ¹0º by Lemma 13.3.8(1) and G 2 M.G/ by Lemma 13.3.6(4), the above gives ŒI; L L. 1/ . Then ŒI; V V \ L. 1/ D L.0/ . This gives V NorL I .1/ . Since L.0/ is a maximal subalgebra, we obtain the contradiction I .1/ C rad L.0/ D L.0/ ;

I .1/ C L:

This proves (b). Let Q denote the restricted subalgebra of L.0/ generated by ŒL. order to obtain (c) we observe that ŒL.

1/ ; rad L.0/

C L.1/ =L.1/ D ŒL.

1/ ; L.1/

1/ ; rad L.0/ .

In

N 0: C L.1/ =L.1/ Š A.G/

Then TR.Q/ 1. If TR.Q/ D 2, then Theorem 1.2.9 shows that T Q. This N 0 in G0 . Then would imply that ˆ.T / were contained in the p-envelope of A.G/ N N TR.A.G/0 ; A.G// D 2, and this would imply TR.S/ D 2 contradicting Lemma 13.3.4(2).

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13


N is simple. Then TR.A.G// N D TR.S/ D 2 (2) Suppose (a)–(c) hold and A.G/ N N N By by Lemma 13.3.6(5). Let A.G/Œp denote the p-envelope of A.G/ in Der A.G/. Jacobson’s formula one has N Œp D span ¹.ad x/pj j x 2 [i A.G/ N i; A.G/

j 0º:

Theorem 1.2.9 shows that N Œp \ Der0 A.G/ N D span ¹.ad x/pj j x 2 A.G/ N 0; ˆ.T / A.G/

j 0º:

Recall that rad L.0/ L.1/ by assumption (a) and the irreducibility of L. 1/ =L.0/ . N 0 . Then T is contained in the Hence ˆ.ŒL.1/ ; rad L.0/ C L.1/ / ŒGN 1 ; GN 1 D A.G/ p-envelope of ŒL. 1/ ; rad L.0/ C L.1/ in L.0/ . As L.1/ acts nilpotently on L, this implies T Q. But then TR.ŒL. 1/ ; rad L.0/ / D 2, and this contradicts (c). N is simple for every choice of L. 1/ or there is exAs a consequence, either A.G/ actly one filtration which satisfies our requirements. In this case m D m.T; ˛; L.0/ / and S0 D S0 .T; ˛; L.0/ / are uniquely determined by the admissible triple under consideration. Theorem 13.3.10. Let L be a simple Lie algebra of absolute toral rank 2 and let N is not simple and S0 .T; ˛; L.0/ / Š .T; ˛; L.0/ / be an admissible triple. Suppose A.G/ sl.2/. Then n.˛/ D 0 and L Š K.3I 1/ or L Š M.1; 1/. Proof. Lemma 13.3.8 states that G 2 D ¹0º, whence GN D G and L D L. 1/ . By assumption on S0 the grading of S is the standard grading (Theorem 11.4.1). (a) In the course of the proof we have to modify the maximal torus a bit. Therefore let us start with any 2-dimensional torus T 0 of L.0/ . Due to Lemma 13.3.8(2) T 0 is non-rigid and therefore CL .T 0 / acts trigonalizably on L. Normalize Q 0 .T 0 / D F .h0 ˝ 1/ C F . 0 ı ˝ 1 C Id ˝ t00 / ˆ Q 0 .T 0 // and normalize as in Proposition 13.3.7. Abbreviate i WD .Si ˝ O.mI 1/; ˆ 0 h by a nonzero scalar to obtain by Equation (13.3.5) 0 D .2ˇ 0 C Fp 0 / [ . 2ˇ 0 C Fp 0 / [ Fp 0 ;

1

D .ˇ 0 C Fp 0 / [ . ˇ 0 C Fp 0 /:

Choose T 0 -invariant vector spaces V 1 , V000 , V0 , such that V

1

X

Lˇ 0 Cj 0 C

j 2Fp

V000 V0

X

L

ˇ 0 Cj 0 ;

j 2Fp

X j 2Fp

L2ˇ 0 Cj 0 C

X j 2Fp

L

2ˇ 0 Cj 0

C

X j 2Fp

Lj 0 ;

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269

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and LDV

1

˚ L.0/ ;

L.0/ D V0 ˚ L.1/ ;

Q0 V000 C L.1/ D ˆ

1

.Der0 S/ ˝ O.mI 1/ :

Properties of the associated graded Lie algebra G ensure (cf. Equation (13.3.7) for the first inclusion) that X X Liˇ 0 Cj 0 \ L.0/ V000 C L.1/ ; i 2Fp j 2Fp (13.3.8) ŒL; L.1/ V000 C L.1/ ; while properties of

ŒV000 ; L.0/ V000 C L.1/ ;

yield

1

ŒV 1 ; V 1 L.0/ : Then ŒV 1 ; V0

X X

Liˇ 0 Cj 0

i 2Fp j 2Fp

DV

1

C

X X

Liˇ 0 Cj 0 \ L.0/ V

1

C V000 C L.1/ ;

(13.3.9)

i 2Fp j 2Fp

and from all this it is not hard to deduce that V 1 C V000 C ŒV 1 ; V 1 C L.1/ is a nonzero ideal of L and therefore must coincide with L, LDV

1

C V000 C ŒV 1 ; V 1 C L.1/ :

P Suppose ŒV 1 ; V 1 V000 C L.1/ . Then we had G D A.G/ C i>0 Gi , whence A.G/ is no longer G-simple (recall that m ¤ 0). This contradiction shows that ŒV 1 ; V 1 6 V000 C L.1/ : (b) Let 1 W G 1 ! V 1 denote the inverse of the canonical linear isomorphism V 1 Š L=L.0/ D G 1 . The Lie multiplication of L gives rise to a skew-symmetric bilinear mapping ƒWG

1

G

1

! G0 ;

ƒ.v; v 0 / WD Œ 1 .v/; 1 .v 0 / C L.1/ :

Note that Œƒ.v; v 0 /; v 00 D ŒŒ 1 .v/; 1 .v 0 / C L.1/ ; 1 .v 00 / C L.0/ D ŒŒ 1 .v/; 1 .v 0 /; 1 .v 00 / C L.0/ for all v; v 0 ; v 00 2 G 1 . The Jacobi identity then yields Œƒ.v; v 0 /; v 00 C Œƒ.v 0 ; v 00 /; v C Œƒ.v 00 ; v/; v 0 D 0 for all v; v 0 ; v 00 2 G 1 :

270

13


Set ƒ2 WD 2 ı 0 ı ƒ where 2 is the projection into W .mI 1/ and 0 is given as in Q 0 as in (13.3.2). More explicitly, (13.3.1) connected to ˆ Q 0 .Œ 1 .v/; 1 .v 0 // C ker.2 ı ˆ Q 0 /: ƒ2 .v; v 0 / D ˆ Q 0 / D V 00 C L.1/ . If ƒ2 D 0, then ŒV 1 ; V 1 V 00 C L.1/ , Note that ker.2 ı ˆ 0 0 contrary to a former statement. So ƒ2 6D 0 holds. (c) The Lie multiplication of L gives rise to an L.0/ . 0 /-invariant bilinear mapping X X X X Lˇ 0 Cj 0 C L ˇ 0 Cj 0 ! L.0/ : Lˇ 0 Cj 0 C L ˇ 0 Cj 0 W j 2Fp

j 2Fp

j 2Fp

j 2Fp

P P Since and ŒL ; L V000 C j 2Fp Lˇ 0 Cj 0 C j2Fp L ˇ 0 Cj 0 \ L.0/ L.1/ P .1/ P 0 L.1/ by Equation (13.3.8) and j 2Fp L˙ˇ 0 Cj 0 ; L.0/ . / j 2Fp L˙ˇ 0 Cj 0 , 0 induces a G0 . /-invariant mapping . 2 W G 1 G 1 ! .Der0 S/ ˝ O.mI 1/ Ì Id ˝ W .mI 1/ .Der0 S/ ˝ O.mI 1/ Š W .mI 1/; Q 0 .Œu; u0 / C .Der0 S/ ˝ O.mI 1/: .u C L.0/ ; u0 C L.0/ / 7! Œu; u0 C .V000 C L.1/ / 7! ˆ Note that (with uN WD u C L.0/ 2 G 1 ) Q 0 .Œ 1 .u/; Q 0 /: 2 .u; N u0 / D ˆ N 1 .u0 // C ker.2 ı ˆ Hence 2 D ƒ2 , and as a consequence ƒ2 is G0 . /-invariant. We have now verified the assumptions of the setting (8.2.1) of Volume 1, p. 452 (with U D S 1 and t D h). Q 0 D ˆ. Q Then (d) Suppose m D 2. Apply the former reasoning for T 0 WD T and ˆ 0 Fp ˛ D Fp by Lemma 13.3.8. Let u; u constitute a basis of S 1 consisting of root vectors with respect to h. Since t0 D .1 C x1 /@1 or t0 2 F x1 @1 C F x2 @2 by Lemma p 1 Q /-root vector. Theorem 8.2.2 gives (after the identification 13.3.7, u ˝ x2 is a ˆ.T A.G/ Š S ˝ O.2I 1/) p 1

ƒ2 .u ˝ x2

;S

1

p 1

˝ O.2I 1// D ƒ2 .u ˝ x2

O.2I 1/; S

1

˝ 1/

W .2I 1/.1/ : p 1

and w 2 L arbitrary. Put w D Let v 2 V 1 \ L be a preimage of u ˝ x2 w 1 C w 0 where w 1 2 V 1 \ L and w 0 2 L.0/ \ L . We mentioned before that L.0/ \ L L.1/ and ŒL; L.1/ V000 C L.1/ . The preceding deliberations yield Q Q Q 000 C L.1/ / \ ˆ Q 0 .H / ˆ.Œv; w/ 2 ˆ.Œv; w 1 / C ˆ.V Q .Der0 S/ ˝ O.2I 1/ C Id ˝ W .2I 1/.1/ \ ˆ.H /:

13.3

271

Filtrations

Q Write ˆ.Œv; w/ D d C Id ˝ D, where d 2 .Der0 S/ ˝ O.2I 1/ and D 2 W .2I 1/.1/ . Q / and d CId˝D 2 Since .Der0 S /˝O.2I 1/ and Id˝W .2I 1/ are invariant under ˆ.T Q Q Q ˆ.H /, one has d 2 ˆ.H /, Id ˝ D 2 ˆ.H /. In particular, d 2 CS0 CF ı .h/ ˝ O.2I 1/ D .F h C F ı/ ˝ O.2I 1/ (Lemma 13.3.6(3)). Write d D h ˝ f0 C ı ˝ f1 . s Q The semisimple part of ˆ.Œv; w/ is a suitable Œp-power, whence equals h˝f0 .0/p C s Q /, and this gives f1 .0/ D 0. ı ˝f1 .0/p for s 0. Since t0 ¤ 0, we have ı ˝1 62 ˆ.T As a result, Q Q Q ˛/ ˆ.Œv; w/ 2 F h ˝ 1 C .Der0 S/ ˝ O.2I 1/.1/ C Id ˝ W .2I 1/.1/ \ ˆ.H / ˆ.H for all w 2 L . This result implies Œv; L H˛ . But v 62 M .˛/ .T / by construction, a contradiction. We now have m D 1. Set D the image of ƒ2 in W .1I 1/. Theorem 8.2.5 (with Q .0/ // (it is nonzero by the earlier G D G0 . /) shows that D is an ideal of 2 .ˆ.L remark ƒ2 ¤ 0 in (b)). Since the latter is a transitive subalgebra of W .1I 1/, D is not contained in W .1I 1/.0/ . Then D itself is a transitive subalgebra of W .1I 1/, and this shows that O.1I 1/ is D-simple. The theorem says that 1 .D/ is a simple algebra, and either 1 .D/ \ W .1I 1/.2/ ¤ ¹0º or p D 5 and 1 .D/ Š sl.2/. In the first case it can only be that D D 1 .D/ D W .1I 1/. In the second one has D D 1 .D/ Š sl.2/. Proposition 7.6.8 then shows that there is an automorphism of W .1I 1/ which maps D onto F @ C F x@ C F x 2 @. This automorphism is induced by an automorphism of O.1I 1/ (Theorem 7.3.2). Applying this normalization we may take D D F @ C F x@ C F x 2 @. (e) Suppose n.˛/ ¤ 0. Take again T 0 D T . Then Fp ˛ D Fp by Lemma 13.3.8. Equation (13.3.5) implies L.˛/ L.0/ . IfP D Š sl.2/, or D Š W .1I 1/ and Q Q .2 ı ˆ/.T / is an improper torus of W .1I 1/, then i 2Fp .2 ı ˆ/.K i˛ / D ¹0º, and this shows that Q i ˛ /; ˆ.K Q Œˆ.K i˛ / Œ.F h C F ı/ ˝ O.1I 1/; .F h C F ı/ ˝ O.1I 1/ D ¹0º: P Then i 2Fp ŒKi ˛ ; K i˛ L.1/ acts nilpotently, whence n.˛/ D 0, a contradiction. P Q Q / is a proper torus, then i 2Fp .2 ı ˆ/.K If D Š W .1I 1/ and .2 ı ˆ/.T i˛ / W .1I 1/.2/ and X Q i˛ /;ˆ.K Q Œˆ.K i˛ / i 2Fp

.F h C F ı/ ˝ O.1I 1/ C Id ˝ W .1I 1/.2/ ; .F h C F ı/ ˝ O.1I 1/ C Id ˝ W .1I 1/.2/

.F h C F ı/ ˝ O.1I 1/.2/ C Id ˝ W .1I 1/.4/ P acts nilpotently on .G/. Again this gives that i 2Fp ŒKi˛ ; K i˛ acts nilpotently on L contrary to the assumption n.˛/ ¤ 0.

272

13


0 (f) Next we will determine L. Begin the construction with T WD T , and derive 1 from (d) that m D 1 and G0 =G0 \ .Der0 S/ ˝ O.1I 1/ has a unique minimal ideal isomorphic to D. Since D is isomorphic to W .1I 1/ or to sl.2/ and therefore has no outer derivations, we conclude that

G0 =G0 \

1

.Der0 S/ ˝ O.1I 1/ Š D:

We now observe that .G0 /\ .Der0 S/˝O.1I 1/ is an ideal of .G0 / which stabilizes S ˝ O.1I 1/.1/ D rad .A.G//. In fact, it is the unique maximal ideal with this property. Then G0 \ 1 .Der0 S/ ˝ O.1I 1/ is the unique maximal ideal of G0 which stabilizes rad A.G/, and this gives a description of D which is independent of the choice of T 0 . We also recall (see Lemma 13.3.6(3)) sl.2/ ˝ O.1I 1/ .G0 / sl.2/ ˝ O.1I 1/ C F ı ˝ O.1I 1/ C Id ˝ D and F @ C F x@ C F x 2 @ D. Since x@ 2 D, .G0 / contains an element of .S0 C F ı/ ˝ O.1I 1/ C Id ˝ x@. Then it contains some ı ˝ fQ C Id ˝ x@, and therefore the p-envelope of .G0 / contains a semisimple element ı ˝ f C Id ˝ x@ (for some some choice of f ). Put Q 0 / D t. t WD F h˝1CF .ı ˝f CId˝x@/ and choose a torus T 0 L.0/ for which ˆ.T Such a torus has the property that ˆ.T 0 / stabilizes the maximal ideal of A.G/. We Q 0 /. Equation (13.3.5) now argue in .G 1 / and .G0 / and observe that h ˝ 1 2 ˆ.T (applied for T ) shows that h ˝ 1 acts invertibly on S 1 ˝ O.1I 1/ and this yields Q 0 // D ¹0º. Moreover, C .G 1 / .ˆ.T Q 0 // CS CF ı .F h/˝1CId˝CD .F x@/ D F h˝1CF ı ˝1CId˝F x@: C .G0 / .ˆ.T Consequently, CL .T 0 /.1/ L.1/ acts nilpotently on L. The normalization of Proposition 13.3.7 for T 0 (which means that we have to engage a different isomorphism 0 ) gives Q 0 .T 0 / D F .h0 ˝ 1/ C F . 0 ı ˝ 1 C Id ˝ x@/: ˆ Q 0 .T 0 /-root vector, which projects onto @ 2 D. Write Let d 1 2 0 .G0 . 0 // be a ˆ (observe Lemma 13.3.6(3)) X X d 1D wa ˝ x a C b ı ˝ x b C Id ˝ @; wa 2 S0 ; b 2 F: a

b

P a 0 Since a wa ˝ x is contained in .G0 /, we may adjust d 1 to obtain d 1 D P b b b ı ˝ x C Id ˝ @. The requirement of d 1 to be an eigenvector gives d 1 D p 1 ı ˝ x p 1 C Id ˝ @. Note that d 1 Œp D .ad Id ˝ @/p 1 .p 1 ı ˝ x p 1 / D . 1/Šp 1 ı ˝ 1. Since G0 does not contain a 3-dimensional torus (as TR.G/ 2) we must have p 1 D 0. Consequently, Id ˝ @ is contained in 0 .G0 . 0 //.

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273

Filtrations

(g) Let u 2 S 1;ˇ 0 .h0 ˝1/ and u0 2 S Theorem 8.2.2 gives

1; ˇ 0 .h0 ˝1/

be nonzero F h0 -root vectors.

ƒ2 .u ˝ x i ; u0 ˝ 1/ D .1

i /x i ƒ2 .u ˝ 1; u0 ˝ 1/ C ix i i.i

1

ƒ2 .u ˝ x; u0 ˝ 1/

1/

Q 2 .u ˝ x; u0 ˝ x/ xi 2ƒ 2 D .1 i /x i ƒ2 .u ˝ 1; u0 ˝ 1/ C ix i 1 ƒ2 .u ˝ x; u0 ˝ 1/ i.i 1/ i 2 C x ƒ2 .u ˝ x; u0 ˝ x/ xƒ2 .u ˝ 1; u0 ˝ x/ 2 xƒ2 .u ˝ x; u0 ˝ 1/ C x 2 ƒ2 .u ˝ 1; u0 ˝ 1/ C

D

.i

1/.i 2/ i x ƒ2 .u ˝ 1; u0 ˝ 1/ C i.2 2 i.i 1/ i 2 C x ƒ2 .u ˝ x 2 ; u0 ˝ 1/ 2

i/x i

1

ƒ2 .u ˝ x; u0 ˝ 1/

for all 0 i p 1. Since ƒ2 .u ˝ x k ; u0 ˝ 1/ is an eigenvector for T 0 , there are sk 2 F and lk 2 ¹0; : : : ; p 1º such that ƒ2 .u ˝ x k ; u0 ˝ 1/ D sk x lk @, whence .i

ƒ2 .u ˝ x i ; u0 ˝ 1/ D

1/.i 2 C i.2

Since ı acts on S

1

2/

s0 x i Cl0 @

i/s1 x

i 1Cl1

@C

i.i

1/ 2

(13.3.10) s2 x

i 2Cl2

@:

as Id we must have 2 0 C k lk

1

Applying Id ˝ @ shows that kƒ2 .u ˝ x k 0 D s0 l0 x l0

1

;

.mod .p//

1 ; u0

s0 x l0 D s1 l1 x l1

˝ 1/ D sk lk x lk 1

;

s0 D s1 l1 ;

(13.3.11) 1 @.

This gives

2s1 x l1 D s2 l2 x l2

Consequently (but one has to take care of the term x lk s0 l0 D 0;

8k:

1

1

:

if lk D 0),

2s1 D s2 l2 :

(13.3.12)

Moreover, l0 D l1

1

or

s0 D 0;

l1 D l2

1

or

s1 D 0:

and If s0 D s1 D s2 D 0, then, as S1 D F u C F u0 , Theorem 8.2.2 shows that ƒ2 D 0, which is not true. It is now immediate from Equation (13.3.12) that 0 2 ¹l0 ; l1 ; l2 º.

274

13


Suppose l1 D 0. Then s0 D 0, 0 D 1 and l2 D 1 (by Equation (13.3.11)). By Equation (13.3.12), 2s1 D s2 . In this case (see Equation (13.3.10)) i.i 1/ i 1 ƒ2 .u ˝ x i ; u0 ˝ 1/ D i.2 i/s1 C s2 x @ D is1 x i 1 @: 2 Pp 2 Since ƒ2 ¤ 0, we have s1 ¤ 0. It follows that D D i D0 F x i @. As p > 3, this Pp 2 implies x 2 @ 2 D. However, i D0 F x i @ is not x 2 @-stable. This contradiction shows that l1 ¤ 0. (h) Suppose l0 D 0. Then 2 0 D 1, whence l1 D 1, l2 D 2. By Equation (13.3.12), s0 D s1 D s2 . Equation (13.3.10) yields .i 1/.i 2/ i.i 1/ i ƒ2 .u ˝ x i ; u0 ˝ 1/ D C i.2 i/ C s0 x @ D s0 x i @ 2 2 (0 i p

1). Adjust u0 to obtain s0 D 1. Then ƒ2 .u ˝ x i ; u0 ˝ 1/ D x i @;

0i p

1:

(13.3.13)

Using the notation of (a) we choose a T 0 -invariant subspace V00 V0 such that V000 V00 and V00 C L.1/ is the inverse image of 0 .G0 / \ .Der0 S/ ˝ O.1I 1/ Ì Id ˝ W .1I 1/.0/ under the homomorphism L.0/ ! L.0/ =L.1/ Š 0 .G0 /, and a T 0 -invariant subspace V 0 1 V 1 such that V 0 1 C L.0/ is the inverse image of S

1

˝ O.1I 1/.1/

under the homomorphism L ! L=L.0/ Š 0 .G 1 /. By definition, V00 CL.1/ is closed under Lie multiplication. By Equation (13.3.13), ƒ2 .u ˝ x r ; u0 ˝ x s / D ƒ2 .u ˝ x rCs ; u0 ˝ 1/ 2 W .1I 1/.0/ whenever r C s 1. Therefore ŒV 0 1 ; V 0 1 V00 C L.1/ : Properties of the associated graded algebra give ŒV 0 1 ; V00 V 0 1 C L.0/ : Next recall that V 01 V

1

X X

Liˇ 0 Cj 0 ;

i D˙1 j 2Fp

V000 V00 V0

X

X

i 2¹˙2;0º j 2Fp

Liˇ 0 Cj 0 :

13.3

275

Filtrations

Therefore (cf. Equation (13.3.8)) ŒV 0 1 ; V00 V 0 1 C

X X

Liˇ 0 Cj 0 \ L.0/

i 2Fp j 2Fp

V 0 1 C V000 C L.1/ V 0 1 C V00 C L.1/ : Finally, Equation (13.3.8) gives ŒV 0 1 ; L.1/ V000 C L.1/ V00 C L.1/ : Consequently, L0.0/ WD V 0 1 C V00 C L.1/ is a subalgebra of L. By construction, L0.0/ is a T 0 -invariant subalgebra of codimension 3. By Equation (13.3.11), 0 D .p C 1/=2. From this it follows that the vectors u ˝ 1 and u0 ˝ 1 belong to the root spaces 0 .G/ˇ 0 .pC1/ 0 and 0 .G/ ˇ 0 .pC1/ 0 , 2 2 respectively. Pick v1 2 Lˇ 0 .pC1/ 0 \ V 1 and v2 2 L ˇ 0 .pC1/ 0 \ V 1 , such that 2

0 .gr

1 v1 /

D u ˝ 1;

2

0 .gr

1 v2 /

D u0 ˝ 1;

Œv1 ; v2 DW w 2 L

0:

Since ƒ2 .u ˝ 1; u0 ˝ 1/ D @ by Equation (13.3.13), one has w 62 L0.0/ , whence L D L0.0/ ˚ F v1 ˚ F v2 ˚ F w: Next, Equation (13.3.13) yields ŒV 1 ; V 0 1 L0.0/ , while Equation (13.3.9) gives ŒV 1 ; V00 V

1

C V000 C L.1/ V

1

C L0.0/ :

This shows that L0.

1/

WD V

1

C L0.0/ D F v1 C F v2 C L0.0/

is L0.0/ -invariant. Since x@ 2 D, there is a vector d 2 CL.0/ .T 0 / which is mapped onto x@ 2 D. Q 0 .d / D h0 ˝ 1 C 0 ı ˝ 1 C Id ˝ x@. If 0 D 0, then ˆ Q 0 .L.0/ / would contain Write ˆ 0 the 3-dimensional torus F h ˝ 1 C F ı ˝ 1 C F Id ˝ x@ (as 0 ¤ 0). But that is not true as TR.L/ D 2. It has already been mentioned above that V 0 1 C L.1/ annihilates L0. 1/ =L0.0/ . Then L0.0/ acts on L0. 1/ =L0.0/ Š F u ˝ 1 ˚ F u0 ˝ 1 as S0 ˝ 1 C F ı ˝ 1 Š gl.2/: Thus L has a standard filtration L D L0.

2/

L0.

1/

L0.0/ L0.sC1/ D ¹0º;

such that the associated graded algebra G 0 satisfies ˚j 0 Gj0 Š ˚j 0 K.3I 1/j .

276

13


Note that L.k/ L.1/ L0.0/ for k 1. As L D L. 1/ L0. 1/ , it is now easy to see by induction (and the definition of standard filtrations) that L.k/ L0.k 1/ for all k 1. Since S Š H.2I 1/.2/ carries the standard filtration and this filtration has height 2p 5, we derive s 2p 6 4. Theorem 5.2.7(3),(4) shows L Š K.3I 1/. (i) Suppose l2 D 0. By Equation (13.3.12), s1 D s0 D 0, so that (see Equation (13.3.10)) ƒ2 .u ˝ x i ; u0 ˝ 1/ D

i.i

1/ 2

s2 x i

2

0i p

@;

1:

(13.3.14)

Pp 3 i Then D D i D0 F x @. Since this is a Lie algebra, we must have p D 5 and 0 D Š sl.2/. Choose a T -invariant subspace V00 V0 such that V000 V00 and V00 CL.1/ is the inverse image of 0 .G0 / \ .Der0 S/ ˝ O.1I 1/ Ì Id ˝ W .1I 1/.0/ under the homomorphism L.0/ ! L.0/ =L.1/ Š 0 .G0 /, and a T 0 -invariant subspace V 0 1 V 1 such that V 0 1 C L.0/ is the inverse image of S

1

˝ O.1I 1/.2/

under the linear mapping L ! L=L.0/ Š 0 .G 1 /. Due to Equation (13.3.14), ƒ2 .u ˝ x r ; u0 ˝ x s / D ƒ2 .u ˝ x rCs ; u0 ˝ 1/ 2 W .1I 1/.0/

whenever r C s 3:

Therefore, ŒV 0 1 ; V 0 1 V00 C L.1/ : A word-by-word repetition of the argument used in (h) yields that L0.0/ WD V 0 1 C V00 C L.1/ is a subalgebra of L. By construction, L0.0/ is T 0 –invariant subalgebra of codimension 5. By Equation (13.3.11), 0 D 1 (as p D 5). From this it follows that the vectors u ˝ 1, u0 ˝ 1, u ˝ x, u0 ˝ x 2 0 .G 1 / belong to the root spaces 0 .G/ˇ 0 C 0 , 0 .G/ ˇ 0 C 0 , 0 .G/ˇ 0 C2 0 , 0 .G/ ˇ 0 C2 0 , respectively. Pick v1 2 Lˇ 0 C 0 \ V 1 , v2 2 L ˇ 0 C 0 \ V 1 , v3 2 Lˇ 0 C2 0 \ V 1 and v4 2 G ˇ 0 C2 0 \ V 1 such that 0 .gr

1 v1 /

D u ˝ 1;

0 .gr

1 v2 /

D u0 ˝ 1;

0 .gr

1 v3 /

D u ˝ x;

0 .gr

1 v4 /

D u0 ˝ x;

and w 2 L.0/;

0

as a preimage of @. Then L D L0.0/ ˚ F v1 ˚ F v2 ˚ F v3 ˚ F v4 ˚ F w:

13.3

277

Filtrations

Applying Equations (13.3.14) and (13.3.8) gives ŒF v3 C F v4 ; V 0 1 C L.1/ V00 C L.1/ L0.0/ : Properties of the associated graded algebra and Equation (13.3.8) yield X X ŒF v3 C F v4 ; V00 F v3 C F v4 C V 0 1 C Liˇ 0 Cj 0 \ L.0/ i 2Fp j 2Fp

F v3 C F v4 C V 0 1 C V000 C L.1/ F v3 C F v4 C L0.0/ : Hence L0. 1/ WD F v3 C F v4 C L0.0/ is L0.0/ -invariant. Arguing as in (h) shows that L0.0/ acts on L0. 1/ =L0.0/ as gl.2/. Next ƒ2 .u ˝ x; u0 ˝ x/ D ƒ2 .u ˝ x 2 ; u0 ˝ 1/ D s2 @: Passing to the corresponding root vectors we conclude that Œv3 ; v4 s2 w .mod L0.0/ /. It is also clear that Œw; v3 v1 ; Œw; v4 v2

.mod L0.0/ /:

Thus L has a standard filtration L D L0.

3/

L0.

2/

L0.

1/

L0.0/ L0.s/ D ¹0º;

such that the associated graded algebra satisfies ˚j 0 grj L Š ˚j 0 M.1; 1/j , where the grading of M.1; 1/ is the natural depth 3 grading. As before one proves s 2p 6 4. Theorem 5.4.1 states that the graded algebra associated with this depth 3 filtration is isomorphic as a graded algebra to a Melikian algebra with natural grading. Theorem 6.7.3 then gives that L itself is a Melikian algebra. As TR.L/ D 2, Corollary 10.5.14 completes the proof. We mention that the exceptional cases L Š K.3I 1/ and L Š M.1; 1/ really do occur. As an example, take the 2-dimensional torus FDK .x1 x2 / ˚ FDK .1 C x3 / in K.3I 1/, define a root direction Fp ˛ by setting ˛.DK .x1 x2 // D 0 and put j

L.0/ WD span ¹DK .x1i x2 x3k / j i C j 2 or i D j D 0º; j

L.l/ WD span ¹DK .x1i x2 x3k / j i C j l C 2º; l > 0: Using Equations (4.2.8) and (4.2.10) it is not hard to show that .T; ˛; L.0/ / is an admissible triple and that this definition establishes a T -invariant standard filtration such that gr L Š H.2I 1/.1/ ˝ O.1I 1/ Ì Id ˝ W .1I 1/ . One can do similarly for M.1; 1/ by taking T D FDH .x1 x2 / ˚ F .1 C x1 @Q 1 /, ˛.1 C x1 @Q 1 / D 0 and L.0/ D

3 X

G

M.1; 1/Œ2i C DH .O.2I 1/.3/ / C DH .O.2I 1/.3/ / C

iD 1

We leave details of these examples to the reader.

X i 7

M.1; 1/Œi :

278

13

13.4


More on Hamiltonian roots

The case S0 Š W .1I 1/ is much harder to handle than the previous in which S0 Š sl.2/. We will need detailed information on Hamiltonian roots and on elementary switchings in the following. In this section we investigate Hamiltonian roots. Suppose L 6Š M.1; 1/ and 2 .L; T / is a Hamiltonian root. Set g WD L./.1/ . Then 0 ¤ TR.g/ TR.L.// 1, g= rad g Š H.2I 1/.2/ (cf. Theorem 11.1.2) and g D g.1/ . Let h be any CSA of g and T0 be the maximal torus of the p-envelope of h in LŒp . Then T 0 WD .T \ ker / C T0 is a torus of LŒp of dimension at least 2, hence is a torus of maximal dimension. Since L is assumed not to be a Melikian algebra, CL .T 0 / is a trigonalizable CSA of L (Corollary 12.5.10). Clearly, CL .T 0 / \ L./.1/ is mapped onto h. Lemma 13.4.1. Suppose L 6Š M.1; 1/. Let 2 .L; T / be a Hamiltonian root and .2/ assume that L./.1/ =C.L./.1/ / Š PH.2I 1/ . Suppose there is 2 .L; T / n .1/ Fp such that the L./ -module i 2Fp LCi has a composition factor V of dimension < p 4 . Then n./ D 0 and dim Li =Ri 4. If, moreover, is proper then Ki D Ri holds for all i 2 Fp . Proof. The introductory remark shows that Theorem 11.5.2 applies. As a consequence, Œg.0/ ; g.1/ C Œg; g.2/ acts nilpotently on V , where g.j / D L./.1/ .j / denotes the preimage of H.2I 1/.2/ .j / ;

0 j 2p

5;

under the canonical homomorphism g ! H.2I 1/.2/ . Note that H \ L./.1/ .1/ ker . Then the above shows that every d 2 H \ ŒL./.1/ .0/ ; L./.1/ .1/ C ŒL./.1/ ; L./.1/ .2/ has the property that d 2 ker \ ker . This means d 2 nil H . For i 2 Fp one has Ki L./.1/ . Remark 13.1.4 yields Ki ; K i L./.1/ .1/ . The above shows that ŒKi ; K i nil H , whence ni D 0 for all i 2 Fp . More exactly, one gets the following. Let D 2 Ki , where i 2 Fp . Let DN be the image of D in L./.1/ =C.L./.1/ / Š H.2I 1/.2/ . If DN 2 H.2I 1/.2/ .2/ , then the above result shows ŒD; L./ i nil H . In this case D 2 Ri holds. Suppose is proper. By Remark 13.1.4 (after a suitable normalization) and the preceding observation Ki .FDH .x13 / C FDH .x23 // \ Li C Ri holds. If DN 2 FDH .x13 / [ FDH .x23 /, then L./.1/ i L./.1/ .0/ and again by the preceding result ŒD; L./ i acts nilpotently on L. This means that D 2 Ri . This proves Ki D Ri if is proper. Suppose is improper. By Remark 13.1.4 (after a suitable normalization) and the preceding observation Ki FDH ..1 C x1 /l x23 / C Ri holds, where l is determined

13.4

279


by the equation 3 l i.DH ..1 C x1 /x2 //. In this case we get dim Ki =Ri 1, whence dim Li =Ri dim Li =Ki C 1 D 4. In the general case when L./ is Hamiltonian but has a non-central radical we obtain the following (cf. Corollary 11.2.4). Lemma 13.4.2. Suppose L 6Š M.1; 1/. Let 2 .L; T / be a Hamiltonian root and let W L./ ! LŒ H.2I 1/ denote the canonical homomorphism. Suppose P there is 2 .L; T / n Fp such that the L./.1/ -module j 2Fp LCj has a composition factor V of dimension dim V < p 3 . The following holds: (1) n./ D 0. (2) If is proper and L./ ¤ L.0/ ./ C rad L./, then .L.0/ .// H.2I 1/.0/ . (3) If, moreover, is proper, L.0/ ./ is solvable and dim Li =L.0/;i D 1 for all i 2 Fp , then p D 5 and .L.0/ .// D H.2I 1/.2/ .1/ C .H /. Proof. (1) Set g WD L./.1/ and define g.k/ (0 k 2p 5) as in the previous proof. The introductory remark shows that Theorem 11.5.4(2) applies. This theorem yields that Œg.0/ ; g.1/ \ H acts nilpotently on V . As in the proof of the preceding Lemma 13.4.1 this gives ni D 0 for all i 2 Fp . (2) Suppose is proper and L./ ¤ L.0/ ./ C rad L./. Then T stabilizes g.0/ and g.1/ and acts on H.2I 1/.2/ as F .x2 @2 x1 @1 / (Theorem 11.2.5). No generality is lost by assuming .x2 @2 x1 @1 / D 1. Then g D L C L C g.0/ . If i ¤ ˙1, then Li g.0/ . By Remark 13.1.4, Ki g.1/ for any i 2 Fp . We mentioned above that Œg.0/ ; g.1/ \ H acts nilpotently on L. Consequently, ŒL i ; Ki acts nilpotently on L whenever i ¤ ˙1. In other words, Ki D Ri L.0/;i for any i 2 Fp n¹0; ˙1º: Suppose L.0/ ./ 6 H.2I 1/.0/ , whence L.0/ ./ 6 g.0/ C H . By symmetry there is a 2 L.0/ such that .a / DH .x1 / .mod H.2I 1/.0/ /. Since T stabilizes L.0/ and H.2I 1/.0/ , we may take a a T -eigenvector, whence a 2 L.0/; and .a / D DH .x1 / C

p X2

i DH .x1iC1 x2i / C p

p 1 /: 1 DH .x2

i D1

There is b 2 L L.0/ ./. Then

2

p 2

such that .b/ D DH .x2

.L.0/ / 3 .ad a /p

3

.b/ .p

/. By the above b 2 K

2/ŠDH .x2 /

2

DR

.mod H.2I 1/.0/ /:

It follows that there is aC 2 L.0/; such that .aC / D DH .x2 / C

p X2 i D1

i DH .x1i x2i C1 / C p

p 1 /: 1 DH .x1

2

280

13

Towards graded algebras p 3 p 1

Similarly we find c˙2 2 L˙2 L.0/ ./ such that .c2 / D DH .x1 x2 / and p 1 p 3 .c 2 / D DH .x1 x2 /. Since all these are root vectors for nonzero roots, they are contained in H.2I 1/.2/ . Let M denote the subalgebra of H.2I 1/.2/ generated by .aC /; .a /; .c2 /; .c 2 /. It is easy to see that p 2 p 1 x2 /

H.2I 1/.2/ D M C FDH .x1

p 1 p 2 x2 /:

C FDH .x1

Then M coincides with H.2I 1/.2/ ([S-F, Theorem 4.7.2]). We deduce that L.0/ ./ C rad L./ contains 1 .H.2I 1/.2/ / g. However, L./ D H Cg and H L.0/ ./. This contradiction shows that L.0/ ./ H.2I 1/.0/ . (3) Note that the assumptions of (2) are satisfied in the present case. Let us normalize .T / D FDH .x1 x2 / and .DH .x1 x2 // D 1. Suppose p > 5 and pick k 2 Fp n ¹0; ˙1; ˙2º ¤ ;. Remark 13.1.4 shows Lk D Kk . By our discussion in (2), Ki D Ri L.0/;i for any i 2 Fp n ¹0; ˙1º. Therefore, Lk D Rk L.0/ . As this contradicts our present assumption on dim Lk =L.0/;k we must have p D 5. Next we intend to show g.1/ L.0/ . Due to Remark 13.1.4 1 .H.2I 1/.1/ / \ L˙2 D K˙2 holds. As mentioned in (1), ŒK

2 ; K2

Œg.1/ ; g.1/ \ H

acts nilpotently on L. Consequently,

1

.H.2I 1/.1/ / \ L˙2 D K˙2 D R˙2 L.0/ :

Pick i; j 2 ¹0; : : : ; 4º with i 6 j .5/ and choose a root vector v.i; j / 2 g such that j .v.i; j // D DH .x1i x2 /. Suppose i j 1 .5/ and i C j > 1. Notice that v.i; j / and v.0; 1/ are in L . By our assumption, L.0/; has codimension 1 in L . So v.i; j / 62 L.0/; implies v.0; 1/ C v.i; j / 2 L.0/; for some 2 F . But then j DH .x2 / D .v.0; 1// 2 .L.0/ .// C FDH .x1i x2 / H.2I 1/.0/ by assertion (2), a contradiction. Thus v.i; j / 2 L.0/ ./. Similarly, v.i; j / 2 L.0/ ./ for all i; j with i j 1 .5/ and i Cj > 1. As a consequence, g.1/ CH L.0/ ./CH D L.0/ ./. On the other hand, L./=.g.1/ C H / is spanned by the images of v.1; 0/, v.0; 1/, v.2; 0/ and v.0; 2/. The assumption that dim Li =L.0/;i D 1 for all i 2 F5 gives L.0/ ./ D g.1/ C H as desired. Lemma 13.4.3. Suppose L 6Š M.1; 1/. Let 2 .L; T / be a Hamiltonian root. Suppose there is 2 .L; T / n Fp such that X dim LiCj < 2p 3 8 i 2 Fp : j 2Fp

Then n./ D 0 or 3 .rad L./.1/ / D ¹0º.

13.4


281

P Proof. Set g WD L./.1/ . If there is i 2 Fp for which j 2Fp LiCj is not irreducible as a g-module, then it has a composition factor of dimension < p 3 . In this case Lemma 13.4.2 yields n./ D 0. P So we may assume that j 2Fp LiCj is g-irreducible for every i 2 Fp . From Theorem 11.5.4 one concludes that there is an irreducible g.0/ -submodule W of P 2 j 2Fp Li Cj of dimension < p with representation W g.0/ ! gl.W /. Set M WD .g.0/ /. Choose a semisimple element h 2 gŒp for which .h/ D DH .x1 x2 /. Set T 0 WD .T \ ker / ˚ F h and h WD Cg.0/ .T 0 /. Observe that h D Cg.0/ .F h/ is self-normalizing. Note that CL .T 0 / acts trigonalizably on L by our assumption L 6Š M.1; 1/. Then .h/.1/ consists of nilpotent transformations. If .h/ acts nilpotently on M , then M D .h/. In this case .g.0/ / D .h/ acts trigonalizably on W . The irreducibility of W forces g.0/ .1/ ker . Otherwise Theorem 11.3.1 applies for M . Note that g.0/ =g.1/ Š sl.2/ is simple. If ker 6 g.1/ , then g.0/ D ker C g.1/ . In this case M is nilpotent, but this is the former case which we do not consider here. Therefore ker g.1/ , rad M D .g.1/ / and M= rad M Š g.0/ =g.1/ Š sl.2/. Case 1 of Theorem 11.3.1 is impossible. In Case 2 of that theorem one has .M=C.M //.1/ Š sl.2/ Š M= rad M . Then Œ.g.0/ /; .g.1/ / D ¹0º, whence

3 .g.1/ / ker . In Case 3 one also has 3 .g.1/ / ker . In Case 4, Theorem 11.3.4 applies. Since M is nonsolvable, only the last case of that theorem can occur and in this case again 3 .g.1/ / ker holds. Recall that rad g g.0/ and therefore rad g rad g.0/ . As a consequence, 3 .rad g/ annihilates W . Therefore P the ideal 3 .rad g/ of g has common eigenvalue 0 on the irreducible g-module j 2Fp LiCj . By Engel’s theorem this ideal annihilates the full module, whence Œ 3 .rad L./.1/ /; LiCj D ¹0º .1/ /; L D ¹0º for all holds for all i 2 Fp and all j 2 Fp . This means Œ P 3 .rad L./ P 2 .L; T /nFp . The simplicity of L gives L D 62Fp L C 62Fp ŒL ; L . Then 3 .rad L./.1/ / C.L/ D ¹0º. This proves the lemma.

There is a way to describe the properness of Hamiltonian roots by looking at the associated 1-section of gr L. The following remarks are well known, but nevertheless quite useful. Let g be a filtered Lie algebra, g D g.

s1 /

© © g.s2 / © ¹0º;

and set 2 gr g D ˚si D

s1 gri

g;

gri g WD g.i/ =g.iC1/

the corresponding graded Lie algebra. Assume that H.2I 1/.2/ g= rad g H.2I 1/;

H.2I 1/.2/ gr g= rad.gr g/ H.2I 1/:

282

13


Consider any subalgebra q of g and any ideal j of q. Clearly, 2 gr q D ˚si D s1 q \ g.i/ C g.iC1/ =g.i C1/ is a subalgebra of gr g, and dim gr g= gr q D dim g=q: Also, gr j is an ideal of gr q satisfying dim gr q= gr j D dim q=j. The quotient gr q= gr j inherits a natural grading from gr q. In addition, gr j is solvable or nilpotent if j is so. This implies gr.rad q/ rad.gr q/: Set W g ! g= rad g H.2I 1/;

gr W gr g ! gr g= rad.gr g/ H.2I 1/

and g.0/ WD

1

H.2I 1/.0/ ;

.gr g/.0/ WD gr 1 H.2I 1/.0/ :

We intend to show that gr g.0/ D .gr g/.0/ :

(13.4.1)

Observe that dim g=g.0/ D dim gr g=.gr g/.0/ D 2; whence dim gr g.0/ D dim g

2 D dim.gr g/.0/ :

As a consequence, gr .gr g.0/ / has codimension 2 in gr .gr g/, and this gives that gr .gr g.0/ / \ H.2I 1/.2/ is a subalgebra of codimension 2. Due to Theorem 4.2.6 only H.2I 1/.2/ gr .gr g.0/ / or gr .gr g.0/ / \ H.2I 1/.2/ D H.2I 1/.2/ .0/ are possible. Next observe that .g.0/ / H.2I 1/.2/ .0/ C H.2I 1/.1/ , and therefore g.0/ = rad g.0/ Š H.2I 1/.2/ .0/ =H.2I 1/.2/ .1/ Š sl.2/: As ker gr D rad.gr g/ rad.gr g.0/ / and gr.rad g.0/ / rad.gr g.0/ /, this gives dim gr .gr g.0/ /= rad gr .gr g.0/ / D dim gr g.0/ = rad.gr g.0/ / dim g.0/ = rad g.0/ D 3: This shows that the first case is not possible. Hence gr .gr g.0/ / \ H.2I 1/.2/ D H.2I 1/.2/ .0/ :

13.4

283


Then it is not difficult to conclude that gr .gr g.0/ / H.2I 1/.0/ . Consequently, gr g.0/ gr 1 .H.2I 1/.0/ / D .gr g/.0/ . As both these algebras have dimension dim g 2, we obtain equality. Proposition 13.4.4. Consider L as a filtered algebra with a filtration associated with an admissible triple .T; ˛; L.0/ / as in §13.3 and put G WD gr L. Let T 0 LŒp be a 2-dimensional torus and 2 .L; T 0 /. Suppose that CL .T 0 / L.0/ and ŒT 0 ; L.i / L.i / holds for all i 2 Z. Assume that L 6Š M.1; 1/ and H.2I 1/.2/ G./= rad G./ H.2I 1/; ® ¯ G0 ./= rad G0 ./ 2 sl.2/; W .1I 1/ : Then is a Hamiltonian T 0 -root of L. Moreover, is a proper root of L if and only if G0 ./= rad G0 ./ Š sl.2/ or else the image of T 0 is contained in W .1I 1/.0/ . Proof. (a) Note that L./ is not solvable, since otherwise so were G./. As L./ is a 1-section of L then one has TR L./ D 1 and there is h 2 CL./ .T 0 / for which s F hŒp Š T 0 =T 0 \ ker (Proposition 11.2.1) for large s. Note that CL .T 0 / acts trigonalizably under the present assumption that L is not a Melikian algebra. Then L./= rad L./ is one of ¹0º; sl.2/; W .1I 1/, or H.2I 1/.2/ L./= rad L./ H.2I 1/ holds (Corollary 11.2.4). Since p2

2 D dim H.2I 1/.2/ dim G./= rad G./ D dim gr L./= rad.gr L.// dim gr L./= gr.rad L.// D dim L./= rad L./;

the first three cases are impossible. Then H.2I 1/.2/ L./= rad L./ H.2I 1/. This shows that is a Hamiltonian T 0 -root of L. (b) The filtration of L induces a grading on G./ and on G./= rad G./ DW M H.2I 1/: The grading of M is given by homogeneous generators u1 ; u2 of degree a1 ; a2 , respectively (Theorem 11.4.1). By assumption on G0 ./= rad G0 ./ the grading of M is as in Cases 2 or 3 of that theorem. We mentioned above that there is h 2 CL./ .T 0 / DW s H 0 for which F hŒp Š T 0 =T 0 \ ker . One of the assumptions of this proposition states that H 0 L.0/ . This gives X s gr .gr0 h/ 2 M0 and CM .gr .gr0 h/Œp / D gr .gr H 0 / Mi : i 0

If the grading of M is as in Case 2 of Theorem 11.4.1, then (as M H.2I 1/) one has M0 Š W .1I 1/. As W WD

p X1 i D0

.i/ .p 2/

FDH .u1 u2

/ M.p

3/a2

284

13


is a restricted irreducible M0 -module of dimension p, 0 is a weight of W with res spect to ad gr .gr0 h/Œp (Theorem 7.6.10). We mentioned before that gr .gr H 0 / P 3/a2 0, whence a2 > 0. i 0 Mi . This shows that .p If the grading of M is as in Case 3 of Theorem 11.4.1, then we conclude similarly to the former case that M0 Š sl.2/. Note that M.2p

6/a2

D

2 X

.p 1 i/ .p 3Ci/ / u2

FDH .u1

i D0 s

is an irreducible M0 -module of dimension 3. Hence 0 is a ad gr .gr0 h/Œp -weight of this module, again yielding a2 > 0. (c) Next we consider the canonical maximal subalgebra H.2I 1/.0/ of codimension 2. Set L./.0/ WD 1 .H.2I 1/.0/ /. Equation (13.4.1) proves gr gr.L./.0/ / D M \ H.2I 1/.0/ : Using the homogeneous generators u1 ; u2 we obtain .i/ .j /

H.2I 1/.0/ D ˚i Cj 2 FDH .u1 u2 /: Moreover, as a2 > 0 by (b), one has X .i/ .kC1/ gr grk L.k/ ./ D Mk D FDH .u1 u2 / H.2I 1/.0/

8k 1;

i

if a1 D 0. This gives L.1/ ./ L./.0/ (in both cases). s (d) By definition is proper if and only if hŒp stabilizes L./.0/ . In other words, if is proper, then .i/ .j /

gr .gr0 h/ 2 H.2I 1/.0/ D ˚i Cj 2 FDH .u1 u2 /; and if this inclusion holds, then gr0 h 2 gr.L./.0/ / and h 2 L./.0/ C L.1/ ./ D L./.0/ : .i/ .j /

Consequently, is proper if and only if gr .gr0 h/ 2 ˚i Cj 2 FDH .u1 u2 /. On the other hand, gr .gr0 h/ 2 M0 . Combining this with the result a2 > 0 from (b) we conclude that is proper if and only if M0 Š sl.2/ or M0 Š W .1I 1/ and .i/ .j /

gr .gr0 h/ 2 M0 \ ˚i Cj 2 FDH .u1 u2 / D

X

.i/

FDH .u1 u2 /:

i 1

This is the claim.

13.5

13.5

285

Switching tori

Switching tori

Similar to the proof of Lemma 9.2.8 we will need to switch tori. To do so we employ the process of toral switching described in §1.5, which description we have roughly repeated in §10.7. For the notations employed in the following we refer to that section. It is important to get control on admissible triples under elementary switchings. Let .T; ˛; L.0/ / denote an admissible triple and let as in section §13.3 L.0/ stand for the p-envelope of L.0/ in LŒp . Define an associated standard filtration and set N Š S ˝ O.mI 1/ with m WD m.T; ˛; L.0/ /. Normalize ˆ.T Q / according to W A.G/ Equation (13.3.3) as Q / D F h ˝ 1 C F .ı ˝ 1 C Id ˝ t0 /: ˆ.T Recall the notion of the roots ˇ; from Equation (13.3.4). For a root define ; N Q 2 Fp by N WD .h ˝ 1/ and Q WD .ı ˝ 1 C Id ˝ t0 /: If n.˛/ ¤ 0 or m D 2, Lemma 13.3.8(5) states Fp ˛ D Fp , and then Proposition Q // Fp ˇ C Fp ˛. In particular, L.˛/ \ L. 1/ D L.0/ .˛/ 13.3.7 gives .GN 1 ; ˆ.T holds. Lemma 13.5.1. Suppose m ¤ 0. (1) If n˛ ¤ 0, then there exists u 2 Ki˛ for i D 1 or i D W .mI 1/.0/ .

Q 1 such that .2 ı ˆ/.u/ 62

(2) Let T 0 L.0/ be contained in T .L/, v 2 [2.L;T 0 / L.0/; and 2 F be such that .v; / 2 X. Then Tv0 2 T .L/. Proof. (1) Recall that K.˛/ L.0/ by construction. As n˛ ¤ 0 and S0 ˝ O.mI 1/ Q Q /-roots (Proposition 13.3.7), ˆ.ŒK carries two Fp -independent ˆ.T ˛ ; K ˛ / acts nonnilpotently on S0 ˝ O.mI 1/. Q Q for Set D 0 WD .2 ı ˆ/.K.˛// W .mI 1/ and D0Q WD ¹D 2 D 0 j Œt0 ; D D Dº 0 Q 2 .L; T /. Then .2 ı ˆ/.K holds. Note that i˛ / D D i ˛Q

Q .0/ / .Der0 S/ ˝ O.mI 1/ D .S0 C F ı/ ˝ O.mI 1/ .ker 2 / \ ˆ.L and, as ˛.h ˝ 1/ D 0, Q ˆ.K.˛// CS0 CF ı .h/ ˝ O.mI 1/ C Id ˝ D 0 : As CS0 CF ı .h/ D F h C F ı is abelian, Q ˆ.ŒK ˛; K

˛ /

.F h C F ı/ ˝ O.mI 1/ C Id ˝ ŒD˛0Q ; D 0 ˛Q

and Q adF h˝O.1I1/ ˆ.ŒK ˛; K

˛ /

D adF h˝O.1I1/ ŒD˛0Q ; D 0 ˛Q :

286

13


Q .0/ .˛//, Since ŒK˛ ; K ˛ H˛ acts nilpotently on L.˛/ and F h ˝ O.1I 1/ ˆ.L 0 0 0 Id ˝ ŒD˛Q ; D ˛Q acts nilpotently on F h ˝ O.mI 1/. Therefore Id ˝ ŒD˛Q ; D 0 ˛Q acts nilpotently also on S0 ˝ O.mI 1/. 0 N Suppose both ŒId˝D˙ ; F h˝ Set O.mI 1/˛N WD ¹f 2 O.mI 1/ j t0 .f / D ˛º. ˛Q O.mI 1/˛N consist of adS0 ˝O.mI1/ -nilpotent elements. Then 0 D˙ ˛Q .O.mI 1/˛N / O.1I 1/.1/ ;

and B WD ŒId ˝ D˛0Q ; .F h C F ı/ ˝ O.mI 1/ [ ŒId ˝ D

0

˛Q ; .F h

˛N

C F ı/ ˝ O.mI 1/˛N [ Id ˝ ŒD˛0Q ; D 0 ˛Q

would consist of adS0 ˝O.mI1/ -nilpotent elements. Clearly, B is a Lie set. Then the Lie subalgebra spanned by B would act nilpotently on S0 ˝ O.mI 1/. But then Q ˆ.ŒK ˛ ; K ˛ / would act nilpotently on S0 ˝ O.mI 1/, a contradiction. Thus there Q is u 2 Ki ˛ with i 2 ¹˙1º, such that ŒId ˝ .2 ı ˆ/.u/; F h ˝ O.mI 1/ i ˛N acts Q 62 W .mI 1/.0/ . non-nilpotently on S0 ˝ O.mI 1/. This implies .2 ı ˆ/.u/ (2) As T 0 L.0/ and v 2 L.0/ , Tv0 respects the filtration of L. Lemma 13.3.8(2) proves that Tv0 is not rigid. Corollary 12.5.10 shows that CL .Tv0 / acts trigonalizably on L. As m ¤ 0, there is a realization Q 0v W L.0/ ! .Der0 S/ ˝ O.mI 1/ Ì Id ˝ W .mI 1/ ; ˆ Q 0v .Tv0 / D F h0v ˝ 1 ˚ F .v0 ı ˝ 1 C Id ˝ tv0 / ˆ Q 0v .t1 / (Proposition 13.3.7 and (13.3.2)). There is an element t1 2 CL .Tv0 / satisfying ˆ 0 D hv ˝ 1. Q 0v /.L.0/ / and D WD .2 ı ˆ Q 0v /.L.0/ / the algebra generated Put D WD .2 ı ˆ by D and associative p-th powers in W .mI 1/. If CD .tv0 / is not p-nilpotent, then Q 0 .t2 / D q C Id ˝ d , where q 2 .S0 C there is an element t2 2 CL .Tv0 / satisfying ˆ s Œps 0 p F ı/ ˝ O.mI 1/, d 2 CD .tv / and d is semisimple (s 0). Then t2 2 Tv0 n F t1 , whence Tv0 is contained in the p-envelope of CL .Tv0 /. This is the claim. Otherwise all elements of CD .tv0 / are p-nilpotent. Set Di WD ¹d 2 D j 0 Œtv ; d D id º (i 2 Fp ). Let d 2 Di for i 2 Fp . Since F tv0 is a maximal torus in D, l

l

d p D tv0 for some l 0. But then 0 D Œd p ; d D id , whence D 0. Hence the Lie set [i 2Fp Di consists of ad-nilpotent elements. The algebra generated by this Lie set and associative p-th powers is D. Then D is nilpotent by the Engel–Jacobson Q 0 /.CL .tv0 //. theorem. This algebra contains tv0 , which gives D D D0 D .2 ı ˆ The present assumption says that D consists of p-nilpotent elements. Let k .D /p denote the algebra generated by k .D / and associative p-th powers. Using Lemma 1.1.1 we obtain that, if inductively 2k .D /p consists of p-nilpotent elements, then so does k .D /p . As D is nilpotent, we obtain that 1 .D /p D D consists of pnilpotent elements. However, the latter contains the non-zero semisimple element tv0 , a contradiction.

13.5

287

Switching tori

Next we show that toral switchings respect the subalgebras M .˛/ .T /. .˛/

Proposition 13.5.2. Let u 2 M .T / where ˛; 2 .L; T / n ¹0º. If 2 .L; T / n .˛/ Fp ˛, suppose in addition that [i 2Fp Mi .T / consists of Œp-nilpotent elements of LŒp . Let 2 F be such that .u; / 2 X and assume that CL .Tu / acts trigonalizably on L. Then E.u;/ .M .˛/ .T // M .˛.u;/ / .Tu /: If u 2 [i 2Fp Ki ˛ .T /, then M .˛.u;/ / .Tu / D E.u;/ .M .˛/ .T // and K.˛.u;/ ; Tu / D K.˛; T /. r

Proof. (a) If 2 Fp ˛, then ˛.uŒp / D 0 for all r > 0. If 62 Fp ˛, then the same holds by our additional assumption. As E.u;/ is an invertible linear transformation 1 on a finite dimensional vector space, E.u;/ is a polynomial in E.u;/ . Since E.u;/ D 1 f .ad u/, there is a polynomial ' 2 F Œ such that E.u;/ D '.ad u/. Note that E.u;/ .H / D CL .Tu / M .˛.u;/ / .Tu /. Now let ¤ 0 and a 2 .˛/ M .T /, b 2 L . Considering root spaces with respect to Tu gives E.u;/ .a/; E.u;/ .b/ D E.u;/ .h/ for some h 2 H . Hence 1 h D E.u;/ E.u;/ .a/; E.u;/ .b/ D '.ad u/ f .ad u/.a/; f .ad u/.b/ 2 H \ span

p X1

®

¯ .ad u/i .a/; .ad u/j .b/ j i; j 0

.˛/

ŒMCi .T /; L

i

H˛ :

i D0

This proves X .˛/ E.u;/ .a/; E.u;/ .b/ D E.u;/ .h/ D f .ad u/.h/ 2 H˛ ˚ Mi .T /: i 2Fp

P .˛/ Let M denote the p-envelope of M.˛; / WD H˛ ˚ i 2Fp Mi .T / in LŒp . As M.˛; / is a subalgebra of L, Jacobson’s formula on p-th powers gives X XX Œpj j .˛/ MD H˛ Œp C Mi .T / : i 2Fp j 0

j 0

Therefore the set [ j 0

H˛ Œp

j

[

[ [ i 2Fp j >0

.˛/

Mi .T /

Œpj

288

13


spans M \ CLŒp .T /. If 2 .L; T / n Fp ˛ then, by our assumption, every element Œpj .˛/ .˛/ acts nilpotently of [i 2Fp Mi .T / is Œp-nilpotent. If 2 Fp ˛, then Mi .T / on L˛ whenever i 2 Fp and j > 0. Therefore each element of the above set acts nilpotently on L˛ . Since it is a Lie set, the Engel–Jacobson theorem applies and gives T \ M T \ .ker ˛/: r

Choose r 2 N such that E.u;/ .h/Œp 2 Tu and write for suitable t 2 T r E.u;/ .h/Œp D tu D t .t/ u C q.u/ : Since u; E.u;/ .h/ 2 M.˛; /, r

t D E.u;/ .h/Œp C .t/.u C q.u// 2 T \ M T \ .ker ˛/: Consequently, ˛.t/ D 0. But then ˛.u;/ .tu / D ˛.t/

˛..u; //.t/ 2

X

i

F ˛.uŒp /.t/ D 0

i>0

by our assumption on u. This, in turn, means that ˛.u;/ E.u;/ .a/; E.u;/ .b/ D ˛.u;/ E.u;/ .h/ D 0; .˛/ yielding ˛.u;/ E.u;/ .M .T //; L

.u;/

D 0. Thus

.˛.u;/ / E.u;/ M.˛/ .T / M.u;/ .Tu / for all 2 .L; T /; as claimed. (b) Suppose u 2 Ki˛.u;/ .T /. Then u D E.u;/ .u/ 2 Ki˛.u;/ .Tu /. In addition, .Tu / u D T holds. Applying the first part of the lemma with Tu , u and a suitable 0 instead of T , u, gives E. u;0 / .M .˛.u;/ / .Tu // M .˛/ .T /. So equality follows from the fact that det .E. u;/ ı E.u;0 / / ¤ 0: As a further consequence, K.˛.u;/ ; Tu / D E.u;/ .K.˛; T //. Since u 2 K.˛; T / the latter coincides with K.˛; T /. Corollary 13.5.3. Let u 2 [i 2Fp Ki˛ .T / and 2 F be such that .u; / 2 X. If .T; ˛; L.0/ / is an admissible triple which satisfies m.T; ˛; L.0/ / ¤ 0, then so is .Tu ; ˛.u;/ ; L.0/ /. Proof. According to Lemma 13.5.1(2) one has Tu 2 T .L/. By assumption, Proposition 13.5.2 applies and shows that M .˛.u;/ / .Tu / D E.u;/ .M .˛/ .T // L.0/ . As a consequence, .Tu ; ˛.u;/ ; L.0/ / is an admissible triple. The number m.T; ˛; L.0/ / only depends on L.0/ . Therefore m.Tu ; ˛.u;/ ; L.0/ / D m.T; ˛; L.0/ / ¤ 0:

13.5

289

Switching tori

In the following we will make a subtle choice of the admissible triple .T; ˛; L.0/ / in order to achieve the main result m.T; ˛; L.0/ / D 0 of this chapter. Set imp .L; T / the set of all (nonzero) improper roots. Definition 13.5.4. A torus T 0 LŒp is called a good torus if (1) T 0 T .L/, (2) jimp .L; T 0 /j is minimal with respect to all tori in T .L/. The question if good tori at all exist will be taken up later. Lemma 13.5.5. If L 6Š M.1; 1/, n.L/ D 0, T is a good torus, .T; ˛; L.0/ / is an admissible triple, m.T; ˛; L.0/ / ¤ 0 and S0 Š W .1I 1/, then jimp .L; T /j p

1:

N Proof. (a) Consider first the case that ˆ.T / stabilizes the maximal ideal of A.G/. Q / D F h ˝ 1 C F .ı ˝ 1 C Id ˝ t0 / where t0 D Then there is a realization ˆ.T P m a x @ ¤ 0 with a i 2 F (Proposition 13.3.7). Observe that D WD .2 ı i D1 i i i Q ˆ/.L.0/ / is T -stable and h ˝ 1 2 S0 ˝ O.mI 1/ acts trivially on D. Hence D D ˚i 2Fp Di Q where Di Q D ¹d 2 D j Œt0 ; d D i d Q º. Suppose Di Q W .mI 1/.0/ for all i 2 Fp . Then Œt0 ; D W .mI 1/.0/ . As D is a transitive subalgebra of W .mI 1/ this implies ai D 0 for all i , a contradiction. Consequently, there is i 2 Fp and Q u 2 L.0/;i such that .2 ı ˆ/.u/ 62 W .mI 1/.0/ . Pdim L MT.L/ 1 Œpi Recall the notion of q.x/ D i D1 x . As F is infinite there is r 2 Q F such that .2 ı ˆ/.ruCq.ru// 62 W .mI 1/.0/ . Adjusting u we may assume r D 1. Choose 2 F such that .u; / 2 X and apply the elementary switching E.u;/ . Q Lemma 13.5.1(2) shows that Tu 2 T .L/. Let t 2 T be such that .2 ı ˆ/.t/ D t0 . This assumption implies i .t/ ¤ 0. One has tu D t i .t/.u C q.u// 62 W .mI 1/.0/ . Q (b) We now turn to the general case. If .2 ı ˆ/.T / 6 W .mI 1/.0/ we set T 0 WD 0 Q / W .mI 1/.0/ then we set T WD Tu . In both cases T 0 2 T , and if .2 ı ˆ/.T 0 / 6 Q T .L/, it respects the filtration of L, acts naturally on GN and satisfies .2 ı ˆ/.T W .mI 1/.0/ . Due to Proposition 13.3.7 there is a realization Q 0 .T 0 / D F h0 ˝ 1 ˚ F Id ˝ .1 C x1 /@1 : ˆ N 0 = rad A.G/ N 0 Š W .1I 1/. Recall that S0 Š A.G/ 0 If h 62 W .1I 1/.0/ , we switch T 0 to yet another 2-dimensional torus T 00 . In order to avoid confusion Pp 1 we use y as the indeterminate to describe W .1I 1/, i.e., we set W .1I 1/ D i D0 F y i @y . There is a normalization (see Theorem 7.5.1) h0 D .1 C Q 0 .v/ D @y ˝ 1 and 0 2 F such y/@y . Choose a root vector v 2 L.0/; such that ˆ 0 that .v; / 2 X. Note that X i X Q 0 .v C q.v// D ˆ Q 0 .v/ C Q 0 .v/Œpi D @y ˝ 1 ˆ ˆ @yp ˝ 1 D @y ˝ 1: i>0

i>0

290

13


Q 0 .t1 / D .1 C y/@y ˝ 1 and ˆ Q 0 .t2 / D Id ˝ .1 C x1 /@1 . Choose t1 ; t2 2 T 0 such that ˆ Then .t1 / D 1 and .t2 / D 0 and Q 0 .t1;v / D .1 C y/@y ˝ 1 ˆ

@y ˝ 1 D y@y ˝ 1;

Q 0 .t2;v / D Id ˝ .1 C x1 /@1 : ˆ Set T 00 WD Tv . If h0 2 W .1I 1/.0/ , then set T 00 WD T 0 . We have in both cases T 00 2 T .L/ (Lemma 13.5.1) and Q 0 .T 00 / D F h00 ˝ 1 ˚ F Id ˝ .1 C x1 /@1 ; ˆ where h00 2 S0;.0/ (the image of W .1I 1/.0/ in S0 ). The respective roots ˇ 00 and 00 are given by (cf. Equation (13.3.4) ˇ 00 .h00 ˝ 1/ D 1;

ˇ 00 .Id ˝ .1 C x1 /@1 / D 0;

00 .h00 ˝ 1/ D 0;

00 .Id ˝ .1 C x1 /@1 / D 1:

It is easy to check that for any i 2 Fp .S ˝ O.mI 1//.ˇ 00 C i 00 / D

X

Sj ˇ 00 ˝ .1 C x1 /ij F Œx2 ; : : : ; xm :

j 2Fp

The natural homomorphism given by xl 7! 0 for all l 1 X .S ˝ O.mI 1//.ˇ 00 C i 00 / ! Sj ˇ 00 j 2Fp

has solvable kernel and the torus is mapped onto a proper torus of W .1I 1/ Š S0 . Then Proposition 13.4.4 says that each ˇ 00 C i 00 with i 2 Fp is a Hamiltonian proper root of L. Consequently, T 00 has at most p 1 improper roots. Since T is a good torus we deduce that T has at most p 1 improper roots as well. The conclusion of this section is the following Theorem 13.5.6. Let L be a simple Lie algebra of absolute toral rank 2 satisfying N is simple for every good n.L/ D 0. Then L Š K.3I 1/ or L Š M.1; 1/ or A.G/ torus T and every admissible triple .T; ˛; L.0/ /. Proof. (a) Let T be a good torus, .T; ˛; L.0/ / be an admissible triple and m WD N Š S ˝ O.mI 1/ m.T; ˛; L.0/ / ¤ 0. Lemma 13.3.4(3) yields m 2. Recall that A.G/ .2/ with S Š H.2I 1/ and S0 Š sl.2/ or S0 Š W .1I 1/ (Lemma 13.3.6(1),(2)). If S0 Š sl.2/, then Theorem 13.3.10 gives L Š K.3I 1/ or L Š M.1; 1/. This is the claim in this case. Suppose S0 Š W .1I 1/ and m D 2. Define roots ˇ, as in Equation (13.3.4). Due to Lemma 13.5.5 there is a proper root 2 .L; T / n Fp . According to

13.5

Switching tori

291

Equation (13.3.6) is contained in .G 1 ; ˆ.T //. Lemma 13.3.8(5) implies that dim G 1; p 5. This gives (by the pentagon (13.1.1), the assumption n.L/ D 0 and Lemma 13.1.2) 5 dim G

1;

dim L =R D dim L =K C dim.RK/ =Ri 4;

a contradiction. (b) We now have S0 Š W .1I 1/ and m D 1. We may assume L 6Š M.1; 1/. Lemma 13.5.5 shows that Fp ˇ C Fp contains a proper root . By Equation (13.3.6), dim GN 1;i ¤ 0 for all i 2 Fp . Then one has 0 ¤ dim GN 1;i dim Li =Ri 2 dim Li =Ki , whence Li ¤ Ki for all i 2 Fp . As is proper, Lemma 13.1.2 shows that is Hamiltonian and p D 5. By the same reasoning we obtain that every proper root in F5 ˇ C F5 is Hamiltonian. (c) Suppose t0 62 W .1I 1/.0/ . In this case we may assume that D 0 and t0 D .1 C x/@ (Proposition 13.3.7). We proved in (b) that there is a proper Hamiltonian root . Lemma 13.3.8(5) establishes the assumptions of Lemma 13.4.2. Part (2) of this lemma shows that either L./ D L.0/ ./Crad L./ or L.0/ ./= rad L.0/ ./ 2 ¹.0/; sl.2/º. By the choice of , neither of these two cases can occur. This contradiction yields Q can be chosen so that t0 2 W .1I 1/.0/ . Then ˆ Q / D F h ˝ 1 ˚ F .ı ˝ 1 C Id ˝ x@/: ˆ.T (d) Let S0;.0/ denote the standard maximal subalgebra of S0 Š W .1I 1/. There is i0 2 F5 such that S0 D S0;.0/ ˚ F S0;i0 ˇN . Rescaling h if necessary we may assume that i0 D 1. For all i 2 F5 and all j 2 F5 , there exist nonzero e 1;i 2 S 1 and e0;j 2 S0 , such that S 1;i ˇN D F e 1;i and S0;j ˇN D F e0;j . For 0 a p 1, the vectors e 1;i ˝ x a ; i 2 F5 ; and e0;j ˝ x a ; j 2 F5 ; form bases of the root spaces .S

1

˝ O.1I 1//iˇ C.a

/

and

.S0 ˝ O.1I 1//jˇ Ca ;

respectively. Then p 1

.S0 ˝ O.1I 1//.ˇ/ D ˚j D0 F e0;j ˝ 1 Š S0 Š W .1I 1/: Since W .1I 1/ has no outer derivations, we conclude L.0/ .ˇ/=rad L.0/ .ˇ/ Š .S0 ˝ O.1I 1//.ˇ/ Š W .1I 1/. Suppose ˇ is proper. Then ˇ is Hamiltonian by (b), hence L.ˇ/ ¤ L.0/ .ˇ/ C rad L.ˇ/. By the preceding remark, L.0/ .ˇ/=rad L.0/ .ˇ/ 62 ¹¹0º; sl.2/º. This contradicts Lemma 13.4.2. Thus ˇ is improper. Since F5 ˇ .L; T / and jimp .L; T /j p 1 D 4 (Lemma 13.5.5), each element in ˇ C F5 must be a proper root of L, hence Hamiltonian by (b).

292

13


For i 2 F5 , let i denote the canonical homomorphism L.ˇ C i / ! L.ˇ C i /= rad L.ˇ C i / ,! H.2I 1/: Set L.ˇ C i /.l/ WD i 1 .H.2I 1/.l/ /. Given i 2 F5 and j 2 F5 choose n.i; j / 2 ¹0; : : : ; 4º such that n.i; j / ij .mod .5//. Then .S0 ˝ O.1I 1//j.ˇ Ci / D F e0;j ˝ x n.i;j / . As a consequence, .S0 ˝ O.1I 1//.ˇ C i / F e0;0 ˝ 1 C S0 ˝ O.1I 1/.1/ N 0 .ˇ Ci / is a homomorphic image is solvable (recall i ¤ 0). Since GN 0 .ˇ Ci /=A.G/ of H and hence is solvable, so is L.0/ .ˇ Ci /. Since dim Lj.ˇ Ci / =L.0/;j.ˇ Ci / D 1 for all i; j 2 F5 , Lemma 13.4.2(3) applies to each 2 ˇ C Fp showing that L.0/ .ˇ C i / D L.ˇ C i /.1/ C H

8 i 2 F5 :

Since ˇCi is proper, T stabilizes L.ˇCi /.0/ , hence each subspace L.ˇCi /.l/ (l 1). From this it is immediate that there is j0 2 F5 depending on i such that L.ˇ C i /.0/ D Lj0 .ˇ Ci / C L

j0 .ˇ Ci /

C L.ˇ C i /.1/ C H

D Lj0 .ˇ Ci / C L

j0 .ˇ Ci /

C L.0/ .ˇ C i /:

It follows that there are ui;˙ 2 L˙j0 .ˇ Ci / n¹0º such that L.ˇ C i /.0/ D F ui;C ˚ F ui; ˚ L.0/ .ˇ C i /: Since H.2I 1/.2/ .1/ is an ideal of H.2I 1/.2/ .0/ , one has Œui;˙ ; L.0/;j.ˇ Ci / L.0/ .ˇ C i / 8 j 2 F5 :

(13.5.1)

Given i; j 2 F5 let l.i; j / 2 ¹0; : : : ; 4º be such that l.i; j / C ij

.mod .5//:

(13.5.2)

Then F e 1;j ˝ x l.i;j / D .S 1 ˝ O.1I 1//j.ˇ Ci / . Clearly, the root vectors ui;˙ can be chosen so that the images of ui;C and ui; in S 1 ˝ O.1I 1/ are e

1;j0

˝ x l.i;j0 /

and

e

1; j0

˝ x l.i;

j0 /

:

(13.5.3)

It follows from Equation (13.5.1) that the two elements considered in (13.5.3) annihilate .S0 ˝ O.1I 1// .ˇ Ci / . Now e0; 1 2 S0; ˇN and e0; 1 62 S0;.0/ Š W .1I 1/.0/ . Since S 1 Š O.1I 1/=F as W .1I 1/-module, we must have annS 1 e0; 1 D S 1;ˇN . Therefore, annS

1 ˝O.1I1/

.e0;

1

˝ x/ D S

1;ˇN

˝ O.1I 1/ C S

1

˝ x4:

13.6

Observe that e0; ann.S

1

˝ x 2 .S0 ˝ O.1I 1//

1 ˝O.1I1//.ˇ

/ .e0; 1

293

Good triples

ˇ C .

˝ x/ D F e

Then 1;1

˝ x l.

1;1/

C Fe

1;r

˝ x4;

where r 2 F5 has the property that l. 1; r/ 4 .mod .5//. The subspace on the right is at most 2–dimensional, hence must coincide with F e 1;j0 ˝ x l. 1;j0 / C F e 1; j0 ˝ x l. 1; j0 / , the span of the images of u 1;˙ . Thus ¹1; rº D ¹˙j0 º forcing r D 1. By Equation (13.5.2), 4 l. 1; 1/ C 1. Thus D 3. Next e0; 1 ˝ x 2 2 .S0 ˝ O.1I 1// ˇ C2 and annS

1 ˝O.1I1/

.e0;

1

˝ x2/ D S

1;ˇN

˝ O.1I 1/ C S

1

˝ x3 C S

1

˝ x4:

As a consequence, ann.S

1 ˝O.1I1//.ˇ

2 / .e0; 1

D Fe

˝ x2/ 1;1

˝ x l.

2;1/

C Fe

1;r1

˝ x3 C F e

1;r2

˝ x4;

where r1 ; r2 2 F5 satisfy l. 2; r1 / 3 .mod .5// and l. 2; r2 / 4 .mod .5//. The subspace on the right should contain e 1;˙j0 ˝ x l. 2;˙j0 / (we set i D 2 in Equation (13.5.3)). But then ¹˙j0 º ¹1; r1 ; r2 º, hence one of the following must hold: (i) r1 D 1; (ii) r2 D 1; (iii) r1 D r2 : If (i) holds, then 3 l. 2; 1/ C 2 (by Equation (13.5.2)). If (ii) holds, then 4 C 2, in a similar fashion. If (iii) holds, then 3 2r1 and 4 C 2r1 . As D 3, each of the three cases leads to a contradiction, completing the proof.

13.6

Good triples

Next we treat the case n.L/ ¤ 0. Here we have to deal with a choice of distinguished admissible triples. Definition 13.6.1. Assume n.L/ ¤ 0. A triple .T; ˛; L.0/ / is called a good triple if (1) T 2 T .L/, (2) ˛ is a nonzero T -root for which n.˛/ D n.L/, (3) dim M .˛/ .T / is maximal with respect to all pairs .T 0 ; ˛ 0 / satisfying (1), (2), (4) .T; ˛; L.0/ / is an admissible triple. The question if good triple at all exist will be taken up later. Lemma 13.6.2. Let .T; ˛; L.0/ / be an admissible triple. Suppose n.˛/ ¤ 0 and m.T; ˛; L.0/ / ¤ 0. Then S0 .T; ˛; L.0/ / Š W .1I 1/ and m.T; ˛; L.0/ / D 1.

294

13


Proof. (a) Due to Theorem 13.3.10, S0 .T; ˛; L.0/ / 6Š sl.2/, whence S0 .T; ˛; L.0/ / Š W .1I 1/. Recall that m WD m.T; ˛; L.0/ / 2. To derive a contradiction we assume m D 2. Then dim L .dim S/p 2 D .p 2 2/p 2 > p 3 , whence L 6Š M.1; 1/. Q Suppose .2 ı ˆ/.T / W .2I 1/.0/ . According to Lemma 13.5.1(1) there is u 2 Q Ki ˛ L.0/ for some i ¤ 0 such that .2 ı ˆ/.u/ 62 W .2I 1/.0/ . As F is infinite Q there is r 2 F such that .2 ı ˆ/.ru C q.ru/// 62 W .mI 1/.0/ . Adjusting u we may assume r D 1. Choose 2 F such that .u; / 2 X and switch by E.u;/ . Q u / 62 W .2I 1/.0/ Then tu D t i˛.t/.u C q.u// for all t 2 T and hence .2 ı ˆ/.t Q if ˛.t / ¤ 0. Consequently, .2 ı ˆ/.T / 6 W .2I 1/ . Corollary 13.5.3 yields that u .0/ .Tu ; ˛.u;/ ; L.0/ / is an admissible triple and m.Tu ; ˛.u;/ ; L.0/ / ¤ 0 holds. Thus in Q what follows we assume that .2 ı ˆ/.T / 6 W .2I 1/.0/ . (b) Proposition 13.3.7 tells us that we may assume Q / D F h ˝ 1 ˚ F Id ˝ .1 C x1 /@1 : ˆ.T Let ˇ, as in Equation (13.3.4). With this normalization one obtains S ˝O.2I 1/.ˇ/ D S ˝ F Œx2 . Set X X ŒLjˇ ; L jˇ D L.ˇ/.1/ ¤ ¹0º: J WD Ljˇ C j 2Fp

j 2Fp

Note that gr J may be regarded as a graded subalgebra of gr L D G. By construction, .gr J /=M.G/ \ .gr J / Š S ˝ F Œx2 . In addition, rad.gr J / is a graded ideal which contains M.G/ \ .gr J /. Since gr.rad J / is a solvable ideal of gr J , one has gr.rad J / rad.gr J /. Therefore there exists the following commutative diagram, where all involved mappings are surjective algebra homomorphisms homogeneous of degree 0 gr J

'2

!

S ˝ F Œx2

# 2

# 3 2

.gr J /= gr.rad J / ! .gr J /= rad.gr J / Š S Š H.2I 1/.2/ : Next observe that rad J is T -invariant (Theorem 11.2.3(2)). Choose T -invariant vector spaces Vi0 rad J such that .rad J / \ L.i/ D Vi0 ˚ .rad J / \ L.i C1/

8i

2

and extend these to T -invariant vector spaces Vi Vi0 satisfying J \ L.i/ D Vi ˚ .J \ L.iC1/ / 8i

2:

These vector spaces give compatible T -invariant gradings of J and rad J as vector spaces (not necessarily as algebras) J D ˚i

2 Vi ;

rad J D ˚i

0 2 Vi :

13.6

295

Good triples

Therefore there is a natural T -invariant grading preserving vector space isomorphism of degree 0 '1 W J D ˚i 2 Vi ! gr J D ˚i 2 Vi C L.iC1/ =L.i C1/ satisfying '1 .Vi0 / D gri .rad J /:

'1 .Vi / D gri J;

Then we can extend the above diagram by T -invariant linear mappings (where all mappings are surjective) J D ˚i

'1

!

2 Vi

# 1 J = rad J Š ˚i

'2

! S ˝ F Œx2

gr J

# 30

# 2 0 2 .Vi =Vi /

0 2

1

! .gr J /= gr.rad J / ! H.2I 1/.2/ :

Since J = rad J is ¹0º or simple of absolute toral rank 1, J = rad J 2 ¹¹0º; sl.2/; W .1I 1/; H.2I 1/.2/ º: We obtain by dimension reasons that J = rad J Š H.2I 1/.2/ , i.e., ˇ 2 .L; T / is Hamiltonian. As 20 ı 1 is bijective, therefore ker 30 D .'2 ı '1 /.ker 1 / D ˚ .'2 ı '1 /.Vi0 / and from this we conclude for i D 1 dim V 0 1;jˇ dim .ker 30 / \ .S 1 ˝ F Œx2 /jˇ D dim S

1;j ˇN

˝

pX1

F x2l p

1 8 j 2 Fp :

lD1

(c) Since V we conclude

0

1;jˇ

.rad J / \ Ljˇ Kjˇ and V 0 1;jˇ \ Rjˇ V dim Kjˇ =Rjˇ dim V 0 1;jˇ p

1

\ L.0/ D ¹0º,

1:

On the other hand, dim Kjˇ =Rjˇ D dim Kjˇ =.RK/jˇ C dim.RK/jˇ =Rjˇ njˇ C dim L

jˇ =K jˇ

njˇ C 3

by definition of njˇ and Lemmas 13.1.1 and 13.1.2. This gives n.ˇ/ D

p X1

njˇ .p

1/.p

4/ ¤ 0:

j D1

(d) Note that '1 induces by restriction a linear mapping X X X '1 W J \ L.0/ D Vi ! Vi = Vi D J \ L.0/ =J \ L.1/ ! gr0 J; i0

i 0

i 1

296

13


which in fact is an algebra homomorphism. We mentioned that '2 is an algebra homomorphism as well. Therefore '2 ı '1 W J \ L.0/ ! S0 ˝ F Œx2 is a surjective algebra homomorphism. We obtained in (b) that .'2 ı '1 / .rad J / \ L.0/ D .'2 ı '1 /.V00 / D .'2 ı '1 /.V00 / D

ker 30 jS0 ˝F Œx2

D S0 ˝

pX1

F x2l :

lD1

From this result it is clear that 3 ..rad J / \ L.0/ / ¤ ¹0º. As a consequence, one has

3 .rad L.ˇ/.1/ / ¤ ¹0º. Lemma 13.3.8(5) shows that X dim L Cjˇ < 4p 2 C .p 1/p 2 < 2p 3 : j 2Fp

These results are in contradiction to Corollary 13.4.3 for D ˇ.

Lemma 13.6.3. Let .T; ˛; L.0/ / be an admissible triple. Suppose n.˛/ ¤ 0 and m.T; ˛; L.0/ / ¤ 0. The following holds: Q (1) dim.2 ı ˆ/.L .0/ / D 2; (2) n.˛/ D 2; (3) dim G

2

1.

Proof. One has m.T; ˛; L.0/ / D 1 and S0 .T; ˛; L.0/ / Š W .1I 1/ by Lemma 13.6.2. Q Q (1) SetP D WD .2 ı ˆ/.L .0/ / W .1I 1/. As .2 ı ˆ/.T / D F t0 ¤ ¹0º and Q .2 ı ˆ/. i 2Fp Ki˛ / ¤ ¹0º by Lemma 13.5.1(1), one has dim D 2. Suppose dim D 3. As D is a transitive subalgebra of W .1I 1/, Proposition 7.6.8 implies that Q D Š sl.2/ or D Š W .1I 1/. If D Š sl.2/ or .2 ı ˆ/.T / is an improper torus of Q Q D Š W .1I 1/, then .2 ı ˆ/.K.˛// D ¹0º, while in case that .2 ı ˆ/.T / is a proper Q torus of D Š W .1I 1/, then .2 ı ˆ/.K.˛// W .1I 1/.0/ . As this contradicts Lemma Q Q 13.5.1(1), dim D D 2, i.e., D D F t0 ˚ F .2 ı ˆ/.u/ with F t0 D .2 ı ˆ/.T / and u 2 Ki ˛ for some i 2 Fp as in Lemma 13.5.1. (2) Recall that Q .0/ / .Der0 S/ ˝ O.1I 1/ D .S0 C F ı/ ˝ O.1I 1/: .ker 2 / \ ˆ.L The result of (1) then implies Q .0/ .˛// .F h C F ı/ ˝ O.1I 1/ C ˆ.T Q / C F ˆ.u/: Q ˆ.L This shows that L.0/ .˛/ is solvable. It also follows immediately from this description that nk˛ D 0 if k ¤ ˙i and n˙i˛ D 1.

13.6

297

Good triples

(3) Suppose G 2 ¤ ¹0º. Lemma 13.3.8(4),(3) tells us first that Fp D Fp ˛ and then that G 2 G.˛/. Since L.0/ .˛/ is solvable, every root vector x 2 G0 .˛/.1/ has the property ˛.x Œp / D 0. Therefore G0 .˛/.1/ acts nilpotently on G.˛/ by the Engel– Jacobson theorem and G 2 contains a common eigenvector v ¤ 0 for G0 .˛/. Let v 2 [i 2Fp Li ˛ be an inverse image of v. Using the fact that Fp D Fp ˛ and applying Equation (13.3.6) we get ŒL.0/ .˛/; v F v C L.˛/ \ L. 1/ D F v C L.0/ .˛/ and ŒF v C L.0/ ; F v C L.

1/

Œv; L. 1/ C ŒL.0/ ; v C L. 1/ X Œv; L C Œv; L.0/ .˛/ C L.

1/

62Fp ˛

X

L C F v C L.0/ .˛/ C L.

1/

F v C L.

1/ :

62Fp ˛

Consequently, F vCL.0/ NorL .F vCL. 1/ /. If F vCL. 1/ ¤ L then NorL .F vC L. 1/ / ¤ L. The maximality of L.0/ now enforces v 2 L.0/ , a contradiction. Consequently, L D F v C L. 1/ and dim G 2 D 1. Lemma 13.6.4. Let .T; ˛; L.0/ / be an admissible triple. Suppose n.˛/ ¤ 0 and m.T; ˛; L.0/ / ¤ 0. Let T 0 L.0/ be a 2-dimensional torus and ˛ 0 2 .L; T 0 / 0 0 N D 0. The following holds: be a nonzero root satisfying ˛ ˆ.T / \ A.G/ (1) L.0/ .˛ 0 / is a solvable subalgebra of L.˛ 0 / of codimension 1 and ˛ 0 is solvable, classical, or proper Witt. Q (2) If there exists u 2 [i 2Fp Ki˛0 such that .2 ı ˆ/.u/ 62 W .1I 1/.0/ , then one has 0 0 0 Œu; L.0/ .˛ / K.˛ / and n.˛ / ¤ 0. Proof. (1) Lemma 13.3.8(2) shows that N2 .L; T 0 / ¤ ; and then Corollary 12.5.10 implies that CL .T 0 / acts trigonalizably on L. Due to Proposition 13.3.7 there is a realization Q 0 .T 0 / D F h0 ˝ 1 ˚ F . 0 ı ˝ 1 C Id ˝ t00 /: ˆ The present assumption on ˛ 0 gives ˛ 0 .h0 ˝ 1/ D 0, whence S0 ˝ O.1I 1/ .˛ 0 / D CS0 .h0 / ˝ O.1I 1/ D F h0 ˝ O.1I 1/: 0 / 2. 0 / is abelian. Lemma 13.6.3(1) yields dim G N 0 .˛ 0 /=A0 .G/.˛ N N Then A0 .G/.˛ 0 0 0 Consequently, L.0/ .˛ / is a T -invariant solvable subalgebra of L.˛ /. The assumption on ˛ 0 in combination with Equation (13.3.4) also yields Fp ˛ 0 D Fp 0 . Then Equation (13.3.6) shows L. 1/ .˛ 0 / D L.0/ .˛ 0 /, while due to Lemmas 13.3.8(1) and 13.6.3(3) we have dim L.˛ 0 /=L. 1/ .˛ 0 / D dim G 2 .˛ 0 / 1. Then L.0/ .˛ 0 / has codimension 1 in L.˛ 0 /. This forces that ˛ 0 is solvable, classical, or proper Witt.

298

13


(2) We refer to Remark 13.1.4. Set K 0 .˛ 0 / WD H˛0 C show ŒK 0 .˛ 0 /; L.0/ .˛ 0 / K 0 .˛ 0 /:

P

i 2Fp

Ki˛0 . We intend to ./

If ˛ 0 is solvable, then trivially () holds. If L.˛ 0 / is classical, then K 0 .˛ 0 / D rad L.˛ 0 /. Consequently, ŒK 0 .˛ 0 /; L.˛ 0 / K 0 .˛ 0 /, in particular ./ holds. If L.˛ 0 / is proper Witt, then L.0/ .˛ 0 / is the canonical subalgebra of codimension 1 and K 0 .˛ 0 / is the inverse image of W .1I 1/.2/ . Again we obtain ŒK 0 .˛ 0 /; L.0/ .˛ 0 / K 0 .˛ 0 /, i.e., ./ holds in all cases. 0 / D F x@. We may assume that u 2 K 0 and . ı ˆ/.u/ Q Q Suppose .2 ı ˆ/.T D 2 ˛ 0 0 0 0 2 Q .w/ D h ˝ x . One immedi@. Choose a T -root vector w 2 L.0/ .˛ / for which ˆ ately observes that Q ˆ.Œu; Œu; w/ 2 2h ˝ 1 C S0 ˝ O.1I 1/.1/ : Then u 2 K ˛0 by definition, Œu; w 2 K˛0 by ./ and ˇ 0 .Œu; Œu; w/ ¤ 0, whence u 62 .RK/ ˛0 . Consequently, n.˛ 0 / ¤ 0. 0 / D F .1Cx/@ and . ı ˆ/.u/ Q Q Otherwise we may assume .2 ı ˆ/.T D .1Cx/j @ 2 0 0 0 Q for some j ¤ 1. Choose a T -root vector w 2 L.0/ .˛ / for which ˆ .w/ D h0 ˝ .1 C x/2 and argue as before. Lemma 13.6.5. Assume L 6Š M.1; 1/. Suppose n.L/ ¤ 0. Let .T; ˛; L.0/ / be a good triple satisfying m.T; ˛; L.0/ / ¤ 0. Then ˆ.T / \ ker ˛ is an improper torus of N Moreover, there is u 2 [i 2F Ki˛ and an elementary switching E.u;/ such A0 .G/. p that (1) .Tu ; ˛.u;/ ; L.0/ / is a good triple, N and the other does not, (2) one of ˆ.T / or ˆ.Tu / does stabilize the radical of A.G/ (3) T \ ker ˛ D Tu \ ker ˛.u;/ . Proof. One has m.T; ˛; L.0/ / D 1 and S0 .T; ˛; L.0/ / Š W .1I 1/ by Lemma 13.6.2. N Then there is (a) Consider first the case that ˆ.T / stabilizes the radical of A.G/. a realization Q / D F h ˝ 1 ˚ F .ı ˝ 1 C Id ˝ x@/: ˆ.T The definition of a good triple yields n.˛/ D n.L/ ¤ 0. Lemma 13.6.3(2) then Q gives n.L/ D 2. Due to Lemma 13.5.1(1) there is u 2 Ki˛ such that .2 ı ˆ/.u/ 62 Q D d ˝ xp 1 W .1I 1/.0/ for some i 2 Fp . Adjusting u we may assume that ˆ.u/ Œp / D . ı ˆ/.u/ p D Q Q Id ˝ @ where Œh; d D 0. Note that .2 ı ˆ/.u @p D 0. 2 Choose 2 F such that .u; / 2 X and apply the elementary switching E.u;/ . We obtain Q u / D .2 ı ˆ/.t/ Q Q .2 ı ˆ/.t i˛.t/.2 ı ˆ/.u/ 8 t 2 T:

13.6

299

Good triples

Q In particular, for t 2 T satisfying ˆ.t/ D ı ˝ 1 C Id ˝ x@ one has i˛.t/ D 1 and Q .2 ı ˆ/.tu / D x@ @ 62 W .1I 1/.0/ . This shows that ˆ.Tu / does not stabilize the N radical of A.G/. Corollary 13.5.3 shows that .Tu ; ˛.u;/ ; L.0/ / is an admissible triple. Proposition 13.5.2 gives E.u;/ .M .˛/ .T // D M .˛.u;/ / .Tu /. In particular, u D E.u;/ .u/ 2 Ki ˛.u;/ . Note that Fp ˛ D Fp as n.˛/ ¤ 0 (Lemma 13.3.8) and ˛.u;/ ..h ˝ 1/u / D ˛.h ˝ 1/

˛..u; //i˛.h ˝ 1/ D 0:

Lemma 13.6.4(2) yields n.˛.u;/ / ¤ 0. Then 2 n.˛.u;/ / n.L/ D 2, hence n.˛.u;/ / D n.L/. This shows that .Tu ; ˛.u;/ ; L.0/ / is a good triple. N Then there is a realization (b) Suppose T does not stabilize the radical of A.G/. Q / D F h ˝ 1 ˚ F Id ˝ .1 C x/@: ˆ.T Proceed word-by-word (substituting x by 1 C x if necessary) as in the former case to N obtain that .Tu ; ˛.u;/ ; L.0/ / is a good triple and T stabilizes the radical of A.G/. (c) Observe that ˛.u;/ .tu / D ˛.t/

˛..u; //i˛.t/ D ˛.t/

8 t 2 T:

From this it is obvious that Tu \ ker ˛.u;/ D .T \ ker ˛/u . Since tu D t i˛.t/.u C q.u// for all t 2 T , we conclude .T \ ker ˛/u D T \ ker ˛. N 0 it by (c) (d) In order to prove that ˆ.T / \ ker ˛ is an improper torus of A.G/ N suffices to take T which does not stabilize the radical of A.G/. Then Q / D F h ˝ 1 ˚ F Id ˝ .1 C x/@: ˆ.T Q / \ ker ˛ D Take ˇ, as in Equation (13.3.4) and observe that Fp D Fp ˛. Then ˆ.T F h ˝ 1. N Since ˇ.Id ˝ .1 C x/@/ D 0, A.G/.ˇ/ Š S ˝ F Š H.2I 1/.2/ , whence ˇ 2 .L; T / is a Hamiltonian root (Proposition 13.4.4). Lemma 13.3.8(5) shows that P 3 j 2Fp dim L Cjˇ < .p C 3/p < p . Since L.0/ .ˇ/= rad L.0/ .ˇ/ Š S0 Š W .1I 1/, ˇ is not a proper root of L (Lemma 13.4.2(2)), and then F h is an improper torus of S0 due to Proposition 13.4.4. This is the claim. Lemma 13.6.6. Assume L 6Š M.1; 1/. Suppose n.L/ ¤ 0. Let .T; ˛; L.0/ / be a good triple satisfying m.T; ˛; L.0/ / ¤ 0. Then M .˛/ .T / D L.1/ C K.˛/: Proof. One has m.T; ˛; L.0/ / D 1 and S0 .T; ˛; L.0/ / Š W .1I 1/ by Lemma 13.6.2. N then we use the original triple (a) If ˆ.T / does not stabilize the radical of A.G/, in the following. Otherwise Lemma 13.6.5 shows that we may switch by some E.u;/

300

13


to obtain a good triple .Tu ; ˛.u;/ ; L.0/ / (and let the filtration of L unchanged) for N Since u 2 L.0/ , E.u;/ .L.1/ / D which ˆ.Tu / does not stabilize the radical of A.G/. L.1/ . Since u 2 K.˛/ L.0/ , Proposition 13.5.2 shows that E.u;/ .M .˛/ .T // D M .˛.u;/ / .Tu /, E.u;/ .K.˛; T // D K.˛.u;/ ; Tu /. So we may assume right from the N Choose a realization beginning that ˆ.T / does not stabilize the radical of A.G/. Q / D F h ˝ 1 ˚ F Id ˝ .1 C x/@. Take ˇ, as in Equation (13.3.4) and observe ˆ.T that Fp D Fp ˛ (Lemma 13.3.8). Lemma 13.6.5 shows that F h is an improper torus of W 1/. We use y as the P.1I p 1 indeterminate to describe W .1I 1/, i.e., we set W .1I 1/ D i D0 F y i @y . Since F h is an improper torus of W .1I 1/, there is a normalization (see Theorem 11.2.5) h D .1 C y/@y . As a consequence there is a simultaneous realization Q / D F .1 C y/@y ˝ 1 ˚ F Id ˝ .1 C x/@ : ˆ.T (b) Recall that Fp ˛ D Fp by Lemma 13.3.8(4). Now suppose that there is D .˛/ .˛/ j˛ Ckˇ with k 2 Fp for which L.0/; D M .T /CL.1/; . Then there is w 2 M such that Q ˆ.w/ D k 1 .1 C y/kC1 @y ˝ .1 C x/j : Choose suitable 2 F and switch T by E.w;/ . Lemma 13.5.1(2) gives Tw 2 T .L/. .˛/ Q Every v 2 [i 2Fp Mi has the property that ˆ.v/ is contained in S0 ˝ O.1I 1/, Q hence annihilates ˆ.L.0/ .˛//=S0 ˝ O.1I 1/. The latter space carries a root l˛ ¤ 0 Q (as .2 ı ˆ/.u/ ¤ 0). Therefore ˛.v Œp / D 0 holds. As v 2 Li for some Œp i 2 Fp , .v / D 0. Then every such v is Œp-nilpotent. Therefore Proposition 13.5.2 applies. As a consequence, E.w;/ .K.˛; T // K.˛.w;/ ; Tw /. In parQ Q ticular, E.w;/ .u/ 2 K.˛.w;/ ; Tw /. As .2 ı ˆ/.E .w;/ .u// D .2 ı ˆ/.u/ … W .1I 1/.0/ , Lemma 13.6.4(2) yields n.˛.w;/ / ¤ 0. Proposition 13.5.2 implies that E.w;/ .M .˛/ .T // M .˛.w;/ / .Tw /. Recall that n.L/ D 2 (Lemma 13.6.3(2)). Then 2 n.˛.w;/ / n.L/ D 2. Hence n.˛.w;/ / D n.L/. Since .T; ˛; L.0/ / is a good triple, Definition 13.6.1 now gives dim M .˛.w;/ / .Tw / dim M .˛/ .T /. Since E.w;/ is an invertible transformation, we get the reverse inequality from a previous inclusion, whence dim M .˛/ .T / D dim M .˛.w;/ / .Tw /. Consequently, M .˛.w;/ / .Tw / D E.w;/ .M .˛/ .T // L.0/ ; whence .Tw ; ˛.w;/ ; L.0/ / is a good triple. (c) Next we determine Tw in more detail. Choose t1 2 T for which Q 1 / D .1 C y/@y ˝ 1: ˆ.t Then Q .w;/ .t1 // D ˆ.t Q 1 C Œw; t1 / D .1 C y/@y ˝ 1 ˆ.E

.1 C y/kC1 @y ˝ .1 C x/j

ky@y ˝ 1 .mod W .1I 1/.1/ ˝ O.1I 1/ C W .1I 1/ ˝ O.1I 1/.1/ /:

13.6

301

Good triples

The semisimple part E.w;/ .t1 /sem is contained in Tw and also satisfies Q .w;/ .t1 /sem / D ˆ.E Q .w;/ .t1 //ps ˆ.E

ky@y ˝ 1

.s 0/

modulo W .1I 1/.1/ ˝ O.1I 1/ C W .1I 1/ ˝ O.1I 1/.1/ . Then ˇ.w;/ is Hamiltonian proper (Proposition 13.4.4). An application of Lemma 13.6.5(3) for the good triple .Tw ; ˛.w;/ ; L.0/ / shows that this is not possible. (d) This contradiction proves L.0/; ¤ M˛ .T / C L.1/; for all D j˛ C kˇ with .˛/ k 2 Fp . Since dim S0 ˝ O.1I 1/ 1, this gives M .T / L.1/; . By Lemma Q / \ ker ˛ and 13.3.8(1),(3), L./ L. 1/ . Observing the fact that F h ˝ 1 D ˆ.T looking at the associated graded algebra one easily obtains L.1/; M .˛/ .T /. As a .˛/ consequence, M .T / D L.1/; . This is the claim for all 62 Fp ˛. Consider the case G 2 D ¹0º. Then L.˛/ D L.0/ .˛/ is solvable (Equation (13.3.6) and Lemma 13.6.4(1)). Then L.˛/ D K.˛/ and therefore M .˛/ .T / D L.1/ C K.˛/ holds. Consider the case G 2 ¤ ¹0º. According to Lemmas 13.6.3(3) and 13.3.8(3) there is d 2 Lk˛ for some k 2 Fp such that G 2 D F .gr 2 d /. Since G 2 D M.G/, one has Œd; L.i / L.i 1/ for all i 1. Note that Œd; L.0/ 6 L.0/ because L.0/ is maximal. As d 62 L. 1/ , it induces a homogeneous outer derivation gr d of G of degree 1. As Œgr d; G 2 G 3 D ¹0º, d induces even more an outer derivation dN 2 Der 1 GN Š .Der 1 S/ ˝ O.1I 1/. The grading of S is given by generators u1 ; u2 .p/ and degrees deg.u1 / D 0, deg.u2 / D 1. Then .Der 1 S/=S 1 Š FDH .u1 / is 1-dimensional. As a result, we may adjust dN by an inner derivation such that .dN / D N q 1 ˝ f for some q 1 2 Der 1 S. Next we observe that GN 1 D ŒGN 1 ; GN 2 A.G/. Therefore Œ .dN /; .GN 1 / D Œq

1

˝ f; S1 ˝ O.1I 1/ S0 ˝ O.1I 1/:

As ˛.h˝1/ D 0 and Œd; L.2/ L.1/ , the above implies Œd; L.1/; k˛ H˛ . Finally, N Consequently, ŒL. 1/;i˛ ; L.1/; i˛ H˛ for all i 2 Fp . As a ŒGN 1 ; GN 1 A.G/. result, L.1/ .˛/ K.˛/, whence M .˛/ .T / D L.1/ C K.˛/. The conclusion of this section is Theorem 13.6.7. Let L be a simple Lie algebra of absolute toral rank 2 satisfying N is simple for every good triple .T; ˛; L.0/ /. n.L/ ¤ 0. Then L Š M.1; 1/ or A.G/ Proof. To derive a contradiction we assume that L 6Š M.1; 1/ and take a good triple N is not simple. Then m.T; ˛; L.0/ / ¤ 0. As n.˛/ D .T; ˛; L.0/ / for which A.G/ n.L/ ¤ 0, Lemma 13.6.2 gives S0 .T; ˛; L.0/ / Š W .1I 1/ and m.T; ˛; L.0/ / D 1, and Lemma 13.6.3(2) states n.L/ D 2. Note that Fp D Fp ˛ (Lemma 13.3.8(4)). (a) According to Lemma 13.6.5 there is a good triple (we take .T; ˛; L.0/ / as such) N Put for which ˆ.T / does stabilize the radical of A.G/. Q / D .F h ˝ 1/ ˚ F .ı ˝ 1 C Id ˝ x@/: ˆ.T

302

13


Due to Lemma 13.6.5 F h is an improper torus of S0 , and then by Theorem 11.4.1(5b) F h is an improper torus of S . Theorem 11.4.2 (applied to the 2-dimensional torus F hCF ı) shows that we may adjust h by a scalar and normalize h D DH ..1Cu1 /u2 /. Let D i ˛ C jˇ, i 2 Fp , j 2 Fp . Then L./ L. 1/ . It follows from Lemma .˛/

13.6.6 that L.2/; R M D L.1/; . Thus to determine R we are to deal with S1 ˝ O.1I 1/. Observe that S1 ˝ O.1I 1/ D S1;j ˇN ˝ x s ; where C s i mod.p/: If s ¤ 0, then ŒS1;j ˇN ˝ x s ; S 1 ˝ O.1I 1/ S0 ˝ O.1I 1/.1/ , and the algebra on the right acts nilpotently on S ˝ O.1I 1/. Consequently, L.1/; D R

whenever i ¤ :

Now suppose D ˛ C jˇ. Recall that F h D DH ..1 C u1 /u2 /. Therefore .S1 ˝ F / contains DH ..1Cu1 /2 j u22 /˝1 and .S 1 ˝F / contains DH ..1Cu1 /j /˝1. Then ˇ.ŒL.1/; ; L / ¤ 0. It follows that L.1/; © R

whenever i D :

Recall that dim L.l/; =L.lC1/; D 1 for l 2 ¹ 1; 0; 1º. As a consequence, for D i ˛ C ˇ, i 2 Fp , one has ´ 2 if ki ¤ ; dim Lk =Rk D 3 if ki D : Next consider the root " WD ˛Cˇ. Observe that " cannot be Hamiltonian improper, because in that case one would have dim Lk" =Rk" dim Lk" =Kk" D 3 for all k 2 Fp . Suppose " is Hamiltonian proper. Lemma 13.4.1 shows that Kk" D Rk" , whence dim Lk" =Rk" D dim Lk" =Kk" 2 for all k 2 Fp . So this case cannot happen j either. Note that DH ..1 C u1 /2j 1 u2 / ˝ 1 2 S ˝ O.1I 1/ .j 1/˛ .j 1/ˇ . Then S ˝ O.1I 1/ ."/ contains p X1

FDH ..1 C u1 /2j

1 j u2 /

˝ 1 Š W .1I 1/:

j D0

Consequently, " is a Witt root. Since one has gr.rad L."// rad.gr L."// and gr L."/= gr.rad L."// Š W .1I 1/ Š L."/= rad L."/, a dimension argument yields gr.rad L."// D rad.gr L."//. But then the above description gives that L.0/ ."/ C rad L."/ is a solvable subalgebra of codimension 1 in L."/, and therefore " is proper Witt. We also derive X X dim Lk" =Kk" C n."/ D 4 C n."/; 2.p 1/ C 1 D dim Lk" =Rk" 2 k2Fp

k2Fp

whence n."/ p > 2 D n.L/. This contradiction proves the theorem.

13.7

13.7

On the existence of good tori and good triples

303


In the last sections we have established the fact that, if n.L/ D 0, T is a good torus, .T; ˛; L.0/ / is an admissible triple, L 6Š K.3I 1/ and L 6Š M.1; 1/, or n.L/ ¤ 0, .T; ˛; L.0/ / is a good triple and L 6Š M.1; 1/, N D S is a simple Lie algebra for every admissible choice of L. 1/ . In this then A.G/ section we will answer the question if such tori and triples exist, and if so, will derive some properties of S . Lemma 13.7.1. (1) If L is not classical and L 6Š H.2I 1I ˆ.//.1/ , then there exists a 2-dimensional non-rigid torus T 0 in LŒp . (2) If L is not classical, then every 2-dimensional torus in LŒp having a proper non-solvable root is non-rigid. (3) If n.L/ ¤ 0, then there exists a 2-dimensional non-rigid torus T 0 in LŒp for which n.T 0 / D n.L/ holds. Proof. (1) Under the present assumptions there exists a non-rigid 2-dimensional torus (Corollary 12.4.7). (2) Let T 0 be a 2-dimensional torus having a proper non-solvable root. Assume that T 0 is rigid. We are going to show that L is classical. Applying Theorem 12.4.6 one gets the following. (i) In Case (1) of that theorem all nonzero roots are Hamiltonian improper. (ii) In Case (2) of that theorem all nonzero roots are improper Witt. (iii) Look at Case (3) of Theorem 12.4.6. In Case (3a) L is classical. In Case (3b) L Š H.2I 1I ˆ.//.1/ . Then all 1-sections are abelian (cf. Theorem 10.3.2(3)). In Case (3c) of Theorem 12.4.6 all non-solvable roots are improper Witt. (3) By definition (see Equation (13.2.1)) there is a 2-dimensional torus T in LŒp and a nonzero T -root ˛ such that CL .T / acts trigonalizably on L and n.˛/ D n.T / D n.L/ holds. We have to show that such a torus exists which in addition is non-rigid. So assume that T is rigid. Applying Theorem 12.4.6 one gets the following. (i) In Case (1) of that theorem H WD CL .T / does not act trigonalizably on L. (ii) In Case (2) of that theorem dim L D 2 for ¤ 0 and dim H 2 hold, and all roots ¤ 0 are improper Witt. Remark 13.1.4 states K rad L./. Therefore Theorem 12.4.6 yields ŒK ; K .rad L.//.1/ D ¹0º. This gives K D .RK/ and n D 0. As this is true for all nonzero roots one obtains the contradiction n.L/ D n.T / D 0.

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13


(iii) Next look at Case (3) of Theorem 12.4.6. In Case (3a) L is classical. Lemma 10.6.2 in combination with [Sel 67, Lemma II.3.2] shows that L.˛/ Š sl.2/ ˚ H˛ , whence K˛ D ¹0º. Then n.˛/ D 0 follows and this case does not occur. In Case (3b) L Š H.2I 1I ˆ.//.1/ . Then all 1-sections are abelian (cf. Theorem 10.3.2(3)), hence L˛ D R˛ . This gives the contradiction n.˛/ D 0. the present case one has dim L D 1 for all ¤ 0. Then P Consider Case (3c). In 2 for every ˇ independent of ˛, and Theorem 13.2.2(4) now dim L < p ˇ Ci˛ i 2Fp proves n.˛/ 2. Note that n.L/ is even by definition. Therefore n.L/ is minimal possible, and it suffices to construct a non-rigid torus T 0 at all, for which n.T 0 / ¤ 0. Suppose that ˛ is solvable. Lemma 12.4.3(3) gives a clear description of L.˛/. Namely, LŒp .˛/ D CLŒp .T / C A ˚ F x , where A is an abelian ideal of LŒp .˛/ and x 2 Lk˛ for some k ¤ 0. Since n.˛/ ¤ 0, we have x ¤ 0 and there is y 2 A k˛ such that Œx; y is not Œp-nilpotent. Write T D F t ˚ F t˛ where t; t˛ are toral, k˛.t / D 1 and F t˛ D T \ ker ˛, adjust y such that Œx; y D t˛ . Set y Œp DW p t˛ for some 2 F . Set T 0 WD T x D F t˛ ˚ F .t C x C q.x// and y 0 WD y t˛ . Then y 0 is a T 0 -eigenvector and satisfies y 0Œp D y Œp p t˛ D 0. It also satisfies .ad y 0 /2 .L.˛// D 0. Then T 0 is not rigid. Clearly, Œx; y 0 D Œx; y D t˛ is not Œp-nilpotent. Then n.T 0 / ¤ 0. If ˛ is not solvable then ˛ is improper Witt. We have Ki˛ D ¹0º for all i 2 Fp , contradicting the assumption n.˛/ ¤ 0. Proposition 13.7.2. Suppose one of the following holds: (a) n.L/ ¤ 0; (b) L is not of classical or Cartan type. Then every 2-dimensional non-rigid torus T , which satisfies n.T / D n.L/ in case (a), is contained in T .L/. In particular, good triples exist if n.L/ ¤ 0, and admissible triples .T; ˛; L.0/ / with a good torus T exist if n.L/ D 0. Proof. (A) Let T LŒp be a 2-dimensional non-rigid torus. Suppose n.T / D n.L/ ¤ 0. By definition there is a nonzero T -root ˛ for which n.˛/ D n.T / D n.L/ ¤ 0. Theorem 13.2.2(1) proves that the p-envelope of CL .T / in LŒp contains T . Then T 2 T .L/. Suppose L is not of classical or Cartan type. Set H WD CL .T /. Let T0 denote the uniquely determined maximal torus of the p-envelope HŒp of H in LŒp . It is clear by definition, that T0 T and TR.H; L/ D dim T0 S dim T D 2. If dim T0 1, then there is ˛ 2 T n ¹0º vanishing on H . Then H [ i 2Fp Li˛ is a nil Lie set, and the Engel–Jacobson theorem shows that L.˛/ is nilpotent. Proposition 11.2.6 implies that L.˛/ is a CSA of L of toral rank 1. Theorem 9.2.11 shows that L is isomorphic to some sl.2/, W .1I n/, H.2I nI ˆ/.2/ , which contradicts our assumption. Therefore dim T0 D 2. Then T0 D T and H is a CSA of L for which T HŒp . Consequently, T 2 T .L/.

13.7


305

(B) Suppose n.L/ ¤ 0. Due to Lemma 13.7.1(3) there is a 2-dimensional nonrigid torus T LŒp for which n.T / D n.L/ holds, and by definition there is a nonzero T -root ˛ for which n.˛/ D n.T /. We mentioned above that T 2 T .L/. Choose the root ˛ such that dim M .˛/ .T / is maximal with respect to all pairs .T 0 ; ˛ 0 / with these properties. Theorem 13.2.2(1) proves that M .˛/ .T / ¤ L holds. Choose a maximal subalgebra L.0/ containing M .˛/ .T /. Then .T; ˛; L.0/ / is a good triple (Definition 13.6.1). (C) Suppose L is not of classical or Cartan type. Lemma 13.7.1(1) shows that LŒp contains a non-rigid 2-dimensional torus T . By (A), T 2 T .L/. Choose T 2 T .L/ with minimal number of improper roots, any nonzero T -root ˛ and observe that M .˛/ .T / ¤ L by Lemma 13.1.3(4). Choose a maximal subalgebra L.0/ containing M .˛/ .T /. Then T is a good torus and .T; ˛; L.0/ / is an admissible triple. As a result, under one of the assumptions of Proposition 13.7.2 there are admissible N is simple. Recall that this latter property does not triples .T; ˛; L.0/ / such that A.G/ depend on the filtration chosen but only on the triple itself (Proposition 13.3.9). Lemma 13.7.3. Suppose L is not of classical or Cartan type and T LŒp is a good torus. Then .L; T / ¤ imp .L; T /. Proof. Let T LŒp be a good torus and assume that all roots are improper. In particular, all roots are Hamiltonian or Witt. Take any such root ˇ and switch T to T 0 WD Tx by a suitable root vector x 2 Lˇ such that L.ˇ/ is a proper T 0 -1-section. It still is Hamiltonian or Witt, respectively. Lemma 13.7.1(2) shows that T 0 is non-rigid. Proposition 13.7.2 shows that T 0 2 T .L/. But then T cannot have minimal number of improper roots, a contradiction. N is Next we consider the case that .T; ˛; L.0/ / is an admissible triple and A.G/ N simple. Then S WD A.G/ is a simple Lie algebra of absolute toral rank 2 (Lemma 13.3.6(5)). Let G and S denote the p-envelope of GN and S in Der S. It is straightforward that C.G / D C.S/ D ¹0º hold. Therefore G is the minimal p-envelope of GN and S is the minimal p-envelope of S . By Jacobson’s formula on p-th powers, G D

X j 0

j

Œp GN 0 C

XX

j

Œp GN i :

i ¤0 j 0 Œpj

By definition of the grading of Der S , GN i Deripj S . Then G naturally carries Œpj N a Z-grading and Gi Gipj holds for all i; j . Comparing degrees shows that the 0-component G0 of G coincides with the p-envelope of GN 0 in G . Similarly, S is naturally graded and the 0-component S0 of S coincides with the p-envelope of S0 in S.

306

13


Recall that L.0/ denotes the p-envelope of L.0/ in LŒp and ˆ W L.0/ G0 is the natural homomorphism of restricted Lie algebras induced by the homomorphism L.0/ ! L.0/ =L.1/ D G0 and the natural action of G0 on S. We mentioned in §13.3 that ˆ.T / is a 2-dimensional torus in G . It is clear that the set of roots .L; T / and N ˆ.T // are essentially identical. However, the property of a root to be proper .G; or solvable, classical, Witt, Hamiltonian depends on if we regard it as a T -root or a ˆ.T /-root. In contrast to earlier sections we do distinguish them here in notation and N ˆ.T //. denote the image of 2 .L; T / by ˆ./ 2 .G; Lemma 13.7.4. (1) S0 acts faithfully on S 1 ; (2) ˆ.T / S0 is a 2-dimensional non-rigid torus of S; .ˆ.˛// .˛/ (3) gr M .L; T / Mˆ./ .S; ˆ.T // for all ¤ 0; (4) ni ˛ ni ˆ.˛/ for all i 2 Fp and n.˛/ n.ˆ.˛//. Proof. (1) Suppose x 2 S0 annihilates S 1 . Recall that the grading of S satisfies (g1) and (g2) of Notation 3.5.2. Then it is easily seen that Œx; Si D ¹0º for all i . This implies x D 0. (2) We mentioned that ˆ.T / is a 2-dimensional torus in G . Since TR.S/ D 2 by Lemma 13.3.6, Theorem 1.2.9 (with KŒp D S and LŒp D G ) shows that ˆ.T / S C C.G / D S. Since ˆ.T / is homogeneous of degree 0 by definition of ˆ, one has ˆ.T / S0 . As T is non-rigid, there is a T -sandwich in L. By Corollary 12.1.3, there is even more a T -sandwich w 2 L for some ¤ 0. By construction and Lemma 13.3.2(1), N Since w 2 L.0/ , whence its image gr w in GN is a nonzero ˆ.T /-sandwich of G. ¤ 0 and ˆ.T / S, one has even more gr w 2 S . Then ˆ.T / is a non-rigid torus of S. .˛/ (3) Let ¤ 0 and x 2 M .L; T /. Then Œx; L H \ ker ˛ L.0/ . Choose i such that x 2 L.i/ nL.i C1/ , set gri x the image of x in GN i . As ¤ 0, one has gri x 2 Si . If j 0 grk H ker ˆ.˛/. This 2 M .ˆ.˛// .S; ˆ.T //. shows that gri x P (4) Note that i 2Fp Ki˛ .L; T / L.0/ . By (3) it follows that gr Ki˛ .L; T / Kiˆ.˛/ .S; ˆ.T //;

i 2 Fp :

Take x 2 Ki ˛ .L; T / \ L.a/ n L.aC1/ and y 2 K i˛ .L; T / \ L.b/ n L.bC1/ such that grb y 2 .RK/ iˆ.˛/ .S; ˆ.T //. Then a; b 0. If a C b > 0, then Œgra x; grb y 2 P l>0 grl H . If a D b D 0, then by the above inclusion Œgr0 x; gr0 y 2 ŒKiˆ.˛/ .S; ˆ.T //; .RK/

iˆ.˛/ .S; ˆ.T //

nil ˆ.H /:

13.7

307


Consequently, Œx; y 2 nil H C L.1/ \ H D nil H in all cases. This gives y 2 .RK/ i ˛ .L; T /, i.e., .RK/

i ˆ.˛/ .S; ˆ.T //

\ gr K

i˛ .L; T /

gr .RK/

:

i˛ .L; T /

Comparing the 0-components one obtains .RK/

i ˆ.˛/ .S; ˆ.T //

As ker ˆ is Œp-nilpotent and .RK/i ˛ .L; T /. Therefore

\ˆ K

P

i 2Fp

i˛ .L; T /

ˆ .RK/

:

i˛ .L; T /

Ki˛ .L; T / L.0/ , .ker ˆ/ \ Ki˛ .L; T /

ni ˛ .L; T / D dim Ki˛ .L; T /=.RK/i˛ .L; T / D dim ˆ.Ki˛ .L; T //=ˆ..RK/i˛ .L; T // dim ˆ.Ki˛ .L; T //=.RK/iˆ.˛/ .S; ˆ.T // \ ˆ.Ki˛ .L; T // dim Kiˆ.˛/ .S; ˆ.T //=.RK/iˆ.˛/ .S; ˆ.T // D niˆ.˛/ .S; ˆ.T //:

Lemma 13.7.5. Suppose n.S/ 2. Let S0 contain a unique minimal ideal I . Then I 6Š H.2I 1/.2/ . Proof. Let us suppose the contrary, I Š H.2I 1/.2/ . (a) Since I ¤ ¹0º is the unique minimal ideal of S0 , H.2I 1/.2/ S0 Der H.2I 1/.2/ D CH.2I 1/: Then S0 is the minimal p-envelope of S0 . As H.2I 1/ is restricted and TR.H.2I 1// D 1 but ˆ.T / S0 , one obtains S0 6 H.2I 1/. Then dim S0 =S0 \ H.2I 1/ D 1. One obtains a nonzero pairing !WS

1

S1 ! S0 =S0 \ H.2I 1/ Š F:

(b) Let E WD ¹x 2 S1 j !.ŒS 1 ; x/ D 0º and g denote the Lie algebra generated by S 1 , S0 \ H.2I 1/, and E. It is clear that g is a graded subalgebra of S . Let gŒp denote the p-envelope of g in S. If TR.g/ D 2, then Theorem 1.2.9 shows that ˆ.T / gŒp . However, it is easy to see that gŒp \ Der0 g is the p-envelope of g \ S0 H.2I 1/. We obtain gŒp \ Der0 g H.2I 1/, which does not contain ˆ.T /. Consequently, TR.g/ D 1 and TR.g= rad g/ 1. We obtain by Theorem 11.1.2 H.2I 1/.2/ Š I S0 ,! g.1/ = rad g.1/ H.2I 1/: Note that S 1 D ŒI; S 1 and ŒI; E are contained in g.1/ . By dimension reasons we deduce S 1 rad g.1/ . But then ŒS 1 ; ŒI; E S0 is a solvable ideal of S0 ,

308

13


and this gives ŒS 1 ; ŒI; E D ¹0º. (g1) of Notation 3.5.2 now yields ŒI; E D ¹0º. Consequently, ! establishes an S0 -module homomorphism S

1

Š .S1 =E/ ;

where ŒI; E D ¹0º:

(c) Due to Theorem 7.5.5 one can normalize ˆ.T / as ˆ.T / D F t1 @1 ˚ F t2 @2 ;

t1 @1 / D 1 and .t2 @2 / D 0. Then

Let denote the ˆ.T /-root given by .t2 @2 S0 \ Si D FDH .t1i

ti 2 ¹xi ; 1 C xi º:

1

t2 /;

1 i p:

A plain computation shows that S0 . / Š W .1I 1/ ˚ ˆ.T / \ ker . Note that F t2 @2 D T \ ker annihilates CS 1 . / .t2 @2 t1 @1 /. We get CS 1 . / .t2 @2 t1 @1 / ˆ.H / \ S 1 D ¹0º. Since the pairing ! is S0 -invariant, it is also invariant under ˆ.T / and hence nondegenerate when restricted to S 1 . / S1 . /=S1 . / \ E. As a conclusion, every composition factor of the restricted S0 . /.1/ -module S 1 . / and S1 . /=S1 . / \ E has only nonzero F .t2 @2 t1 @1 /-weight. Theorem 7.6.10 implies that every such composition factor has dimension p 1. As E is annihilated by S0 . /.1/ , both S˙1 . / have no p-dimensional S0 . /.1/ -composition factor. (d) Note that S. / cannot be solvable or classical. Suppose it is Witt. Then one has S0 \ rad S. / ˆ.T / \ ker and therefore dim S0 . /=S0 \ rad S. / p D dim S. /= rad S. /. Consequently S˙1 . / rad S. /. That gives us the inclusion S˙1 . / K. /. Recall that the pairing ! is nondegenerate when restricted to S 1 . / S1 . /=S1 . / \ E. That gives us S 1 \ .RK/. / D ¹0º and S1 \ .RK/. / E. Then 2 dim S 1 . / D dim S 1 . / C dim S1 . /=S1 . / \ E n.S/ 2; hence dim S 1 . / 1. But then S0 . /.1/ annihilates S 1 . /. This contradicts (c) and yields that S. / is Hamiltonian. (e) The grading of S induces a grading on S. /, rad S. / and SŒ . Recall that S Œ 0 Š W .1I 1/ ˚ C.SŒ 0 /. Then Theorem 11.4.1 shows that the grading is given by giving suitable generators u1 the degree 0 and u2 the degree a2 ¤ 0. If ja2 j > 1, then S˙1 . / rad S. / and one argues as in the former case to obtain a contradiction. Hence a2 2 ¹˙1º. Looking at Theorem 11.4.1 it is not hard to compute that S Œ a2 contains a p-dimensional irreducible W .1I 1/-module. But then Sa2 . / has a p-dimensional composition factor contrary to (c). N Lemma 13.7.6. Let V be a composition factor of i .M.G//= i C1 .M.G// as a Gmodule and W GN ! gl.V / the corresponding representation. Then there exists a restricted representation N W G ! gl.V / whose restriction to GN coincides with .

13.7


309

O be the universal p-envelopes of G and G, N respectively. The Proof. Let GO and G O ensures that there is a commutative diagram universal property of GO and G GO " G

1

O ! G " ! GN

2

! & O

G

!

gl.V /

where is the canonical homomorphism and all 1 , 2 , O are restricted homomorN 2 is phisms. Since is surjective, so is 1 . Since G is a minimal p-envelope of G, surjective. Let D 2 ker.2 ı 1 /. By definition, Œ.2 ı 1 /.D/; S P D ¹0º. Since S 1 D N 1 D GN 1 , this implies ŒD; G 1 M.G/. As M.G/ A.G/ Gi , easy ini 2P duction on i based on (g1), (g2) of Notation 3.5.2 shows that ŒD; Gi j 0 Lj . We may take w homogeneous, whence w 2 L2k ; i . However, in the present case we have Œv; w 2 I \ H H˛ , the final contradiction. Combining Lemmas 14.1.3–14.1.6 one gets a complete proof of Theorem 14.1.1.

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14

14.2

The toral rank 2 case

S is not of Cartan type

In order to prove the Classification Theorem for Lie algebras of absolute toral rank 2 let L denote a finite dimensional simple Lie algebra of absolute toral rank 2 which is not of classical, Cartan or Melikian type. Proposition 13.7.2(2) shows that admissible triples .T; ˛; L.0/ / exist for L for which T is a good torus. Choose such a triple and consider the associated standard filtration, the graded algebra G, and the minimal N Š S of GN D G=M.G/, which is a simple Lie algebra (Theorem 13.5.6) ideal A.G/ of absolute toral rank 2. Proposition 13.7.7 yields that S is not of classical type, while Proposition 13.7.8 shows that S is not a Melikian algebra. It is the goal of this section to prove that S is not of Cartan type, that is to show that S is a counterexample to the Classification Theorem as well. Recall that L.0/ denotes the p-envelope of L.0/ in LŒp , G is the p-envelope of GN in Der S, and S is the p-envelope of S in Der S. The grading of GN induces a grading of G and S. The 0-components G0 and S0 are the p-envelopes of GN 0 and S0 in Der S , respectively. Note that T 2 T .L/ is non-rigid (Theorem 12.3.7). In particular, H D CL .T / acts trigonalizably on L (Corollary 12.5.10). Some subtle estimates on dim L =R will be needed for suitable roots . A first step towards that is the result of Theorem 14.1.1 that no exceptional roots exist. This gives K D .RK/ and dim L =R dim L =K C dim L

=K :

The latter number can be checked in the 1-section L./ alone, while the definition of R involves nil H and therefore needs computations outside L./. Another step in that direction is to obtain a sufficient number of proper roots. Lemma 14.2.1. (1) Let T 0 L.0/ be a 2-dimensional torus such that CL .T 0 / acts trigonalizably on L. If 2 .L; T 0 / is a nonzero proper root, then ˆ./ 2 .S; ˆ.T 0 // is proper as well. In particular, jimp .L; T 0 /j jimp .S; ˆ.T 0 //j: (2) Let t S0 be a 2-dimensional torus such that CS .t/ acts trigonalizably on S and 2 .S; t/ be a nonzero root. Then is a proper root if and only if t acts on S0 ./ as a proper torus. Proof. Let 2 .L; T 0 / be any nonzero root. (a) Suppose is Hamiltonian, ˆ./ 2 .S; ˆ.T 0 // is Hamiltonian and S0 .ˆ.// is of Hamiltonian type.

327

14.2 S is not of Cartan type

N Apply Theorem 11.1.2 for g WD G.ˆ.// and t D ˆ.T 0 /=ˆ.T 0 / \ ker ˆ./. By assumption on S0 .ˆ.// we have g=rad g ¤ sl.2/; W .1I 1/ and therefore H.2I 1/.2/ g= rad g H.2I 1/ holds. Again by the assumption on S0 .ˆ.// the grading of H.2I 1/ induced by the grading of g= rad g is trivial. Then N N rad G.ˆ.// H.2I 1/; H.2I 1/.2/ G.ˆ.//= N N N G.ˆ.//= rad G.ˆ.// Š GN 0 .ˆ.//=GN 0 .ˆ.// \ rad G.ˆ.//: N In particular, GN 0 .ˆ.//=GN 0 .ˆ.// \ rad G.ˆ.// is semisimple. Hence N rad GN 0 .ˆ.// D GN 0 .ˆ.// \ rad G.ˆ.//: Then there is a surjective homomorphism N L.0/ ./ L.0/ ./= rad L.0/ ./ Š GN 0 .ˆ.//=GN 0 .ˆ.// \ rad G.ˆ.//: Next apply Theorem 11.1.2 for g WD L./ and t D T 0 =T 0 \ ker . One obtains similarly H.2I 1/.2/ L./= rad L./ H.2I 1/: Combining these results one gets L./ D L.0/ ./CCL .T 0 /Crad L./. As CL .T / L.0/ , Theorem 10.7.1 (applied to the L.0/ -module L=L.0/ ) shows that CL .T 0 / L.0/ . Then L./ D L.0/ ./ C rad L./ and, as L./= rad L./ is semisimple, L./= rad L./ Š L.0/ ./=L.0/ ./ \ rad L./ Š L.0/ ./= rad L.0/ ./: As a result, there are natural isomorphisms .2/ N N D GN 0 .ˆ.//=GN 0 .ˆ.// \ rad G.ˆ.// H.2I 1/.2/ Š GŒˆ./ .2/ Š L.0/ ./=L.0/ ./ \ rad L./ Š LŒ.2/ :

.2/

Consequently, T 0 stabilizes the natural maximal subalgebra of H.2I 1/.2/ of codimension 2 if and only if ˆ.T 0 / does so. For such one has that is proper if and only if ˆ./ is proper. (b) Suppose is Hamiltonian, ˆ./ 2 .S; ˆ.T 0 // is Hamiltonian and S0 .ˆ.// is not of Hamiltonian type. If S0 .ˆ.// is solvable or classical, then ˆ./ is proper due to Theorem 11.4.1. So assume that S0 .ˆ.// is Witt. Note that S 6Š M.1; 1/ by Proposition 13.7.8. The construction gives CL .T / L.0/ , whence Theorem 10.7.1 then implies CL .T 0 / L.0/ . Due to Proposition 13.4.4 is proper if and only if ˆ.T 0 / acts as a proper torus on S0 .ˆ.//= rad S0 .ˆ.//. Applying Theorem 11.4.1 to SŒˆ./ now shows that this is true if and only if ˆ./ is proper. Consequently, if is proper, then so is ˆ./; and ˆ./ is proper if and only if ˆ.T / acts as a proper torus on S0 .ˆ.//.

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14


(c) Suppose is proper Hamiltonian and ˆ./ 2 .S; ˆ.T 0 // is improper Witt. S Œˆ./ carries a grading inherited by the grading of S . If the grading is nontrivial, it is given by a nonzero degree of a suitable indeterminate x. Then all homogeneous spaces are of the form F x i @, in particular they are 1-dimensional. Since ˆ.T 0 / is homogeneous of degree 0, it stabilizes all homogeneous components of S . But then all x i @ are eigenvectors with respect to ˆ.T 0 /. In this case the maximal subalgebra of codimension 1 in S.ˆ.// is ˆ.T 0 /-invariant. Then ˆ./ is proper, which is not true in the present case. Consequently, the grading is trivial, whence S.ˆ.// D S0 .ˆ.// C rad S.ˆ.//; W .1I 1/ Š S.ˆ.//= rad S.ˆ.// Š S0 .ˆ.//=S0 .ˆ.// \ rad S.ˆ.//: We may normalize the image of T 0 in S.ˆ.//= rad S.ˆ.// Š W .1I 1/ as F .1 C x/@. Choose root vectors wi 2 L.0/ with image wi WD .1 C x/i C1 @ in W .1I 1/. Next we consider L./ and observe that (Corollary 11.2.4) H.2I 1/.2/ LŒ H.2I 1/: Let wQi denote the image of wi in H.2I 1/. Since is proper, there is a normalization of the image of T 0 as FDH .u1 u2 /, where u1 ; u2 denote the indeterminates which describe H.2I 1/ (in order to avoid confusion we use the character u instead of x). Suppose wQi 2 H.2I 1/.1/ for some i 2 Fp . Then .ad wQi /2 .H.2I 1// H.2I 1/.1/ , and this means that .ad wi /2 .L.// acts nilpotently on L./= rad L./. Consequently, .ad wi /2 .L.0/ .// acts nilpotently on L.0/ ./=L.0/ \ rad L./, and this gives that .ad wi /2 .W .1I 1// acts nilpotently on W .1I 1/. However, .ad.1 C x/i C1 @/2 ..1 C x/p 2iC1 @/ D 6i 2 .1 C x/@, a contradiction. Consequently, p D dim W .1I 1/ dim H.2I 1/=H.2I 1/.1/ D 5. This gives p D 5 and 4 X F wQi C H.2I 1/.1/ : H.2I 1/ D i D0

Next we observe that, as n.L/ D 0, there is k 2 Fp for which dim Lk =L.0/;k dim Lk =Rk 2. Since every root space of H.2I 1/.2/ for a nonzero root has dimension p, we have dim L.0/;k =L.0/;k \ rad L./ dim Lk =Lk \ rad L./

2Dp

2 D 3:

On the other hand, L./= rad L./ injects into H.2I 1/ and dim H.2I 1/k =H.2I 1/.1/;k D 1 holds. Therefore there exists an eigenvector w 2 L.0/;k with wQ 2 H.2I 1/.1/ n ¹0º. P Let M denote the algebra generated by 4i D0 F wQi CF wQ and gr M the graded algebra


329

associated with the filtration of M induced by the natural filtration of H.2I 1/. Then H.2I 1/ 1 C H.2I 1/0 ¨ gr M . Since the gradation is separating, one has .gr M / \ H.2I 1/1 ¤ ¹0º. Also, H.2I 1/1 is H.2I 1/0 -irreducible (Lemma 5.2.2(2)). [S-F, Theorem 4.7.2] yields H.2I 1/.2/ gr M . But then the subalgebra of L.0/ generated P by 4i D0 F wi C F w is of Hamiltonian type, and this implies that S0 .ˆ.// is of Hamiltonian type as well. But ˆ./ is Witt in the present case. So this case does not occur. (d) Suppose that is Witt. If ˆ./ is solvable or classical, it is proper. Since gr.rad L.// rad gr L./, one has dim LŒ dim SŒˆ./. Then ˆ./ cannot be Hamiltonian. So assume that ˆ./ 2 .S; ˆ.T 0 // is Witt. Then SŒˆ./ Š W .1I 1/ carries a grading inherited by the grading of S . If the grading is nontrivial, it is given by a nonzero degree of a suitable indeterminate x. Then all homogeneous spaces are of the form F x i @, in particular they are 1-dimensional. Since ˆ.T 0 / is homogeneous of degree 0, it stabilizes all homogeneous components of S . But then all x i @ are eigenvectors with respect to ˆ.T 0 /. In this case the maximal subalgebra of codimension 1 in S.ˆ.// is ˆ.T 0 /-invariant. Then ˆ./ is proper. So assume that the grading of SŒˆ./ is trivial. Then S.ˆ.// D S0 .ˆ.// C rad S.ˆ.//. In this case S0 .ˆ.//= rad S0 .ˆ.// Š S.ˆ.//= rad S.ˆ.// Š W .1I 1/ holds. Observe that L.0/ Œ Š GN 0 Œˆ./ and GN 0 .ˆ.// D S0 .ˆ.// C ˆ.CL .T 0 //. Since CL .T 0 / acts trigonalizably on L, one has that rad S0 .ˆ.// is ˆ.CL .T 0 //invariant (Theorem 11.1.2). Then rad S0 .ˆ.// D S0 \ rad GN 0 ./ and hence S0 .ˆ.//=S0 .ˆ.// \ rad GN 0 .ˆ.// D S0 .ˆ.//= rad S0 .ˆ.// Š W .1I 1/: As ˆ./ is proper if and only if ˆ.T 0 / respects the maximal subalgebra of codimension 1, we obtain that ˆ./ is proper if and only if ˆ.T 0 / acts on S0 .ˆ.// as a proper torus. By the above, there is an injection W .1I 1/ Š S0 .ˆ.//=S0 .ˆ.// \ rad GN 0 .ˆ.// ,! L.0/ ./= rad L.0/ ./; and the image of W .1I 1/ is an ideal of L.0/ ./= rad L.0/ ./. The Witt algebra has no outer derivations. Therefore L.0/ ./= rad L.0/ ./ Š W .1I 1/ holds. Then L./ D L.0/ ./ C rad L./, whence L./= rad L./ Š L.0/ ./= rad L.0/ ./ Š S0 .ˆ.//= rad S0 .ˆ.//: Consequently, is proper if and only if ˆ./ is so. (e) If is solvable or classical, then ˆ./ is at most classical, in particular is proper. This proves statement (1). (f) An easy adaptation of (a) and (b) prove statement (2) in case that 2 .S; t/ is Hamiltonian. If 2 .S; t/ is Witt, then only minor changes are necessary in (d) to prove (2). There is nothing to prove if is classical or solvable.

330

14


Lemma 14.2.2. Suppose S is one of the Cartan type Lie algebras W .2I 1/, S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/, H.2I .1; 2//.2/ , H.2I 1I ˆ.1//. Then M.G/ D ¹0º. N Proof. Suppose M.G/ ¤ ¹0º and let V be a composition factor of the (nonzero) Gmodule M.G/= 2 .M.G//. By Lemma 13.7.6, V is a restricted G -module. Recall that the restricted subalgebra S of G Der S is the minimal p-envelope. Since S is restricted in cases W .2I 1/, S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/, we then have S D S. By Lemma 13.7.4 ˆ.T / is a 2-dimensional torus in S. Suppose S Š W .2I 1/ or S Š H.2I 1I ˆ.1//. Due to Theorem 10.7.3 all nonzero i ˛ C jˇ ¤ 0, i; j 2 Fp , are weights on V . By definition of L.0/ this implies Li ˛Cjˇ ¤ Ri˛Cjˇ . As n.L/ D 0 by Theorem 14.1.1, Lemma 13.1.1 shows that Li ˛Cjˇ ¤ Ki˛Cjˇ or L .i˛Cjˇ / ¤ K .i˛Cjˇ / . By definition of a good torus and Proposition 13.7.3 there is a proper root D i˛ C jˇ ¤ 0, and then Lemma 13.1.2 gives that is Hamiltonian and dim L˙k =K˙k D 1 holds for some k 2 Fp . Then dim Lk =Rk 2. Also observe that Rk L.0/ , whence dim Rk dim GN k . Since all root spaces of H.2I 1/.2/ with respect to a 1-dimensional torus in Der H.2I 1/.2/ are p-dimensional, we obtain p dim Lk dim Lk =Rk C dim GN k ; whence dim GN k 3. Recall that all derivations of W .2I 1/ are inner (Theorem 7.1.2). In particular, GN D S Š W .2I 1/. All root spaces in W .2I 1/ with respect to any 2-dimensional torus are 2-dimensional. So this case is not possible. Similarly, all root spaces in Der H.2I 1I ˆ.1// with respect to any 2-dimensional torus are at most 2-dimensional (Theorem 7.1.3(4)). So this case is not possible either. Suppose S is one of S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/, H.2I .1; 2//.2/ . Theorem 10.7.2 yields annV ˆ.T / ¤ ¹0º contradicting the inclusion CL .T / L.0/ . Proposition 14.2.3. S 6Š W .2I 1/. Proof. Suppose S D W .2I 1/. Let S.i/ D span ¹x1a x2b @j j a C b 1 i º denote the i-th component of the natural filtration of S . (a) Lemma 14.2.2 shows that M.G/ D ¹0º. As all derivations of W .2I 1/ are inner (Theorem 7.1.2), we have G D GN D S D S. Then S0 D S0 . Lemma 13.7.4(2) shows ˆ.T / S0 . According to Theorem 7.4.1 there exists a generating set u1 ; u2 2 O.2I 1/.1/ and a1 ; a2 2 Z such that the grading of S has type .a1 ; a2 / with respect to u1 ; u2 . To simplify notation we assume (without loss of generality) that ui D xi , i D 1; 2. Then j

Sm D span ¹x1i x2 @k j ia1 C ja2

ak D m;

0 i; j < p; k D 1; 2º:

As S ¤ S0 , either S a1 ¤ 0 or a2 ¤ 0. Suppose 0 ¤ a1 ¤ a2 ¤ 0. Since @1 ; @2 ; x1 @2 ; x2 @1 2 i ¤0 Si , we must have S0 F x1 @1 C F x2 @2 C S.1/ . But then S0 D G0 is solvable, contrary to Corollary 9.1.10 and the choice of L.

331


Now suppose 0 ¤ a1 D a2 . Then S0 D span ¹x1 @1 ; x2 @2 ; x1 @2 ; x2 @1 º is isomorphic to gl.2/. As S 1 ¤ ¹0º, a1 2 ¹˙1º. If a1 D 1, then one has S 1 D span ¹x12 @1 ; x1 x2 @1 ; x22 @1 ; x12 @2 ; x1 x2 @2 ; x22 @2 º. The S0 -submodule generated by x12 @2 is 4-dimensional, hence S 1 is a reducible S0 -module. Since S0 D G0 and G satisfies (g3) of Notation 3.5.2, this is impossible. Therefore, a1 D 1, so that S 1 D F @1 ˚ F @2 is a 2-dimensional irreducible S0 -module and S k D ¹0º for k 2. Then L.0/ =L.1/ Š gl.2/, dim L=L.0/ D 2 and L.2/ ¤ ¹0º. Applying Corollary 5.5.3 we get L Š W .2I n/ for some n D .n1 ; n2 /. As dim L D dim S D dim W .2I 1/, we must have L Š W .2I 1/ contrary to our choice of L. As a consequence, either a1 D 0, a2 ¤ 0 or a1 ¤ 0, a2 D 0. (b) Renumbering x1 and x2 if necessary we may assume that a1 D 0. Then S0 D span ¹x1i @1 ; x1i x2 @2 j 0 i < pº: As S 1 ¤ ¹0º, a2 2 ¹˙1º. If a2 D 1, then S 1 D span ¹x1i x2 @1 ; x1i x22 @2 j 0 i < pº is a reducible S0 -module. As before this is impossible. Therefore, a2 D 1 and S

1

D

p X1

F x1i @2 ;

S

k

D ¹0º for k 2:

i D0

Define t WD F x1 @1 ˚ F x2 @2 S.0/ and 1 ; 2 2 t by setting 1 .x1 @1 / D 1;

1 .x2 @2 / D 0;

2 .x1 @1 / D 0;

2 .x2 @2 / D 1:

It is easy to see that .S 1 ; t/ D ¹i1 2 j i 2 Fp º and .S0 ; t/ D Fp 1 . Moreover, dim S 1; D 1 for any 2 .S 1 ; t/. Let T 0 be a 2-dimensional torus in L.0/ which maps onto F x1 @1 ˚ F x2 @2 under the homomorphism ˆ W L.0/ ! Der0 S (in the case under consideration this homomorphism is surjective, as S0 D Der0 G D G0 D L.0/ =L.1/ ) and denote by 10 ; 20 2 T 0 the respective preimages. By our earlier observations, for any 2 .L; T 0 /nFp 10 , the subalgebra L.0/ . / 0 T C L.1/ is solvable, T 0 -invariant, and has codimension 1 in L. /. Then each 2 .L; T 0 / n Fp 10 is solvable, classical or proper Witt. Besides, L.10 / D L.0/ .10 / and L.0/ .10 /=L.1/ .10 / Š S0 . With this in mind it is easily seen that 10 is a proper Witt root of L. Thus all roots in .L; T 0 / are proper and there is a proper non-solvable root. Propositions 13.7.1(2) and 13.7.2 show that T 0 2 T .L/. As T is a good torus, all roots in .L; T / are proper as well. (c) Note that C.S0 / D F x2 @2 ˆ.T /, by the maximality of ˆ.T /. Let t be a nonzero toral element of ˆ.T / such that ˆ.T / D F t ˚F x2 @2 , and define 2 ˆ.T / by setting .t/ D 1; .x2 @2 / D 0 and 0 2 T by ˆ. 0 / D . If 2 .S0 ; ˆ.T //, then .x2 @2 / D 0. Hence S0 D S0 ./. Since x2 @2 acts on S 1 as Id, we also have that .S 1 ; ˆ.T // .S; ˆ.T //nFp . As a consequence, L./ D L.0/ ./ and L.0/ ./=L.1/ ./ Š S0 . Let

332

14


‰ denote the natural Lie algebra epimorphism S0 ! W .1I 1/ whose kernel is spanned by ¹x1i x2 @2 j 0 i < pº. As all roots in .L; T / are proper, Lemma 14.2.1(2) shows that the subalgebra W .1I 1/.0/ is ‰.ˆ.T //-stable. This implies ‰.t/ 2 W .1I 1/.0/ forcing ˆ.T / S0 \ S.0/ . Using our discussion in (b) and Theorem 10.7.1 one observes that each weight p 1 in .S 1 ; ˆ.T // has multiplicity 1. Note that S 1 \ S.p 2/ D F x1 @2 . Since S 1 \ S.p 2/ is S0 \ S.0/ -invariant, there is 2 .S; ˆ.T // n Fp such that S 1; D p 1 F x1 @2 . Since x2 @2 acts on S˙1 as ˙Id, we have that .S 1 ; ˆ.T // \ Fp D ¹ º and .S1 ; ˆ.T // \ Fp ¹ º. Therefore, ŒG 1 . /; G1 . / D ŒS 1 . /; S1 . / D ŒS

1; ; S1;

ˆ.T / \ S.p

3/ :

However, ˆ.T /\S.p 3/ acts semisimply and nilpotently on S. Then ˆ.T /\S.p 3/ D ¹0º. Let 0 2 .L; T / be the root for which ˆ. 0 / D . The above shows that ŒL. 0 /; L.1/ . 0 / L.1/ . 0 /. So L.1/ . 0 / is an ideal of L. 0 /. Since L. 0 / D L 0 C T C L.1/ . 0 /, we obtain ŒL 0 ; L 0 L.1/ , whence L 0 D R 0 L.0/ . This contradiction completes the proof of the proposition. Next we show that S is not isomorphic to one of S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/. Each of these cases is quite involved and requires detailed information on gradings and representations of the respective Cartan type Lie algebras. The relation x a D a1 Š an Š x .a/ ;

0 ai p

1;

links our present truncated-polynomial notation with the divided-power notation used in §4.2 of Volume 1. Proposition 14.2.4. We have S 6Š S.3I 1/.1/ . Proof. (a) Suppose S D S.3I 1/.1/ . By Lemma 14.2.2 G can be identified with a subalgebra of Der S containing S . Note that in this case S is a restricted subalgebra of the p-envelope G of G in Der S (Theorem 7.2.2). By Theorem 7.1.2, Der S.3I 1/.1/ D CS.3I 1/. Since TR.G/ TR.L/ D 2 and TR.S/ D 2, it can only be that G =S is Œp-nilpotent. On the other hand, S.3I 1/ is a restricted subalgebra of Der S.3I 1/.1/ (Lemma 7.2.1(1),(2)) and Der S.3I 1/.1/ =S.3I 1/ is a 1-dimensional torus. Then it can only be that G is a subalgebra of S.3I 1/. In the proof of [S-F, Proposition 4.3.3, Theorem 4.3.7] we showed that S.3I 1/ D

X 1i<j 3

FDi;j .O.3I 1// C

3 X j D1

F

Y

p 1

xi

@j :

i ¤j

According to Theorem 7.4.1, the grading of S has type .a1 ; a2 ; a3 / for some a1 ; a2 ; a3 2 Z and some generating set u1 ; u2 ; u3 2 O.3I 1/.1/ . To simplify notation we assume ui D xi ; 1 i 3.

333


Since S is a restricted subalgebra of Der S, so is S0 . Then ˆ.T / is a 2-dimensional torus of S contained in S0 (Lemma 13.7.4(2)). Let t WD F .x1 @1

x2 @2 / ˚ F .x2 @2

x3 @3 /:

Then t also is a 2-dimensional torus of S contained in S0 . Define 1 ; 2 2 t by setting 1 .x1 @1

x2 @2 / D 1;

1 .x2 @2

x3 @3 / D 0;

2 .x1 @1

x2 @2 / D 0;

2 .x2 @2

x3 @3 / D 1:

Since S0 is a restricted subalgebra of Der S , ˆ 1 .S0 / is a restricted subalgebra of L.0/ . There is a 2-dimensional torus in L.0/ which maps onto t under the natural epimorphism ˆ 1 .S0 / ! S0 . In what follows we identify this torus with t and view the root system .L; t/ as a subset of Fp 1 C Fp 2 . Since H L.0/ , Theorem 10.7.1 (applied to the L.0/ -module L=L.0/ ) shows that CL .t/ L.0/ . Therefore CS .t/ ˚i 0 Si . As D1;2 .x12 x22 x3 / 2 CS .t/ \ Sa1 Ca2 Ca3 , a1 C a2 C a3 0 holds. Let S.3I 1/ denote the k-th component of the standard grading, i.e., the grading of type .1; 1; 1/ with respect to x1 ; x2 ; x3 . Observe that every Dk;l .x1b1 x2b2 x3b3 / and Q p 1 every i ¤j xi @j is homogeneous with respect to both the .a1 ; a2 ; a3 /-grading and the .1; 1; 1/-grading. Therefore, G D ˚i;k2Z Gi \ S.3I 1/ : (b) Suppose a1 a2 a3 ¤ 0. Then P @i 62 G0 for 1 i 3. This gives G0 \ S.3I 1/< 1> D ¹0º, whence G0 k0 S.3I 1/ . There is j 2 Z such that S 1 \ S<j > ¤ ¹0º and S 1 \ S D ¹0º for kP> j . Recall that S 1 D G 1 is an irreducible and faithful G0 -module. As G0 k0 S.3I 1/ , the subspace S 1 \ S<j > is a G0 –submodule of G 1 . Therefore, G 1 D S 1 \ S<j > . P P SupposePj 0. By property (g2) of Notation 3.5.2, i , whence S 2 D ŒS 1 ; S 1 S< 2> D ¹0º and kx / 2 S S k D ¹0º for k 2. Note that Dm;n .xm n .k 1/am ¤ ¹0º whenever m ¤ n, k D 2; 3. Since our grading has only one component of negative degree we therefore have am > 0 for all m. As @m 2 S am and am < 0 we deduce a1 D a2 D a3 D 1. In other words, Gk D G for all k. As a consequence, sl.3/ Š S0 G0 S.3I 1/ Š sl.3/:

334

14


We have proved that L D L. 1/ , L.2/ ¤ ¹0º, L.0/ =L.1/ Š sl.3/ and L. 1/ =L.0/ is a 3-dimensional irreducible .L.0/ =L.1/ /-module. Theorem 5.2.7 now yields L Š S.3I nI ‰/.1/ . As TR.L/ D 2, Theorem 10.6.3 says L Š S.3I 1/.1/ . This contradicts our choice of L. Thus a1 a2 a3 D 0. (c) From now on we may assume (without loss of generality) that a1 D 0. Suppose a2 a3 ¤ 0. Let S.3I 1/Œk denote the k-th component of the .0; 1; 1/-grading. By the same reasoning as in (a), S D ˚i;k2Z Si \ SŒk and G D ˚i;k2Z Gi \ S.3I 1/Œk : Observe that t S0 \ SŒ0 . It is easily seen that SŒ

1

D span ¹x1i @2 ; x1i @3 j 0 i < pº

and

SŒ

k

D ¹0º for k 2:

GŒ

1

SŒ

One also observes that SŒ

1

D S.3I 1/.1/ Œ

1

D S.3I 1/Œ

1

1 ;

whence GŒ

1

D SŒ

1 ;

GŒ

k

D SŒ

k

D ¹0º 8 k > 1:

Now x1i @2 2 G a2 andPx1i @3 2 G a3 yielding SŒ 1 \ G0 D ¹0º. From this it is immediate that G0 k0 GŒk . Let j 2 Z be such that S 1 \ SŒj ¤ ¹0º and S 1 \ SŒk D ¹0º for all k > j . By our previous remark, S 1 \ SŒj is a G0 submodule of G 1 . The Pirreducibility of G 1 forces G 1 D S 1 \ SŒj . As before, if j 0, then G i k0 SŒk for any i > 0. Since @2 ; @3 2 SŒ 1 , we then have a2 0 and a3 0. However, a2 C a3 D a1 C a2 C a3 0 and a2 a3 ¤ 0. This contradiction shows that j < 0. Then j D 1. kx / 2 Arguing as above one obtains S k D ¹0º for k 2. Then Dm;n .xm n S.k 1/am ¤ ¹0º whenever m ¤ n, k D 2; 3. This proves am > 0 for m D 2; 3. As @m 2 S am and am > 0, this gives a2 D a3 D 1. We deduce that Si D SŒi and Gi D GŒi for all i 2 Z. (d) Recall that G0 S.3I 1/ \ span ¹x1i @1 ; x1i xk @k ; x1i x3 @2 ; x1i x2 @3 j 0 i < p; k D 2; 3º D span ¹x1i @1

ix1i 1 x2 @2 ; x1i .x2 @2

x3 @3 /; x1i x3 @2 ; x1i x2 @3 j 0 i < pº

D S0 : Let S0;C denote the subalgebra of S0 spanned by all x1i @1

ix1i 1 x2 @2 ; .i 1/I x1i .x2 @2

Obviously, S0 D S0;C ˚ F @1 .

x3 @3 /; x1i x2 @3 ; x1i x3 @2

.0 i < p/:

335


By the above, L D L. 1/ , and G 1 D S 1 is spanned by x1i @k where 0 i 0

i>0

and V0 ˚j 2¹0;˙2º; i2Fp Lj2 Ci1 L.0/ ; V0 \ L.1/ D ¹0º; gr0 V0 D SŒ0;C : Put M WD V

1

C V0 C L.1/ . Note that L.0/ D V0 C L

1

C L.1/ :

Properties of the associated graded algebra G ensure that ŒV0 ; V0 V0 C L.1/ M

(14.2.1)

ŒV 1 ; L.1/ V0 C L.1/ M:

(14.2.2)

and Considering roots we also have X V 1 \ L.0/ L.1/ ;

Lk2 Ci1 \ L.0/ V0 C L.1/ :

(14.2.3)

k2Fp ; i2Fp

Set G 0 1 WD gr 1 V 1 . Then G 0 1 is SŒ0;C -stable. As a consequence, ŒV 1 ; V0 V 1 C L.0/ . So, by (14.2.2) and (14.2.3), hX i X ŒV 1 ; V0 Lk2 Ci1 \ .V 1 C L.0/ / L˙2 Ci1 ; i 2Fp

k2¹0;˙2º; i2Fp

Lk2 Ci1 \ .V

X

1

C L.0/ /

k2Fp ; i2Fp

V

1

C V0 C L.1/ D M:

(e) We claim that M is a subalgebra of L. In view of the preceding computations it suffices to show that ŒV 1 ; V 1 M . Note that X X ŒV 1 ; V 1 Lk2 Ci1 L.0/ k2¹0;˙2º i 2Fp

and ŒV 1 ; V

1

\ L.0/ ŒV 1 ; L.1/ V0 C L.1/

336

14

by (14.2.2) and (14.2.3). As gr a bilinear mapping


Š V 1 =V

1V 1

1 \L.0/ , the Lie product in L induces

W G 0 1 G 0 1 ! L.0/ =.V0 C L.1/ / Š S0 =S0;C Š F @1 : Since L.0/ =.V0 C L.1/ / is an S0;C .1 /-module, is an S0;C .1 /-module homomorphism. Since x1 @2 ; x1i @3 ; @1 have weights 21 2 ; i1 C 2 ; 1 with respect to t and t S0;C .1 /, we get .x1 @2 ; x1i @3 / D 0 unless i D p 3: Also, p 4

0 D x1 .x2 @2 D

x3 @3 / .x1 @2 ; x1 p 4

.x12 @2 ; x1

0 D .x12 @1

@3 / p 3

@3 / C .x1 @2 ; x1

p 4

2x1 x2 @2 / .x1 @2 ; x1 p 4

D 3.x12 @2 ; x1

@3 /;

@3 / p 3

@3 / C .p

4/.x1 @2 ; x1

@3 /:

Therefore, .x1 @2 ; x1i @3 / D 0 for i 1. Since x1i @2 has t-weight .i C 1/1 2 we have .x1 @2 ; x1i @2 / D 0 as well. But then .x1 @2 ; G 0 1 / D 0. Applying x1i 1 .x2 @2 x3 @3 / with i 1 to both sides of the latter equality we get .x1i @2 ; G 0 1 / D 0 for i 1. Playing the same game with x1i @3 yields .x1i @3 ; G 0 1 / D 0. This implies D 0, i.e., ŒV 1 ; V 1 V0 C L.1/ . The claim follows. (f) Let v1 2 L.0/ n L.1/ be a root vector with respect to t such that gr0 v1 D @1 . Let v2 ; v3 2 L. 1/ n L.0/ be root vectors with respect to t such that gr 1 v2 D @2 and gr 1 v3 D @3 . Let vN i denote the image of vi in W WD L=M . Then ¹vN 1 ; vN 2 ; vN 3 º is a basis of W . Note that for .i; j / 2 ¹.1; 2/; .1; 3/; .3; 2/; .2; 3/; .2; 1/; .3; 1/º there are t-root vectors ui;j 2 M such that grk ui;j D xi @j and k D

1 if i D 1; k D 0 if i; j ¤ 1; k D 1 if j D 1:

Let denote the natural representation of M in gl.W /. It follows immediately from the existence of the ui;j that sl.W / .M /. As a consequence, L=M is an irreducible M -module. This, in turn, implies that M is a maximal subalgebra of L. Put M.0/ D M and let M.k/ denote the k-th component of the standard filtration of L associated with the pair .M.0/ ; L/. Then L D M. 1/ and ¹0º ¤ L.3/ M.2/ (for L.1/ M.0/ ). Since M.1/ D ker and TR.L/ D 2, M.0/ =M.1/ Š sl.3/ necessarily holds. Repeating the argument presented at the end of (b) we conclude that L Š S.3I 1/.1/ . Since this contradicts our choice of L, the case we consider does not occur. Thus a1 D a2 a3 D 0. (g) No generality is lost by assuming that a1 D a2 D 0. As G 1 D S 1 ¤ ¹0º by definition of S and a1 C a2 C a3 0, we have that a3 D 1. Then j

S0 D span ¹ix1i 1 x2 x3 @3

j

j 1

x1i x2 @1 ; jx1i x2

x3 @3

j

x1i x2 @2 j 0 i; j < pº

337

14.2 S is not of Cartan type j

is isomorphic to W .2I 1/ under an isomorphism which maps ix1i 1 x2 x3 @3 j j j 1 j onto x1i x2 @1 and jx1i x2 x3 @3 x1i x2 @2 onto x1i x2 @2 . Also, G

1

DS

j

D span ¹x1i x2 @3 j 0 i; j < p; .i; j / ¤ .p

1

1; p

j

x1i x2 @1

1/º;

and G k D ¹0º for k 2. Note that t S0 and .G 1 ; t/ D Fp 1 ˚ Fp 2 n ¹0º. Moreover, dim G 1; D 1 for any 2 .G 1 ; t). By (a), ˆ.T / is a 2-dimensional torus of S0 . So it follows from Theorem 7.5.1 that there is an isomorphism ‰ W S0 ! W .2I 1/ such that ‰.ˆ.T // D F z1 @y1 ˚ F z2 @y2 where .z1 ; z2 / 2 ¹.1 C y1 ; 1 C y2 /; .1 C y1 ; y2 /; .y1 ; y2 /º P j (in order to avoid confusion we write W .2I 1/ D F y1i y2 @yk ). Define 1 ; 2 2 0 ‰.ˆ.T // by setting i .zj @yj / D ıi;j and i 2 T by ‰.ˆ. i0 // D i , where i; j 2 ¹1; 2º. By the above description of G 1 and Theorem 10.7.1 it follows that .L=L.0/ ; T / D Fp 10 ˚ Fp 20 n ¹0º. Let 2 .L; T / be a nonzero proper but non-Hamiltonian root. Since n.L/ D 0, there is k 2 Fp for which Lk D Rk M .˛/ .T / L.0/ . This contradicts the former statement. We derive that any root in .L; T / is either Hamiltonian or improper Witt. (h) Let D a 1 C b 2 ¤ 0 and ‰.ˆ. 0 // D . Under the above mentioned isomorphism we have j

S0 . / Š span ¹z1i z2 @y1 ; z1k z2l @y2 j i raC1; j rb; k ra; l rbC1; r 2 Fp º: Suppose z1 D 1 C y1 , z2 D 1 C y2 and b ¤ 0. Then S0 . / D

p X1 i D0

F z1ia z2ibC1 @y2 C

p X1

F z1ia z2ib .bz1 @y1

az2 @y2 /

i D0

Pp 1 Pp 1 and i D0 F z1i a z2ibC1 @y2 Š W .1I 1/ and i D0 F z1ia z2ib .bz1 @y1 az2 @y2 / Š O.1I 1/ is an abelian ideal. Moreover, T acts on W .1I 1/ as the torus F z2 @y2 . Hence ˆ.T / acts on S0 . / as an improper torus. One argues similarly if a ¤ 0. Due Lemma 14.2.1 then all T -roots of L are improper. But T is a good torus. So this is impossible. This contradiction shows that we may take z2 D y2 . Putting D 2 yields that S0 Œ 2 Š W .1I 1/ and ˆ.T / normalizes the standard maximal subalgebra of W .1I 1/. (i) Suppose 2 is Witt. Since rad G. 2 / is a graded ideal of G. 2 / D ˚i Gi . 2 / and G0 Œ 2 is of Witt type, we must have the inclusion ˚i ¤0 Gi . 2 / rad G. 2 /. If ŒS 1 . 2 /; G1 . 2 / ¤ ¹0º, then it is a nonzero ideal of S0 . 2 / contained in rad S0 . 2 / Š O.1I 1/. Then there is k 2 Fp such that ¹0º ¤ ŒS 1; k 2 ; G1;k 2 D .F z1 @y1 C F z2 @2 / \ ker 2 . Since we have G1;k 2 D Œ‰.ˆ.T //; G1;k 2 S1;k 2 , it follows that ı.ŒS 1; k 2 ; S1;k 2 / ¤ 0 for some ı 2 .S; ‰.ˆ.T ///. However, the inclusion S˙1 . 2 / rad S. 2 / implies S˙1 . 2 / K˙1 . 2 /. Then ŒK

k 2 .S; ‰.ˆ.T ///; Kk 2 .S; ‰.ˆ.T ///

ŒS

1; k 2 ; S1; k 2

338

14


acts non-nilpotently on S. But this gives n.S; ˆ.T // ¤ 0 contradicting Theorem 14.1.1 applied to the simple algebra S . This contradiction shows ŒG 1 . 2 /; G1 . 2 / D ŒS 1 . 2 /; G1 . 2 / D ¹0º. As L D L. 1/ , we derive ŒL. 20 /; L.1/ . 20 / L.1/ . 20 /. In other words, L.1/ . 20 / is an ideal of L. 20 /. Recall that Der S.3I 1/.1/ D CS.3I 1/, so that dim.Der S/=.ad S/ D 4 (see [S-F, (4.8.6)]). Hence dim L. 20 /= rad L. 20 / dim L. 20 /=L.1/ . 20 / D dim S 1 . 2 / C dim G0 . 2 / .p

1/ C dim S0 . 2 / C 4

D 3.p C 1/ < p 2

2:

So 20 is not Hamiltonian. Therefore 20 2 .L; T / is Witt. Then LŒ 20 Š S0 Œ 2 Š W .1I 1/: As a consequence, L. 20 / D L.0/ . 20 / C rad L. 20 /. As 2 is a proper root on S0 , there is j 2 Fp such that ŒS0; j 2 ; S0;j 2 ker 2 . Then 20 .ŒL j 20 ; Lj 20 / D 0 holds. This implies that 20 is a proper root. Then 20 cannot be Witt. This contradiction shows that 2 2 .S; ‰.ˆ.T /// is Hamiltonian. We mentioned that S0 . 2 / is proper Witt. Proposition 13.4.4 now shows that 20 2 .L; T / is proper Hamiltonian. (j) The Lie algebra S.3I 1/ is the kernel of the map div W W .3I 1/ ! O.3I 1/;

3 X

fi @i 7!

i D1

3 X

@i .fi /:

i D1

It is easily seen that the map is t-invariant and has the property that div .W .3I 1/ / D O.3I 1/ for any nonzero t-root . As dim O.3I 1/ D p and dim W .3I 1/ D 3p for any 2 .O.3I 1/; t/ we have that dim S.3I 1/ D 2p. As t and ˆ.T / are both tori of maximal dimension in S.3I 1/.1/ and G D S.3I 1/.1/ for any nonzero , Theorem 10.7.1 and the definition of G imply that dim L D 2p for any 2 .L; T / n ¹0º. P Let 0 2 .L; T / n Fp 20 and M. 0 ; 20 / WD j 2Fp L0 Cj 20 . Then M. 0 / is a 2p 2 -dimensional L. 20 /-module. Now 20 is proper Hamiltonian and L.0/ Œ 20 Š G0 Œ 2 Š W .1I 1/: It follows that L.0/ . 20 / C rad L. 20 / ¤ L. 20 /. It also follows that L.0/ . 20 / does not map into H.2I 1/.0/ under the epimorphism L. 20 / ! LŒ 20 . This contradicts Lemma 13.4.2(2) finally proving the proposition.

339


.1/ .1/ Next we intend to show that S Š 6 H.4I P 1/ . Recall that H.4I 1/ has basis a1 a2 a3 a4 ¹DH .x1 x2 x3 x4 / j 0 ai < p; 0 < ai < 4.p 1/º: Moreover,

H.4I 1/ D H.4I 1/.1/ ˚ FDH ..x1 x2 x3 x4 /p p 1

˚ .F x1

1

p 1

@3 C F x2

/ p 1

@4 C F x 3

and Der H.4I 1/.1/ D H.4I 1/ ˚ F .

4 X

p 1

@1 C F x 4

@2 /

x i @i /

i D1

(see [S-F, Theorem 4.8.7]). It follows from this description (and Jacobson’s identity) that any Lie subalgebra of H.4I 1/ containing H.4I 1/.1/ is restricted. Lemma 14.2.5. Let g be a Lie algebra satisfying H.4I 1/.1/ g H.4I 1/, and t a toral element of g such that H.2I 1/.2/ Cg .t/= rad Cg .t/ H.2I 1/: Then dim Cg .t /= rad Cg .t/ p 2 1 and .rad Cg .t//.2/ ¤ ¹0º. Proof. As g=H.4I 1/.1/ is Œp-nilpotent, t 2 H.4I 1/.1/ . If t 2 H.4I 1/.1/ .0/ then Theorem 7.5.8 shows that there is 2 Aut H.4I 1/.1/ such that .t/ D DH .x1 x3 /C DH .x2 x4 / where ; 2 Fp . Otherwise we have to look into the proof of that theorem: set t1 WD t; in part (e) of the proof we state that there is (denoted there by ˆ ) such that .t/ D DH ..1 C x1 /x3 /. Now induces an automorphism .1/ of Der H.4I 1/.1/ , hence an automorphism of H.4I 1/ D Der H.4I 1/.1/ . So replacing g by its isomorphic copy .g/ we may assume in proving the lemma that either t D DH ..1 C x1 /x3 / or t D DH .x1 x3 / C DH .x2 x4 / where ; 2 Fp . If in the second case ; ¤ 0, then an easy computation shows that Cg .t/ H.4I 1/.0/ . Then Cg .t /= rad Cg .t/ is a homomorphic image of a subalgebra of sp.4/. Since this contradicts our assumption on Cg .t/, either D 0 or D 0 holds. No generality is lost by assuming D 1 and D 0. Thus t D DH .z1 x3 / where z1 2 ¹x1 ; 1 C x1 º. Since Œt; DH .z1a x2b x3c x4d / D .c whenever 0 a; b; c; d p

a/DH .z1a x2b x3c x4d /

1, we have

I WD CH.4I1/.1/ .t/ D span ¹DH .z1a1 x2a2 x3a1 x4a4 / j 0 ai < p; 0 < 2a1 C a2 C a4 < 4p

4º:

Note that I .1/ D span ¹DH .z1a1 x2a2 x3a1 x4a4 / j 0 ai < p; C.I / D span ¹DH .z1a x3a / j 0 < a < pº I .1/ ;

X

ai > 0; a2 C a4 < 2p

2º;

340

14


and I .1/ =C.I / Š H.2I 1/.2/ ˝ O.1I 1/ under the isomorphism which maps DH .z1a1 x2a2 x3a1 x4a4 / onto DH .x2a2 x4a4 / ˝ x a1 . As H.4I 1/.1/ is an ideal of g, I and C.I / are ideals of Cg .t/. Put I0 WD span ¹DH .z1a1 x2a2 x3a1 x4a4 / j 0 ai < p; a1 ¤ 0; 2a1 C a2 C a4 < 4p

4º

and observe that this is a nilpotent subalgebra which is invariant under I . p 1 Suppose I0 is not invariant under Cg .t/. Then there is an element d D r1 z1 @3 C p 1 p 1 p 1 r2 x2 @4 C r3 x3 @1 C r4 x4 @2 C r5 DH ..z1 x2 x3 x4 /p 1 / 2 Cg .t/ with r1 ¤ 0 and z1 D 1 C x1 . In this case I .1/ =C.I / is Cg .t/-simple. But then I .1/ \ .rad Cg .t// D C.I / and .p 2

2/p D dim I .1/ =C.I / dim Cg .t/= rad Cg .t/ dim H.2I 1/ D p 2 C 1;

a contradiction. Thus I0 is Cg .t/-stable, hence is contained in rad Cg .t/. Therefore dim Cg .t /= rad Cg .t/ dim I =I0 D dim span ¹DH .x2a2 x4a4 / j 0 ai < p; 0 < a2 C a4 º D p2

1:

Moreover, 4DH .z14 x34 / D ŒŒDH .z1 x22 x3 /; DH .z1 x3 x4 /; ŒDH .z1 x2 x3 /; DH .z1 x3 x42 / .2/

2 I0

.rad Cg .t//.2/ :

So .rad Cg .t //.2/ ¤ ¹0º as claimed.

Proposition 14.2.6. We have S 6Š H.4I 1/.1/ . Proof. (a) Suppose S D H.4I 1/.1/ . By Lemma 14.2.2 G can be identified with a subalgebra of Der S containing S . Note that in this case S is a restricted subalgebra of the p-envelope G of G in Der S (Theorem 7.2.2). By Theorem 7.1.2, Der H.4I 1/.1/ D CH.4I 1/. Since TR.G/ TR.L/ D 2 and TR.S/ D 2, it can only be that G =S is Œp-nilpotent. On the other hand, H.4I 1/ is a restricted subalgebra of Der H.4I 1/.1/ (Lemma 7.2.1(3),(4)) and Der H.4I 1/.1/ =H.4I 1/ is a 1-dimensional torus. Then it can only be that G is a subalgebra of H.4I 1/. According to Theorem 7.4.1, there is an automorphism 2 Autc O.4I 1/ such that ˆ .S / D S , the grading of S has type .a1 ; a2 ; a3 ; a4 / for some ai 2 Z satisfying a1 C a3 D a2 C a4 and some generating set .xi / DW ui 2 O.4I 1/.1/ , i D 1; : : : ; 4. Observing Theorem 7.3.6 we may adjust these to obtain ¹ui ; uj º D ¹xi ; xj º for all

341


P i; j so that ˆ .DH .f // D DH ..f // D .i/@ui ..f //@ui 0 holds. To keep the notation simple we shall assume that ui D xi , 1 i 4. Observe that DH .x1a x2b x3c x4d / D ax1a

1 b c d x2 x3 x4 @3

cx1a x2b x3c

1 d x4 @1

C bx1a x2b

1 c d x3 x4 @4

dx1a x2b x3c x4d

1

@2

has degree aa1 C ba2 C ca3 C da4

.a1 C a3 /:

./

Since S is a restricted subalgebra of Der S , so is S0 . Then ˆ.T / is a 2-dimensional torus of S contained in S0 (Lemma 13.7.4(2)). Let t D FDH .x1 x3 / ˚ FDH .x2 x4 /: Then t also is a 2-dimensional torus of S contained in S0 . Define 1 ; 2 2 t by setting 1 .DH .x1 x3 // D 1;

1 .DH .x2 x4 // D 0;

2 .DH .x1 x3 // D 0;

2 .DH .x2 x4 // D 1:

Obviously, DH .x1a x2b x3c x4d / is a weight vector for t corresponding to weight .c a/1 C.d b/2 . This shows that dim S D p 2 for any nonzero 2 .S; t/. Since S0 is a restricted subalgebra of Der S , ˆ 1 .S0 / is a restricted subalgebra of L.0/ . There is a 2-dimensional torus in L.0/ which maps onto t under the natural epimorphism ˆ 1 .S0 / ! S0 . In what follows we identify this torus with t and view the root system .L; t/ as a subset of Fp 1 C Fp 2 . Since H L.0/ , Theorem 10.7.1 (applied to the L.0/ -module L=L.0/ ) shows that CL .t/ L.0/ . Therefore CS .t/ ˚i 0 Si . As DH .x12 x32 / 2 CS .t/ \ Sa1 Ca3 , one has 0 a1 C a3 D a2 C a4 :

(14.2.4)

As before, we let H.4I 1/ denote the k-th component of the standard grading, i.e., the grading of type .1; 1; 1; 1/ with respect to x1 ; x2 ; x3 ; x4 It is easily seen that p 1

F x1 and

p 1

@3 C F x2

p 1

@4 C F x 3

p 1 p 1 p 1 p 1 x2 x3 x4 /

DH .x1

p 1

@1 C F x4

@2 H.4I 1/ :

Therefore G D ˚i;k2Z Gi \ H.4I 1/ : (b) Suppose a1 a2 a3 a4 ¤ 0. Then P@i 62 G0 for 1 i 4. This gives G0 \ H.4I 1/< 1> D ¹0º, whence G0 k0 H.4I 1/ . There is j 2 Z such that S 1 \ S<j > ¤ ¹0º and S 1 \ S D ¹0º for k > j . Recall that S 1 D G 1 is

342

14


P an irreducible and faithful G0 -module. As G0 k0 H.4I 1/ , the subspace S 1 \ S<j > is a G0 –submodule of G 1 . Therefore, G 1 D S 1 \ S<j > . P P SupposePj 0. By property (g2) of Notation 3.5.2, i , whence S 2 D ŒS 1 ; S 1 S< 2> D ¹0º and S k D ¹0º for k 2. Since DH .x12 x3 / 2 Sa1 and DH .x13 x3 / 2 S2a1 , we get Sa1 ¤ ¹0º, S2a1 ¤ ¹0º. Then a1 > 0, since there is only one graded component of negative degree. By symmetry we deduce ai > 0 for all i . Looking at @i 2 S ai , 1 i 4, we eventually obtain ai D 1 for all i . In other words, Gk D G for all k. Therefore, sp.4/ Š S0 G0 H.4I 1/ Š sp.4/: We have proved that L D L. 1/ , L.2/ ¤ ¹0º, L.0/ =L.1/ Š sp.4/ and L. 1/ =L.0/ is isomorphic to the natural 4–dimensional sp.4/-module. Theorem 5.2.7 yields L Š H.4I nI ‰/.2/ . As TR.L/ D 2, Corollary 10.1.8 gives L Š H.4I 1/.1/ contrary to the choice of L. Hence a1 a2 a3 a4 D 0. (c) Renumbering the ai if necessary we may assume that a1 D 0. Suppose a2 a3 a4 ¤ 0. Let H.4I 1/Œk denote the k-th component of the .0; 1; 2; 1/-grading with respect to x1 ; x2 ; x3 ; x4 . By the same reasoning as in (a), G D ˚i;j 2Z Gi \ H.4I 1/Œj : Observe that t S0 \ SŒ0 . It is easily seen that SŒ0 D span ¹DH .x1k x3 /; DH .x1k x22 /; DH .x1k x2 x4 /; DH .x1k x42 / j 0 k < pº; SŒ

1

D span ¹DH .x1k x2 /; DH .x1k x4 / j 0 k < pº;

SŒ

2

D span ¹DH .x1k / j 1 k < pº p 1

p 1

and SŒ i D ¹0º for i > 2. It is straightforward that x1 @3 2 DerŒ 2 S, x3 @1 2 p 1 p 1 DerŒ2p 2 S , x2 @4 ; x4 @2 2 DerŒp 2 S, and DH ..x1 x2 x3 x4 /p 1 / 2 DerŒ4p 6 S . P Hence DerŒ0 S D SŒ0 C F 4i D1 xi @i and DerŒ

1 S

D SŒ

1 ;

DerŒ

2 S

D SŒ

2

DerŒ

i S

D ¹0º

p 1

˚ F x1

@3 ;

for i > 2:

Note that, by (), DH .x1k / 2 S a3 , DH .x1k x2 / 2 S a4 , DH .x1k x4 / 2 S a2 for 0 P p 1 k < p, and x1 @3 2 Der a3 S . This gives G0 \ k j . Our preceding remark shows that S 1 \ SŒj is a nonzero G0 -submodule of G 1 D S 1 . The irreducibility of G 1 forces G 1 D S 1 \ SŒj . P As before, if j 0, then G i k0 SŒk for all i > 0. As DH .x2 /; DH .x4 / 2 SŒ 1 we obtain that DH .x2 / and DH .x4 / have positive degrees. Then a4 > 0 and a2 > 0 contrary to the equation a2 C a4 0. So j < 0. If j D 2, then S i SŒ 2 for all i > 0. Again this implies that a4 > 0 and a2 > 0. This case being impossible we must have j D 1. Then G 2 SŒ 2 and G 3 D ¹0º. Since DH .x2 / 2 S a4 n SŒ 2 and DH .x4 / 2 S a2 n SŒ 2 we have that a4 1 and a2 1. Since 0 a2 C a4 D a3 ¤ 0 and a2 a4 ¤ 0 one obtains a2 D a4 D 1 and a3 D a2 C a4 D 2. But then Si D SŒi for any i 2 Z and (since G 1 D S 1 by definition) Gi D Si D SŒi for i < 0. Therefore, S0 is isomorphic to a semidirect product of S00 WD span¹DH .x1k x3 / j 0 k < pº Š W .1I 1/ and the ideal I0 WD span ¹DH .x1k x22 /; DH .x1k x2 x4 /; DH .x1k x42 / j 0 k < pº: Note that I0 Š sl.2/ ˝ O.1I 1/ under the isomorphism which maps the elements DH .x1k x22 /; DH .x1k x2 x4 /; DH .x1k x42 / onto DH .x22 / ˝ x k ; DH .x2 x4 / ˝ x k ; DH .x42 / ˝ x k : Under this isomorphism S00 Š W .1I 1/ acts as derivations on the second tensor factor of I0 . Moreover, G 1 D SŒ 1 Š V .1/ ˝ O.1I 1/, ŒI0 ; SŒ 2 D 0, and G 2 D SŒ 2 Š O.1I 1/=F as .S0 =I0 /-modules (here V .1/ denotes the natural 2-dimensional sl.2/-module). Since S0 G0 H.4I 1/Œ0 D H.4I 1/.1/ Œ0 D S0 , we have that G0 D S0 . We note G0 Š sl.2/ ˝ O.1I 1/ Ì F Id ˝ W .1I 1/ ; G

1

Š V .1/ ˝ O.1I 1/;

G

2

Š O.1I 1/=F:

As in former cases (see Proposition 13.3.7) there is an automorphism ‰ of sl.2/ ˝ O.1I 1/ such that .‰ ı ˆ/.T / D F h ˝ 1 C F Id ˝ t0 ; where F h is a torus of sl.2/ and t0 2 ¹x@; .1 C x/@º. (d) Using our discussion in (c) it is easy to observe that .G 2 ; t/ D Fp 1 , .G 1 ; t/ D Fp 1 ˙ 2 and .S0 ; t/ D Fp 1 [ .Fp 1 ˙ 22 /.

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14


Let 2 .L; t/ n .Fp 1 [ Fp 2 /. Then S0 . / is a solvable subalgebra of codimension 2 in S. / and therefore L.0/ . / C rad L. / is a solvable subalgebra of codimension 2 in L. /. By Theorem 4.2.6, is not Hamiltonian. If it is Witt, Proposition 7.6.8 shows that this subalgebra is contained the standard maximal subalgebra of codimension 1. As L.0/ . / contains T , by definition then is proper. Hence such is proper in any case. Next we observe that span ¹DH .x1a x3b / j 0 a; b < p; 0 < a C b < 2p

2º

is a subalgebra of G.1 / isomorphic to H.2I 1/.2/ . As a consequence, GŒ1 D G.1 /= rad G.1 / 62 ¹¹0º; sl.2/; W .1I 1/º. Then H.2I 1/.2/ P GŒ1 H.2I 1/ p 1 FDH .x1k x2 x4 / (Corollary 11.2.4). Besides, t stabilizes both rad G0 .1 / D kD0 Pp 1 P p 1 and the standard maximal subalgebra kD1 FDH .x1k x3 / of kD0 FDH .x1k x3 / Š G0 .1 /=rad G0 .1 /. Applying Proposition 13.4.4 we now derive that 1 is a proper Hamiltonian root of L. Similarly, span ¹DH .x2a x4b / j 0 a; b < p; 0 < a C b < 2p 2º is a subalgebra of G.2 / isomorphic to H.2I 1/.2/ . Arguing as before we derive that 2 is proper. Thus all roots in .L; t/ are proper. Proposition 13.7.1(2) and Proposition 13.7.2 show that t 2 T .L/. As T is a good torus, all roots in .L; T / are proper as well. (e) Define a root 2 .G; .‰ ı ˆ/.T // by

.h ˝ 1/ D 0;

.Id ˝ t0 / D 1;

and 0 2 .L; T / by .‰ ı ˆ/. 0 / D . Then G 2 D G 2 . /, G0 D G0 . / C I0 . It is immediate that I0 . / Š F h ˝ O.1I 1/ is an abelian ideal of G0 . / and G0 . /=I0 . / Š W .1I 1/. Suppose ŒG 2 . /; G2 . / I0 . /. Then ŒG 2 ; G2 D ŒG 2 . /; G2 I0 . / C

X

S0; D I0 :

… Fp

However, ŒDH .x1 /; DH .x1k x32 / D 2DH .x1k x3 / 62 I0 ;

0kp

1:

This contradiction shows that ŒG 2 . /; G2 . / C I0 . / D S0 . / holds. Therefore ŒG 2 . /; G2 . / is a non-solvable subalgebra of G. /, whence G 2 . / 6 rad G. /. As rad G. / is a graded ideal of G. / and G0 \ rad G. / I0 . /, we obtain dim GŒ > p. Since gr.rad L. 0 // rad G. /, we obtain dim LŒ 0 > p. Hence

0 2 .L; T / is a Hamiltonian root. Recall that all T -roots are proper and L.0/ . 0 /= rad L.0/ . 0 / Š G0 . /= rad G0 . / Š W .1I 1/:

345


The latter yields that L. 0 / ¤ L.0/ . 0 / C rad L. 0 /. Since all t-root spaces of S for nonzero roots are p 2 -dimensional, Theorem 10.7.1 implies that for any 0 2 .L; T /n Fp 0 , X X dim L0 Cj 0 D dim SCj D p 3 : j 2Fp

j 2Fp

P Lemma 13.4.2(2) now says that every L. 0 /-module M. 0 ; 0 / WD j 2Fp L0 Cj 0 with 0 2 .L; T / n Fp 0 is irreducible. As G 2 D G 2 . / Š O.1I 1/=F , it is an irreducible S0 . /-module. We mentioned earlier that ŒGP2 ; G2 . / 6 I0 . / D rad G0 . /. From this it follows that rad G. / I0 . / C i>0 Gi . /. As gr.rad L. 0 // is a solvable ideal of gr L. 0 / D G. / and I0 . / is abelian, we must have .rad L. 0 //.1/ L.1/ . 0 /:

rad L. 0 / L.0/ . 0 /;

As a consequence, .rad L. 0 //.1/ acts nilpotently on L. But then .rad L. 0 //.1/ annihilates each M. 0 ; 0 /. Therefore, .rad L. 0 //.1/ annihilates L (Proposition 1.3.5). This means that rad L. 0 / is abelian. Next we need some subtle dimension estimates. Due to Lemma 14.2.5 we have dim G. /= rad G. / p 2

1 D dim H.2I 1/

2 dim LŒ 0

2

and this gives dim rad G. / dim rad L. 0 / C 2. Then dim rad G. /= gr.rad L. 0 // 2: By Corollary 11.2.4 (applied to G) rad G. / is nilpotent. The dimension estimate then implies that rad G. /= gr.rad L. 0 // is abelian. One the other hand, we have shown that rad L. 0 / is abelian, whence gr.rad L. 0 // is abelian. Then .rad G. //.2/ D ¹0º contrary to Lemma 14.2.5. (f) Suppose a1 D a2 a3 a4 D 0. First we assume that a2 a4 ¤ 0. Then a3 D 0 and a2 C a4 D a1 C a3 D 0, i.e., .a1 ; a2 ; a3 ; a4 / D .0; m; 0; m/ for some nonzero m 2 Z. It follows that Der0 S D span ¹DH .x1a x2k x3b x4k / j 0 a; b; k < p; a C b C k > 0º p 1

C F x1

p 1

@3 C F x3

@1 C F

4 X

xi @i :

iD1

From this it is immediate that p 1 b p 1 x3 x4 /

J WD span ¹DH .x1a x2

j 0 a; b < p; 0 < a C b < 2p

2º

is an ideal of Der0 S contained in S0 . Since J consists of nilpotent elements of S , it must annihilate the irreducible G0 -module S 1 . This, however, is impossible as S 1 is a faithful G0 -module.

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14


Hence either a2 D 0 or a4 D 0. Renumbering x2 and x4 if necessary we may assume that a2 D 0. Then .a1 ; a2 ; a3 ; a4 / D .0; 0; n; n/ where n 2 Z. As S 1 ¤ ¹0º, n 2 ¹˙1º. As a1 C a3 0, n D 1, i.e., .a1 ; a2 ; a3 ; a4 / D .0; 0; 1; 1/. So we have S0 D span ¹DH .x1a x2b x3 /; DH .x1a x2b x4 / j 0 a; b < pº; S

1

D span ¹DH .x1a x2b / j 0 a; b < p; a C b > 0º;

and S k D ¹0º for k 2. Then S0 is isomorphic to W .2I 1/ under an isomorphism which maps DH .x1a x2b x3 / D ax1a

1 b x2 x3 @3

C bx1a x2b

1

x3 @4

x1a x2b @1 onto

x1a x2b @1 ;

DH .x1a x2b x4 / D ax1a

1 b x2 x4 @3

C bx1a x2b

1

x4 @4

x1a x2b @2 onto

x1a x2b @2 :

Also, L D L. 1/ (as G k D S k D ¹0º for k > 1) and .G 1 ; t/ D .Fp 1 ˚ Fp 2 / n ¹0º. Moreover, dim G 1;ı D 1 for any ı 2 .G 1 ; t/. By Corollary 7.5.2, there is an isomorphism ‰ W S0 ! W .2I 1/ such that ‰.ˆ.T // D F z1 @y1 ˚ F z2 @y2 where .z1 ; z2 / 2 ¹.1 C y1 ; 1 C y2 /; .1 C y1 ; y2 /; .y1 ; y2 /º P j (in order to avoid confusion we write W .2I 1/ D F y1i y2 @yk ). Define 1 ; 2 2 ‰.ˆ.T // by setting i .zj @j / D ıi;j and i0 2 T by ‰.ˆ. i0 // D i , where i; j 2 ¹1; 2º. By the above description of G 1 and Theorem 10.7.1 it follows that .L=L.0/ ; T / D Fp 10 ˚ Fp 20 n ¹0º. Let 2 .L; T / be a nonzero proper but non-Hamiltonian root. Since n.L/ D 0, there is k 2 Fp for which Lk D Rk M .˛/ .T / L.0/ . This contradicts the former statement. We derive that any nonzero root in .L; T / is either Hamiltonian or improper Witt. Next we apply the arguments presented in part (h) of the proof of Proposition 14.2.4 word-by-word and obtain that we may take z2 D y2 and that ˆ.T / acts on S0 Œ 2 Š W .1I 1/ as a proper torus. Then we apply part (i) of that proof word-byword with the only change that dim.Der H.4I 1/.1/ /=H.4I 1/.1/ D 6. We obtain that

20 2 .L; T / is proper Hamiltonian. By the above discussion, dim G 1 . 2 / D p 1. Combining this with Theorem 11.4.1 one deduces that G 1 \ rad G. 2 / D ¹0º. A straightforward compup 1 p 1 p 1 p 1 tation shows that x1 @3 ; x2 @4 2 Der 1 S , x3 @1 ; x4 @2 2 Derp 1 S, and p 1 p 1 p 1 p 1 DH .x1 x2 x3 x4 / 2 Der2p 3 S.PTherefore G0 D S0 . As a consequence, one obtains rad G. 2 / rad S0 . 2 / C i>0 Gi . We observe that rad S0 . 2 / D ‰ 1 .span ¹z1 y2i @y1 j 0 i < pº/ is abelian. As a consequence, .rad G. 2 //.1/ P 0 0 i >0 Gi . As gr.rad L. 2 // is a solvable ideal of gr.L. 2 // D G. 2 / we now obtain rad L. 20 / L.0/ . 20 /;

.rad L. 20 //.1/ L.1/ . 20 /:

Arguing as in (e) one derives for every 0 2 .L; T / n Fp 20 that the L. 20 /-module M. 0 ; 20 / is irreducible, and then that .rad L. 20 //.1/ D ¹0º. The word-by-word analogue of the final part of (e) shows that the present case is impossible, and this contradiction finally proves the proposition.

347


Since 6 6 0 .mod .p//, the Lie algebra K.3I 1/ is simple and Der K.3I 1/ Š K.3I 1/ (see [S-F, Theorems 4.5.5 and 4.8.8]). In particular, K.3I 1/ is a restricted Lie algebra. Recall that K.3I 1/ is the image of O.3I 1/ under the linear isomorphism DK W O.3I 1/ ! K.3I 1/ given by DK .x a / D .a1 x1a1 C.

1 a2 a3 x2 x3

C a3 x1a1 x2a2 C1 x3a3

a2 x1a1 x2a2 1 x3a3

C

1

/@2

a3 x1a1 C1 x2a2 x3a3 1 /@1

for all a D .a1 ; a2 ; a3 / with 0 ai p is easy to deduce that

C .2

a2 /x a @3

a1

1. Using Equations (4.2.10) and (4.2.8) it

ŒDK .x1a1 x2a2 x3a3 /; DK .x1b1 x2b2 x3a3 / D .a1 b2

a2 b1 /DK .x1a1 Cb1

C .a3 .b1 C b2

2/

1 a2 Cb2 1 a3 Cb3 x3 / x2

b3 .a1 C a2

(14.2.5)

2//DK .x1a1 Cb1 x2a2 Cb2 x3a3 Cb3

1

/:

Proposition 14.2.7. We have S 6Š K.3I 1/. Proof. (a) Suppose S D K.3I 1/. By Lemma 14.2.2, G can be identified with a subalgebra of Der S D S containing S, whence G D S . According to Theorem 7.4.1, there is an automorphism 2 Autc O.3I 1/ such that ˆ .S/ D S, the grading of S has type .a1 ; a2 ; a3 / for some ai 2 Z satisfying a1 C a2 D a3 and some generating set .xi / DW ui 2 O.3I 1/.1/ , i D 1; 2; 3. Observing Theorem 7.3.8 we may adjust these to obtain ˆ .DK .f // D DK ..f //. To simplify notation we may and shall assume that ui D xi , 1 i 3. So in the present grading of S , deg DK .x c / D .c1 C c3

1/a1 C .c2 C c3

1/a2 :

(14.2.6)

Let t WD FDK .x1 x2 / ˚ FDK .x3 /. This is aP2-dimensional torus in S contained P in S0 . Let h WD CS .t/. Since CS .ˆ.T // i 0 Si we have that h i 0 Si (by Theorem 10.7.1). Using the commutator relation (14.2.5) and Equation (14.2.6) pC1

pC1

pC1

we obtain DK .x1 2 x2 2 x3 2 / 2 h \ Sp.a1 Ca2 / . This implies that a1 C a2 0. Renumbering x1 and x2 if necessary we may assume that a1 a2 . Hence a1 ja2 j 0. Obviously, .a1 ; a2 / ¤ .0; 0/. As a consequence, a1 > 0. (b) Suppose a2 D 0. As S 1 ¤ ¹0º, we must have a1 D 1. Then S0 D span ¹DK .x1 x2i /; DK .x2i x3 / j 0 i < pº and S

1

D span ¹DK .x2i / j 0 i < pº;

S

k

D ¹0º if k > 1:

Using the commutator relations mentioned above one readily verifies that S00 WD span ¹DK .x1 x2i / j 0 i < pº

348

14


is a subalgebra of S0 isomorphic to W .1I 1/, I0 WD span ¹DK .x2i x3

x1 x2i C1 / j 0 i < pº

is an abelian ideal of S0 isomorphic to O.1I 1/ as a module over S00 , and S0 D S00 Ë I0 . In particular, C.S0 / D FDK .x1 x2 x3 /. Define 1 ; 2 2 t by setting 1 .DK .x1 x2 // D 1;

1 .DK .x3 // D 0;

2 .DK .x1 x2 // D 0;

2 .DK .x3 // D 1:

p 1

Then S 1 D ˚i D0 S 1; i.1 C2 / 22 with dim S 1; P i.1 C2 / 22 D 1 for each i 2 Fp . Moreover, S0 D S0 .1 C 2 / and S.1 C 2 / i 0 Si . Since S0 Š Der0 S is a restricted subalgebra of S , the restricted Lie algebra L.0/ contains a 2-dimensional torus which maps isomorphically onto t under the natural homomorphism L.0/ ! S0 . As before, we identify this torus with t. By our remarks earlier in the proof, L.0/ D L.1 C 2 / C L.1/ and 1 C 2 is a proper Witt root in .L; t/. Then t 2 T .L/ (see Proposition 13.7.1). Furthermore, Li.1 C2 /Cj2 L.1/ if j 62 ¹0; 2º and dim Li.1 C2 /

22 =L.0/; i.1 C2 / 22

D 1:

As a consequence, for any 2 .L; t/ n Fp .1 C 2 /, the subalgebra L.0/ ./ is t-invariant, solvable, and has codimension 1 in L./. This implies that any root in .L; t/ n Fp .1 C 2 / is proper. As 1 C 2 is proper Witt, all t-roots of L are proper. Since T is a good torus, all roots in .L; T / are proper as well. By the maximality of T , C.S0 / ˆ.T /. In other words, ˆ.T / D F t ˚ FDK .x1 x2 x3 / where X X i DK .x2i x3 x1 x2i C1 /: tD i DK .x1 x2i / C i 0

i 0

Define 2 .S; ˆ.T // by setting .DK .x1 x2 x3 // D 0 and 0 2 .L; T / by ˆ. 0 / D . Since DK .x1 x2 x3 / acts invertibly on S 1 D L=L.0/ and trivially on S0 D L.0/ =L.1/ , L. / D L.0/ . /

and

L.0/ . /=L.1/ . / Š S0 :

Let S denote the k-th component of the natural grading of S Š K.3I 1/ with P respect to x1 ; x2 ; x3 (it has type .1; 1; 2/) and S.k/ D i k S denote the k-th component of the induced filtration. Since is proper, ˆ.T / stabilizes the preimage of span ¹DK .x1 x2i / j 1 i < pº under the canonical projection S0 D S00 ˚ I0 ! S00 . This implies 0 D 0. This gives t 2 S0 \ S.0/ , hence ˆ.T / S0 \ S.0/ . Note that p 1 p 1 S 1 \ S.p 3/ D DK .x2 /. It follows that DK .x2 / is a root vector of S relative to ˆ.T /. Consequently, there exist 2 .L; T / and u 2 L such that gr0 u D p 1 DK .x2 /. We mentioned before that all t-root spaces of S 1 are 1-dimensional.

349


Then so are all T -root spaces of L=L.0/ . Therefore .L=L.0/ / D F .u C L.0/ /. Since DK .x1 x2 x3 / acts on S 1 D L=L.0/ as 2Id, we also have L./ D F u C H C p 1 L.1/ ./, so that L L.1/ . This means that ˆ.Œu; L / D ŒDK .x2 /; S1 / ŒS.p 3/ ; S.0/ S.p 3/ (as S1 S.0/ ). This gives u 2 R M .˛/ .T / L.0/ . This contradiction shows that a2 ¤ 0. (c) Suppose a1 ¤ ja2 j. In this case none of DK .1/, DK .x1 /, DK .x2 /, DK .x12 /, DK .x22 / has degree 0. Then S0 FDK .x1 x2 / C FDK .x3 / C S.1/ . This implies that S0 Š G0 is solvable, contrary to Corollary 9.1.10 and the general assumption. Thus either a1 D a2 or a1 D a2 . (d) Suppose a1 D a2 . Then S0 D span ¹DK .x1i x2i x3k / j 0 i; k < pº: Note that j

j

ŒDK .x1i x2i x3k /; DK .x1 x2 x3l / D 2.k.j Hence

1/

p 1 p 1 i x2 x3 /

J0 WD span ¹DK .x1

i Cj

l.i

1//DK .x1

i Cj

x2

x3kCl

1

/:

j 0 i < pº

is an ideal of S0 . As J0 S.6/ , it acts nilpotently on S. By Engel’s theorem, J0 must act trivially on S 1 , contrary to the fact that S 1 is a faithful S0 -module. So the case a1 D a2 is impossible. (e) It remains to consider the case a1 D a2 > 0. Since S 1 ¤ ¹0º we have a1 D 1, that is Si D S for all i . But then Theorem 5.2.7 gives L Š K.3I nI ‰/.1/ , a contradiction. Proposition 14.2.8. There holds S 6Š H.2I .1; 2//.2/ . Proof. (a) Suppose S D H.2I .1; 2//.2/ . By Lemma 14.2.2, one has M.G/ D ¹0º. As a consequence, we can identify G with a subalgebra of Der S containing S. According to Theorem 7.4.1, there is an automorphism 2 Autc O.2I .1; 2// such that ˆ .S / D S , the grading of S has type .a1 ; a2 / for some ai 2 Z and some generating set .xi / DW ui , i D 1; 2. Observing Theorem 7.3.6 we may adjust these to obtain ¹ui ; uj º D ¹xi ; xj º for all i; j so that ˆ .DH .f // D DH ..f // holds. For simplicity of notation we may and shall assume that ui D xi , i D 1; 2. By Theorem 7.1.2, .p 1/ .p 2 1/ x2 /

Der S D H.2I .1; 2//.2/ C FDH .x1

.p 2 /

C FDH .x2

.p/

C FDH .x1 /

p

/ C F @2 C F .x1 @1 C x2 @2 /:

We denote by Derk S (resp., Der S) the k-th component of the .a1 ; a2 /-grading (resp., the natural .1; 1/-grading) of Der S .

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14

Suppose a1 D a2 . As S

1


¤ ¹0º, we then have that a1 2 ¹˙1º and

.2/

.2/

S0 D FDH .x1 / C FDH .x1 x2 / C FDH .x2 / D S Š sl.2/: Hence S0 is a restricted subalgebra of Der S . Then ˆ.T / S0 Š sl.2/ (Lemma 13.7.4), a contradiction. Suppose a1 ¤ a2 and 0 62 ¹a1 ; a2 º. Then DH .x1 /, DH .x2 /, DH .x12 /, DH .x22 / have nonzero degrees in Der S . This implies that S0 FDH .x1 x2 /CH.2I .1; 2//.2/.1/ is solvable. Since ˆ.T / S0 , we have G0 D S0 C CG0 .ˆ.T //. Then G0 is solvable, and so is L.0/ contrary to Corollary 9.1.10 and the general assumption. .i/

Suppose a1 D 0. Then S0 D span ¹DH .x1 x2 / j 0 i < pº Š W .1I 1/. Note p that DH .x2 /p D @1 D 0 and .i/

span ¹DH .x1 x2 / j 1 i < pº D .Der0 S/ \ H.2I .1; 2//.2/ .0/ : Since the latter is a restricted subalgebra of Der S (see Lemma 7.1.1(3)), it follows from Jacobson’s formula that S0 is a restricted subalgebra of Der S . Then again ˆ.T / S0 , by Lemma 13.7.4. But W .1I 1/ has no 2-dimensional tori. (b) It follows that a2 D 0. As S 1 ¤ ¹0º, we must have a1 2 ¹˙1º. In any event, .i/

S0 D span ¹DH .x1 x2 / j 0 i < p 2 º Š W .1I 2/: According to Theorem 7.6.3(1) there is a semisimple element u0 2 W .1I 2/ which generates a 2-dimensional torus in W .1I 2/Œp . Let t DW DH .x1 q/ be its image in S0 , Pp 2 1 .i/ where q D i D0 ci x2 . Note that u0 62 W .1I 2/.0/ and therefore c0 ¤ 0. At this point it is not quite clear if t itself is semisimple in S , however, a suitable Œp-th power s t Œp is so and generates a 2-dimensional torus t. Then annS 1 t annS 1 t. As S 1 is a restricted t-module, Theorem 10.7.1 shows that annS 1 t D ¹0º. Suppose a1 D 1. Then DH ..x1 q/2 / 2 S 1 n ¹0º (for c0 ¤ 0) and Œt; DH ..x1 q/2 / D ŒDH .x1 q/; DH ..x1 q/2 / D 0: This gives annS 1 t ¤ ¹0º, which contradicts the preceding result. .i/ Thus a1 D 1, so that S 1 D span ¹DH .x2 / j 1 i < p 2 º and S k 2. It follows that L D L.

1/

and

S

1

k

D ¹0º for

Š O.1I 2/=F as S0 -module:

Theorem 10.7.3 then shows that j.S 1 ; ˆ.T //j D p 2 1. As a consequence, .L=L.0/ ; T / Fp for any nonzero 2 .L; T /. Therefore, 0 ¤ dim Li =Ri 2 dim Li =Ki for all i 2 Fp by Lemma 13.1.1. We deduce that any root in .L; T / is either Hamiltonian or improper Witt.

351


(c) Let 2 .G; ˆ.T // be such that S0 .// Š W .1I 1/. Define 0 2 .L; T / by ˆ. 0 / D . If ŒS 1 ./; G1 ./ ¤ ¹0º, then ŒS 1 ./; G1 ./ D S0 ./ (because S0 ./ is simple), whence S 1 ./ 6 rad G./. As rad G./ is a graded ideal of G./, we then have dim GŒ > dim S0 ./ D p. Using Corollary 11.2.4 we derive that H.2I 1/.2/ GŒ H.2I 1/. Now Proposition 13.4.4 yields that 0 is Hamiltonian and, moreover, 0 is proper if and only if ˆ.T / acts on S0 ./ as a proper torus. If ŒS 1 ./; G1 ./ D ¹0º, then L.1/ . 0 / is an ideal of L. 0 / (for L. 0 / D L. 1/ . 0 / by (b)). As S has codimension 5 in Der S (see (a)), dim L. 0 /= rad L. 0 / dim L. 0 /=L.1/ . 0 / D dim S 1 ./ C dim G0 ./ .p

1/ C dim S0 ./ C 5 D 2p C 4 < p 2

2:

In view of the final remark in (b), 0 is improper Witt. Then .L.0/ . 0 / C rad L. 0 //= rad L. 0 / Š S0 ./ Š W .1I 1/ Š LŒ 0 D L. 0 /= rad L. 0 /; hence L. 0 / D L.0/ . 0 / C rad L. 0 /. Therefore, the equality ŒS 1 ./; G1 ./ D ¹0º implies that S0 ./ is an improper section of S0 . (d) We now view ˆ.T / as a 2-dimensional torus in S0 Š W .1I 2/Œp . If it is as in Theorem 7.6.2, then j.S0 ; ˆ.T //j D p 2 and all nonzero roots are improper. But then all T -roots of L improper (Lemma 14.2.1) which is not true. The torus cannot be as in Theorem 7.6.4 by dimension reasons. Therefore it is as in Theorem 7.6.5. Then dim S0;ˆ./ D 1 for any ˆ./ 2 .S0 ; ˆ.T // and each ˆ./ 2 .S0 ; ˆ.T // satisfying ˆ./.ˆ.T / \ S0 / ¤ 0 is proper Witt. Let ˆ./ 2 .S0 ; ˆ.T // be such. By (c) is a Hamiltonian proper T -root of L. P Let 2 .L; T / n Fp and let W be a composition factor of the L./-module j 2Fp LCj . As ˆ.T / S0 , dim W dim S D p 3 2. As L.0/ ./= rad L.0/ ./ Š G0 .ˆ.//= rad G0 .ˆ.// contains an ideal isomorphic to W .1I 1/ and is Hamiltonian, we have that L./ ¤ L.0/ ./ C rad L./. It also follows that L.0/ ./ does not map into H.2I 1/.0/ under the epimorphism L./ ! LŒ. This contradicts Lemma 13.4.2(2) and thereby proves the proposition. Next we are going to rule out the possibility that S Š H.2I 1I ˆ.1//. The restricted subalgebra S of Der S generated by S is S D S C F .x1 @1 C x2 @2 / D Der S X D F .a1 x1a1 1 x2a2 @2 a2 x1a1 x2a2 0a.1/

C F .x1 @1 C x2 @2 / W .2I 1/ (Theorems 6.3.10(2), 7.1.3(4)).

1

@2 /

p 1 a1 a2 x1 x2 @2

x1

352

14


Proposition 14.2.9. There holds S 6Š H.2I 1I ˆ.1//. Proof. (a) Suppose S D H.2I 1I ˆ.1//. We mentioned above that Der S D S and t WD F x1 @1 ˚ F x2 @2 is a 2-dimensional self-centralizing torus such that S D t C S. It is easily seen that all root spaces of S relative to t are 1-dimensional and j.S; t/j D p 2 . Also, dim S D p 2 C 1. Keeping all this in mind and using Theorem 10.7.1 we deduce that S D ˆ.T / C S, dim S D 1 for any 2 .S; ˆ.T // and j.S; ˆ.T //j D p 2 . N that is By Lemma 14.2.2, G Š G, S G Der S D S D ˆ.T / C S: Since ˆ.T / is a Cartan subalgebra of S, we have that ˆ.H / D CG .ˆ.T // ˆ.T /. If ˆ.T / S , then S D S would be restricted, which is not true. Hence there is h 2 T \ L such that ˆ.T / \ S D F ˆ.h/. (b) Let ı denote the degree derivation of the graded Lie algebra S D ˚i Si . Since Œı; ˆ.T / D ¹0º and ˆ.T / is a maximal torus of S D Der S we must have ı 2 ˆ.T /. Choose independent roots ; 2 ˆ.T / satisfying .ı/ D 0 and .ı/ D 1. Then .1/ G0 D G0 ./ and Gi ˚j 2Fp GiCj for any i 2 Z. If ˆ.T / \ S0 F ı, then .1/

S0 is nilpotent by the Engel–Jacobson theorem. But then L.0/ would be solvable contradicting the general assumption. Hence we have ˆ.T / D F ı ˚ F ˆ.h/;

Œˆ.h/; S0 ¤ ¹0º;

.1/

ˆ.h/ 2 S0 :

Let i 2 Z be such that Gi ¤ ¹0º and i 6 0 .mod p/. Since dim G D 1 for each ˆ.T /-root and .Gi ; ˆ.T // i C Fp , we have that dim Gi p. Since .1/ h 2 S0 , it follows that 0 D trace adGi h D i.h/ dim Gi . Then dim Gi D p (as .h/ ¤ 0). We obtain that .Gi ; ˆ.T // D i C Fp . (c) Suppose G 2 ¤ ¹0º. Then .G 1 ; ˆ.T // D C Fp and .G 2 ; ˆ.T // D 2 C Fp . Since dim L D dim G D 1 for any root , .L; T / consists of non-Hamiltonian roots. Let 2 .L; T / n Fp . Then Fp intersects with both .G 1 ; ˆ.T // and .G 2 ; ˆ.T //. Since dim Lj =Rj 2 dim Lj =Kj for all j 2 Fp and R L.0/ , one has dim Lj =Kj ¤ 0. This is only possible if all roots

2 .L; T / n Fp are improper Witt. Since G1 ¤ ¹0º, .G1 ; ˆ.T // D C Fp . We claim that G2 D ¹0º. If this is not the case, then .G2 ; ˆ.T // D 2 C Fp . Let 2 .G2 ; ˆ.T //. Then ˙ 21 2 .G˙1 ; ˆ.T //. Since is improper Witt, .ad L 1 /p 3 .L / ¤ ¹0º. Since dim Lj D 1 for any j 2 Fp , .ad L 1 /p 3 .L / D 2

2

L 1 . But then L 1 L.1/ , whence 21 62 .G 1 ; ˆ.T // (for dim L 1 D 1/. 2 2 2 This contradiction proves the claim. Since L is simple, L.2/ D ¹0º, whence L.1/ D ˚j 2Fp L.1/;Cj . Since R.L; T / contains a nonzero T -homogeneous sandwich (Lemma 13.3.2) there is s 2 Fp such

353


that .ad LCs /2 .L s / D ¹0º (as dim LCj D 1 for all j 2 Fp ). This, however, is impossible because C s is improper Witt. P (d) Thus G 2 D ¹0º, i.e., L D L. 1/ . It follows that L D . j 2Fp L Cj / C L.0/ and L.0/ D L./CL.1/ . Recall that CG .ˆ.T // ˆ.T /. Therefore, H \L.1/ D ¹0º. The simplicity of L implies (Proposition 1.3.5) X X X H D ŒL ; L D ŒL Cj ; L j C ŒL.1/; ; L.1/;

62Fp

D

X

ŒL

j 2Fp Cj ; L j

62Fp

C L.1/ D F h C L.1/ :

j 2Fp

As a consequence, T \ L D H D F h is 1-dimensional. Let t0 L.0/ be a 2-dimensional torus which maps onto t D F x1 @1 ˚ F x2 @2 under ˆ. Theorem 10.7.1 implies that CL .t0 / D t0 \ L is 1-dimensional. Then it is easy to see that ˆ.t0 \ L/ D F .x2 @2 x1 @1 /. Choose t 0 2 t0 \ L for which ˆ.t 0 / D x2 @2 x1 @1 . Obviously, t 0Œp t 0 2 ker ˆ, and this means that t 0Œp t 0 acts nilpotently on L. Then all roots 2 .L; t0 / satisfy .t 0 / 2 Fp , and from this we conclude that there is 2 .L; t0 / for which .t 0 / D 0. But then L. / is a nilpotent 1-section, whence a CSA of toral rank 1. Due to Theorem 9.2.11 and the general assumption this is impossible. Proposition 14.2.10. We have S 6Š H.2I 1I ˆ.//.1/ and S 6Š W .1I 2/. Proof. (a) Suppose S Š H.2I 1I ˆ.//.1/ . Due to Theorem 10.3.2 every nonzero root vector u 2 S with respect to any 2-dimensional torus in S satisfies uŒp ¤ 0 and uŒp is semisimple. Since ˆ.T / is homogeneous of degree 0, this observation shows that S D S0 is trivially graded. But this is not true. (b) Suppose S Š W .1I 2/. If ˆ.T / S is as in Theorem 7.6.2, then we argue as in (a) to derive a contradiction. Hence ˆ.T / is as in Theorem 7.6.5. Then dim W .1I 2/ D 1 for all 2 .L; T / and this implies j.L; T /j D dim W .1I 2/ D p 2 . Moreover, there is a root satisfying ˆ.T / \ W .1I 2/ D 0, and (cf. Theorem 7.6.5(4)) every non-zero root vector in S. / acts non-nilpotently. This can only be if S. / S0 and ˆ.T / 6 S . The grading of S gives rise to a toral derivation of S , which annihilates ˆ.T /. Note that ˆ.T / is a maximal torus in S Š Der S (Theorem 7.2.2). Therefore there is a toral element t 2 ˆ.T / which satisfies adSi t D i IdSi for all i . Then ˆ.T / D F t ˚ˆ.T /\W .1I 2/ by dimension reasons, and this implies that S0 D S. /. Next we observe that GN Der S D S C ˆ.T /. Then G0 Š S0 C ˆ.T / D S. / C ˆ.T /. Due to Theorem 7.6.5(4) the latter is solvable. But then L.0/ is solvable. This contradicts Corollary 9.1.10 and the general assumption. We summarize: Let L be a simple Lie algebra of absolute toral rank 2 which is not of classical, Cartan or Melikian type. There exist admissible triples for L in

354

14


which the torus is good. Then gr L=M.gr L/ has a unique minimal ideal S and S is a simple Lie algebra of absolute toral rank 2. It is neither classical (Proposition 13.7.7) nor Melikian (Proposition 13.7.8). The classification of Cartan type Lie algebras of absolute toral rank 2 is given in Theorem 10.6.3. In the present section we have dealt with these candidates and showed that S cannot be of this form. As a consequence, if L is a counterexample to the Classification Theorem, then so is S. It will turn out to be a big relief dealing with graded rather than filtered counterexamples.

14.3

Graded counterexamples

In this final section we investigate graded counterexamples of the Classification Theorem. Set S1 the set of all simple Lie algebras g of absolute toral rank 2 which satisfy (a) g is not of classical, Cartan or Melikian type, (b) dim g is minimal with respect to algebras satisfying (a); S2 the set of all pairs .g; t/, where (a) g 2 S1 , (b) t is a 2-dimensional non-rigid torus in gŒp , (c) jimp .g; t/j is minimal with respect to all pairs satisfying (a), (b); S3 the set of all triples .g; t; ˛/, where (a) .g; t/ 2 S2 , (b) ˛ 2 .g; t/ is proper, (c) dim M .˛/ .g; t/ is maximal with respect to all triples satisfying (a), (b); S4 the set of all quadruples .g; t; ˛; g.0/ /, where (a) .g; t; ˛/ 2 S3 , (b) g D ˚ gi is a graded algebra and this grading satisfies (g1)–(g4) of Notation 3.5.2, (c) t gŒp is homogeneous of degree 0, P (d) g.0/ D i 0 gi is a maximal subalgebra, (e) M .˛/ .g; t/ g.0/ . Lemma 14.3.1. There holds S4 ¤ ;. Proof. By the general assumption S1 ¤ ;. Take g 2 S1 arbitrary. Proposition 13.7.1(1) states that there is a 2-dimensional non-rigid torus. This implies that S2 ¤ ;. Proposition 13.7.2 shows that t 2 T .L/. Definition 13.5.4 shows that t is a good

14.3

355


torus. Since every good torus has at least 1 proper root, S3 ¤ ; as well. Choose a triple .L; T; ˛/ in S3 . Since T 2 T .L/, Lemma 13.1.3(4) shows that M .˛/ .L; T / ¤ L. Choose a maximal subalgebra L.0/ containing M .˛/ .L; T /. Then .T; ˛; L.0/ / is an admissible triple. N be the items introduced Let G WD gr L, GN WD G=M.G/ and S ˝ O.mI 1/ Š A.G/ in §13.3. Theorem 13.6.7 yields m WD m.T; ˛; L.0/ / D 0. The summary of §14.2 states that S is not of classical, Cartan or Melikian type. As L has minimal dimension, we obtain S 2 S1 and dim S D dim L, i.e., GN D S D gr L. Lemma 13.7.4 implies that ˆ.T / is a 2-dimensional and non-rigid torus of S. Moreover, Lemma 14.2.1(1) proves jimp .L; T /j jimp .S; ˆ.T //j. The minimality of jimp .L; T /j implies jimp .L; T /j D jimp .S; ˆ.T //j, whence .S; ˆ.T // 2 S2 . Next Lemma 14.2.1(1) shows that ˆ.˛/ is proper. Lemma 13.7.4(4) in combination with the fact that (recall GN D S ) gr CL .T / CGN .ˆ.T // D CS .ˆ.T // M .ˆ.˛// .S; ˆ.T // yields gr M .˛/ .L; T / M .ˆ.˛// .S; ˆ.T //. Recall that GN D gr L and therefore gr M .˛/ .L; T / D gr M .˛/ .L; T / holds. Since .L; T; ˛/ 2 S3 , M .˛/ .L; T / has maximal possible dimension. This shows that gr M .˛/ .L; T / D M .ˆ.˛// .S; ˆ.T //, whence .S; ˆ.T /; ˆ.˛// 2 S3 . Since GN D S , the construction gives X X GN i D M .ˆ.˛// .S; ˆ.T // D gr M .˛/ .L; T / Si : i 0

i 0

P

Set S.0/ WD i 0 Si . Proposition 3.5.3(4) yields that (g1)–(g4) hold for the respective grading of GN D S. The above shows that .S; ˆ.T /; ˆ.˛/; S.0/ / is contained in S4 if S.0/ is a maximal subalgebra. Otherwise repeat this process. Suppressing the notion of ˆ we now take .S; T; ˛; S.0/ / 2 S4 . The fact that S is graded gives strong information on the set of roots. Lemma 14.3.2. For any 2 .S; T /, the following is true: (1) Both K.S; T / and R.S; T / are homogeneous subalgebras of S , i.e., X X K.S; T / D H ˚ Ki;j ; i 2Z j 2Fp

R.S; T / D H ˚

X X

Ri;j :

i 2Z j 2Fp

(2) For any i 2 Z, dim Si; =Ri; 2 dim Si; =Ki; . In particular, if Si; rad S./, then Si; S.0/ . P Proof. (1) Let x 2 Kj .S; T / where j 2 Fp . Then x D i 2Z xi where xi 2 Si;j (because each Si is T -stable). If k ¤ i , the subspace Œxi ; Sk Si Ck consists of

356

14


ad–nilpotent elements of S , and Œxi ; S

i; j

X

Œxk ; S i \ H C Œx; S

i; j

k¤i

nil H C Œx; S

j

ker :

Hence xi 2 Ki;j for all i , i.e., K.S; T / is homogeneous. Arguing in a similar fashion one obtains that R.S; T / is homogeneous as well. (2) Let .RK/i; D Ki; \ .RK/.S; T / and 2 .S; T /nFp . Since one has Œ.RK/i; ; K i; nil H , composing with the Lie product of S induces a linear map of .RK/i; into Hom .S i; =K i; ; F /. As Œ.RK/i; ; Sk consists of adnilpotent elements of S for k ¤ i, the kernel of this map is ¹x 2 .RK/i; j Œx; S

nil H º D Ri; :

This gives dim.RK/i; =Ri; dim S i; =K i; . Arguing in a similar fashion one observes that and the Lie product of S induce a non-degenerate pairing between Si; =Ki; and S i; =K i; . By Theorem 14.1.1 applied to S, K .S; T / D .RK/ .S; T / forcing Ki; D .RK/i; . Therefore, dim Si; =Ri; D dim Si; =Ki; C dim.RK/i; =Ri; dim Si; =Ki; C dim S

i; =K i;

D 2 dim Si; =Ki; as desired. If Si; rad S./, then Si; D Ki; and Si; D Ri; S.0/ by the previous dimension estimate. Lemma 14.3.3. Let 2 .S; T / be a proper Hamiltonian root. The following is true: (1) If S0 ./= rad S0 ./ H.2I 1/.2/ , then S./ S.0/ . (2) S0 ./= rad S0 ./ 6Š W .1I 1/. (3) Suppose S0 ./= rad S0 ./ Š sl.2/. Then Œ rad S0 ./ i ; rad S0 ./ nil H for all i 2 Fp . There are i0 2 Fp and a 2 N such that

i

(a) Si S.0/ for all i ¤ ˙i0 ; (b) dim Si =S.0/ \ Si 2; (c) S./ D S

a; i0

CS

a; i0

C S.0/ ./.

(4) If S0 ./ is solvable, then there are i0 2 Fp and a1 ; a2 2 N such that a1 > a2 a1 a2 , and (a) Si S.0/ for all i ¤ ˙i0 ; 2i0 ; (b) dim Si =S.0/ \ Si 2 for all i 2 Fp ; (c) S./ D S

a1 ; i0

CS

a2 ; i 0

C S a2

a1 ; 2i0

C S.0/ ./.

14.3

357


Proof. Set M WD S./= rad S./. As rad S./ is a graded ideal of S./, the Lie algebra M is Z-graded: M D ˚i 2Z Mi where Mi Š Si ./=Si ./ \ rad S./. As is Hamiltonian, H.2I 1/.2/ M H.2I 1/ (Theorem 11.1.2). According to Theorem 11.4.1 the grading of M is induced by the .a1 ; a2 /-grading with respect to generators u1 ; u2 of O.2I 1/. To keep the notation simple we assume that ui D xi , i D 1; 2. By Corollary 11.2.4 T stabilizes rad S./, .T C S.//=.T \ ker C rad S.// Š M; and the image of T in M is spanned by a nonzero toral element t. As T acts on S./ as homogeneous derivations, t 2 M0 . Let ı denote the degree derivation of M . Then t WD F t ˚ F ı is a 2-dimensional torus. Due to Theorem 11.4.2 we may assume that t D F .ı1 C x1 /@1 ˚ F x2 @2 , where ı1 2 ¹0; 1º. As t is contained in H.2I 1/, it then has the form t D r.x2 @2 .ı1 C x1 /@1 / where r 2 Fp . As 2 .S; T / is proper, F t is a proper torus of M . Applying Theorem 11.2.5 and rescaling t and if necessary we have ı1 D 0 and may assume r D 1 and .x2 @2 x1 @1 / D 1. (1) Suppose H.2I 1/.2/ M0 =rad M0 . Let W S./ ! S./= rad S./. Then .S.// D .S0 ./ C CS .T //, giving S./ D S.0/ ./ C CS .T / C rad S./ S.0/ ./ C K./: As K./ is a graded subalgebra of S./, this implies dim Si;j =Ki;j D ¹0º whenever i < 0 and j 2 Fp . By Lemma 14.3.2(2) we derive Si;j R S.0/ for all i < 0 and all j 2 Fp . Hence S./ S.0/ in this case. (2) Suppose M0 = rad M0 Š W .1I 1/. Then it follows from Theorem 11.4.1 that a1 D 0 and a2 ¤ 0. Since CS .T / S.0/ ./, the theorem also shows that a2 > 0. Pp 2 Pp 1 Note that i D0 F x1i @2 M a2 i D0 F x1i @2 . The vectors x1i @2 have pairwise distinct weights relative to t . Then all weight spaces of M a2 relative to F t p 2 are 1-dimensional. In particular, M a2 ; D F x1 @2 . But then ŒM a2 ; ; M CM .t / \ W .2I 1/.1/ consists of ad-nilpotent endomorphisms of M . From this it is immediate that S a2 ; D K a2 ; . By Lemma 14.3.2 we get S a2 ; D R a2 ; S.0/ . As a2 > 0, this is impossible. (3) Suppose M0 = rad M0 Š sl.2/. Then it follows from Theorem 11.4.1 that a1 D a2 ¤ 0 and a2 > 0. This theorem also gives M D M a2 C M.0/ and M a2 D F @1 ˚ F @2 . This implies that M a2 D M a2 ; C M a2 ; , and as in (1) one concludes Sj;i D Kj;i D Rj;i S.0/ whenever j < 0 and i 2 Fp n ¹˙1º. But then S./ S

a2 ;

CS

a2 ;

C S.0/ ./:

This establishes (a) and (c). For (b), observe that in view of Lemma 14.3.2(2) dim S

a2 ;˙

2 dim S

a2 ;˙ =K a2 ;˙

2 dim M

a2 ;˙

D 2:

358

14


Finally, ŒS0;i \ rad S0 ./; S

i

ŒS0;i \ rad S0 ./; S0;

i

C nil H H :

Therefore S0;i \ rad S0 ./ K0;i D .RK/0;i . This gives ŒS0;i \ rad S0 ./; S0;

i

\ rad S0 ./ nil H:

(4) Suppose M0 is solvable. Then it follows from Theorem 11.4.1 that 0 ¤ a1 ¤ a2 ¤ 0. Note that DH .x1 / 2 M a2 , DH .x2 / 2 M a1 , DH .x12 / 2 Ma1 a2 , DH .x22 / 2 Ma2 a1 , DH .x12 x2 / 2 Ma1 , and DH .x1 x22 / 2 Ma2 , whose respective P j t-eigenvalues are 1, 1, 2, 2; 1, 1. Set V WD 0i;j 2; 0 a2 . Suppose a2 < 0. Then a1 > 0 and DH .x2i / 2 M.i 1/a2 a1 for 1 i p 1. Since M has no more than 3 negative components, this is impossible. Hence a2 0. Consequently, a1 > a2 > 0 and a2 a1 < 0. Then M D FDH .x1 / ˚ FDH .x2 / ˚ FDH .x22 / ˚ M.0/ . The respective eigenvalues of ad t are 1; 1; 2. Then M D M a2 ; C M a1 ; C Ma2 a1 ;2 C M.0/ . As in (1) this gives Sj;i D Kj;i D Rj;i S.0/ unless .j; i / 2 ¹. a2 ; 1/; . a1 ; 1/; .a2

a1 ; 2/º. In view of Lemma 14.3.2,

dim Sj;i 2 dim Sj;i =Kj;i D 2 dim Mj;i =K.M; F t/j;i 2 for all i 2 Fp and j 2 ¹ a2 ; a1 ; a2 a1 º. Finally, DH .x23 / 2 M2a2 a1 ;3 holds. The pair .2a2 a1 ; 3/ is not contained in ¹. a2 ; 1/; . a1 ; 1/; .a2 a1 ; 2/º. Hence 2a2 a1 0. This completes the proof of the lemma. Lemma 14.3.4. Let 2 .S; T / be a proper Witt root. The following is true: (1) If S0 ./= rad S0 ./ Š W .1I 1/, then S./ S.0/ . (2) S0 ./= rad S0 ./ 6Š sl.2/.

14.3


359

(3) Suppose S0 ./ is solvable. Then there are i0 2 Fp and a > 0 such that (a) Si S.0/ for all i ¤ i0 ; (b) dim Si =Si \ S.0/ 2 for all i 2 Fp ; (c) S./ D S

a; i0

C S.0/ ./.

Proof. Set M WD S./= rad S./. As rad S./ is a graded ideal of S./, the Lie algebra M is Z-graded. Namely, M D ˚i 2Z Mi where Mi Š Si ./=Si ./ \ rad S./. (a) Suppose S0 ./= rad S0 ./ Š W .1I 1/. As rad S0 ./ contains S0 ./\rad S./ and M Š W .1I 1/ by assumption, we must have M D M0 . It follows that S./ D S0 ./ C rad S./ S0 ./ C K./. Hence Si;j D Ki;j for all i < 0 and all j 2 Fp . As before this gives Si;j R S.0/ whenever i < 0 and j 2 Fp . Statement (1) follows. (b) From now on suppose that S0 ./= rad S0 ./ 6Š W .1I 1/. This implies that M ¤ M0 . The grading of M induces a nontrivial grading of W .1I 1/. By Theorem 7.4.1, there is a 2 Z n ¹0º and an isomorphism W M ! W .1I 1/ such that .Mia / D F x i C1 @ for all i D 1; 0; 1; : : : ; p 2. As a consequence, M0 D F x@, so that dim S0 ./=S0 ./ \ rad S./ D 1. Then S0 ./ is solvable proving (2). The torus T acts on all Mia , hence it acts as F x@. Also, M.p 2/a D F x p 1 @ K.M; F x@/; hence S.p 2/a;i D K.p 2/a;i for all i 2 Fp . As before, S.p 2/a;i R S.0/ . This gives a > 0. From this it is clear that X M DM a˚ Mia and M a D F @: i 0

As T acts on M as the torus F x@, all weight spaces of M relative to T are 1dimensional. Choose i0 2 Fp such that Mi0 D F @. Then S./ D S a;i0 C S.0/ ./ C K./. Applying Lemma 14.3.2 we now deduce that S./ D S

a;i0

C S.0/ ./

and, moreover, dim Si0 =Si0 \ S.0/ 2: Statement (3) follows completing the proof. Lemma 14.3.5. Let 2 .S; T / be a classical root. The following is true: (1) If S0 ./= rad S0 ./ Š sl.2/, then S./ S.0/ and Œ rad S0 ./ i ; rad S0 ./ i nil H for all i 2 Fp :

360

14


(2) Suppose S0 ./ is solvable. Then there are i0 2 Fp and a > 0 such that (a) Si S.0/ for all i ¤ i0 ; (b) dim Si =Si \ S.0/ 2 for all i 2 Fp ; (c) S./ D S

a; i0

C S.0/ ./.

The proof of this lemma is very similar to the proof of Lemma 14.3.3 and will be omitted. Lemma 14.3.6. Let 2 .S; T / be a solvable root. Then S./ D S.0/ ./. Proof. Since is solvable, Si D Ki for any i 2 Fp . Now apply Lemma 14.3.2 to obtain Si D Ki D .RK/i D Ri S.0/ . We summarize. Lemma 14.3.7. Suppose all T -roots are proper. Let 2 .S; T /. The following is true: P (1) k 0, T 6 .Der s/ ˝ O.mI n/ and s 2 ¹sl.2/; W .1I 1/; H.2I 1/.2/ º; (c) H.2I 1/.2/ S0 CH.2I 1/; (d) H.2I 1I ˆ. //.1/ ˝ O.mI n/ S0 Der H.2I 1I ˆ.//.1/ ˝ O.mI n/ ; (e) s S0 Der s, where s is simple of absolute toral rank 2. We now take a closer look at these cases.

368

14


(a) Let V denote a minimal submodule of the .s1 ˚s2 /-module S 1 . As anns1 ˚s2 V is an ideal of s1 ˚ s2 , either Œsj ; V D ¹0º for some j 2 ¹1; 2º or V is a faithful .s1 ˚ s2 /-module. In the first case V 0 WD ¹v 2 S 1 j Œsj ; v D ¹0ºº is an S0 -module. But S 1 is an irreducible and faithful S0 -module. Thus V is faithful over s1 ˚ s2 . As Œs1 ; s2 D ¹0º, there are irreducible restricted faithful si -modules Vi , such that V Š V1 ˝ V2 as .s1 ˚ s2 /-modules. Choose proper tori F ti si with toral elements ti , set t WD F t1 ˚ F t2 . Given j 2 Fp and i 2 ¹1; 2º let Vi;j be the eigenspace for ti 2 End Vi belonging to the eigenvalue j . Each weight space V , where 2 t , has the form V D V1;m ˝ V2;n for some m; n 2 Fp . Suppose s1 Š H.2I 1/.2/ . By Theorem 11.5.6, dim V1 p 2 2. It follows that there is s 2 Fp such that dim V1;s > 2. Then some weight space of S 1 relative to t has dimension > 2 and hence by Theorem 10.7.1 there is ı 2 .S; T / such that dim S 1;ı > 2. This contradicts Lemma 14.3.7(2d) and thereby proves that s1 ; s2 2 ¹sl.2/; W .1I 1/º. Then Der si Š si , i D 1; 2, showing that S0 D s1 ˚ s2 . Recall that V is a restricted s1 ˚ s2 -module. Representation theory of sl.2/ and Theorem 7.6.10 imply that Vi;j ¤ ¹0º if and only if Vi; j ¤ ¹0º. Since V1 and V2 are faithful modules over s1 and s2 , respectively, there are m1 ; m2 2 Fp such that Vi;mi ¤ ¹0º for i D 1; 2. Let 2 t be such that .ti / D mi , i D 1; 2. The preceding remark shows that S 1; and S 1; are both nonzero. Notice that .t1 / .t2 / D 0 for any 2 .S0 ; t/. As a consequence, Fp \ .S0 ; t/ D ¹0º, so that S0 ./ D t. By Theorem 10.7.1 there is 2 T with S 1;˙ ¤ ¹0º and S0 ./ D T . Lemma 14.3.7(2) shows that this is impossible. (b) Suppose in Case (b) that s Š H.2I 1/.2/ . Corollary 10.6.6 shows that there is a simultaneous realization s ˝ O.mI n/ Š H.2I 1/.2/ ˝ O.m0 I 1/; S

1

Š U ˝ O.m0 I 1/; m0 ¤ 0;

where U is a faithful H.2I 1/.2/ -module, and T Š F .h ˝ 1/ ˚ F .d ˝ 1 C Id ˝ t0 /; where F h is a torus of H.2I 1/.2/ and F d is a torus of Der H.2I 1/.2/ , t0 ¤ 0. Due to Proposition 9.2.3, annU F h ¤ ¹0º. Then CS .T / \ S 1 ¤ ¹0º, which contradicts our construction. Suppose in Case (b) that s 2 ¹sl.2/; W .1I 1/º. As in the former case there is a simultaneous realization S 1 Š U ˝ O.m0 I 1/; where U is a faithful s-module, and T Š F .h ˝ 1/ ˚ F .Id ˝ t0 /; where F h is atorus of s and t0 ¤0. Since Der s D s, we have s ˝ O.m0 I 1/ S0 s ˝ O.m0 I 1/ Ì Id ˝ W .m0 I 1/ . Then there is a subalgebra D such that S0 D s ˝ O.m0 I 1/ Ì Id ˝ D:

14.3

369


Choose a root ˇ 2 .S 1 ; T / and any nonzero root 2 .S0 ; T / satisfying .h ˝ 1/ D 0. As t0 ¤ 0 and S 1 Š U ˝ O.mI 1/, all ˇ C Fp are roots in .S 1 ; T /. Since 0 62 .S 1 ; T /, one has ˇ 62 Fp , and this shows ˇ.h ˝ 1/ ¤ 0. Let Ui denote the h-eigenspace of U for the eigenvalue i . Representation theory of sl.2/ and Theorem 7.6.10 imply that Ui ¤ ¹0º if and only if U i ¤ ¹0º. Note that S0 Œˇ Š s is non-solvable. Lemma 14.3.7(2b) implies. As a result, .S 1 ; T / D ˙ˇ C Fp : Lemma 14.3.7 shows that dim S to ˇ we may assume ˇ.h ˝ 1/ ¤ 0;

1

4p. Therefore m0 D 1. Adding a multiple of

ˇ.Id ˝ t0 / D 0;

.h ˝ 1/ D 0;

.Id ˝ t0 / ¤ 0:

Suppose s ˝ O.1I 1/ does not annihilate S 2 . As S 2 D ŒS 1 ; S 1 , one has the inclusion .S 2 ; T / .˙2ˇ C Fp / [ Fp . Let W be a nontrivial s ˝ O.1I 1/composition factor. There is a weight D i0 ˇ C j0 2 .W; T / with i0 2 ¹˙2º. Arguing as before there is a s-module W0 and a module isomorphism W Š W0 ˝ O.1I 1/. From this one derives that all C Fp are weights. In particular, i0 ˇ is a root on S 2 . Lemma 14.3.7 implies that S0 .ˇ/ is solvable, which is not true as S0 Œˇ Š s. Consequently, s ˝ O.1I 1/ annihilates S 2 . Put Ui WD ¹u 2 U j Œh; u D iuº; S

1;i

WD ¹w 2 S

1

i 2 Fp ;

j Œh ˝ 1; w D iwº;

i 2 Fp ;

g WD CS0 .h ˝ 1/ D F h ˝ O.1I 1/ C Id ˝ D; and observe that S

1;i

D Ui ˝ O.1I 1/. We obtain

Œs ˝ O.1I 1/; ŒS

1;i ; S 1;i

Œs ˝ O.1I 1/; S 2 D ¹0º;

but Œh˝1; ŒS 1;i ; S 1;i D 2iŒS 1;i ; S 1;i holds. Consequently, ŒS 1;i ; S 1;i D ¹0º for i 2 Fp . For w 2 S1; i and v; v 0 2 S 1;i one has ŒŒw; v; v 0 D ŒŒw; v 0 ; v. Therefore the ad-representation gives rise to a linear homomorphism from L1; i into the space of Cartan prolongations L1;

i

! C.S

1;i ; g/C

Š C.Ui ˝ O.1I 1/; F h ˝ O.1I 1/ C Id ˝ D/C :

Theorem 8.1.6(1),(3a) yields the alternative F h ˝ O.1I 1/ or D is semisimple and O.1I 1/ invariant: P If the first alternative holds for all i 2 Fp , then S0 D ŒS1 ; S 1 k¤0 S0;k CF h˝ O.1I 1/ D F s ˝ O.1I 1/, which is not true. Therefore D is O.1I 1/-invariant. As a result, D Š W .1I 1/. Since all T -roots are proper, we may normalize F t0 D F x@. ŒS1;i ; S

1; i

370

14


Choose WD ˇ C . Then ˙ 2 .S 1 ; T / and S0 ./ s˝O.1I 1/.1/ is solvable. This contradicts Lemma 14.3.7(2). (c) Lemma 13.7.5 shows that the case H.2I 1/.2/ S0 CH.2I 1/ is impossible. (d) The case H.2I 1I ˆ.//.1/ ˝O.mI n/ S0 Der H.2I 1I ˆ.//.1/ ˝O.mI n/ is impossible by Lemma 14.3.8(2). (e) Assume TR.s/ D 2 and m D 0. The minimality of dim S implies that s is of classical, Cartan or Melikian type. Suppose s is classical. Since it has absolute toral rank 2, it is of type A2 , B2 , or G2 (Lemma 10.6.2). As p > 3, the Killing form on s is nondegenerate. Then Der s D ad s forcing S0 D s. This case is listed in the lemma. Suppose that s is of Cartan or Melikian type. Due to Lemma 14.3.8 only the Lie algebras s 2 ¹W .2I 1/; W .1I 2/; H.2I 1I ˆ.1//º can occur. They have the property that j.S 1 ; T /j p 2 1 (Theorem 10.7.3). Lemma 14.3.7 show that j.S 1 ; T /j 3.p C 1/, a contradiction. We denote by S 0 the subalgebra P of S generated by S 1 C S0 C S1 . As S satisfies 0 contains 0 (g2) of Notation 3.5.2, SP i 1 Si . Let M.S / denote the unique maximal 0 0 0 ideal of S contained in i 2 Si . Set g WD S =M.S /. Lemma 14.3.11. The following holds: (1) M.S 0 / is a nonzero graded ideal of S 0 . (2) The graded Lie algebra g is isomorphic to a classical Lie algebra of type A2 , B2 or G2 with one of its standard gradings. (3) S0 Š gl.2/. Moreover, the adjoint action of S0 on S induces a restricted representation of gl.2/ in gl.S/. (4) All irreducible constituents of the g-module i .M.S 0 //= i C1 .M.S 0 //, where i > 0, are restricted g-modules. Proof. Recall that by the previous lemmas either S0 Š gl.2/ or S0 is classical simple of type A2 , B2 or G2 . It is well known that CS0 .T / D T in all cases. Let 2 .S 1 ; T / and x 2 S 1; n ¹0º. Then Œx; S1; ¤ ¹0º, for otherwise x 2 R contrary to the definition of S . As a consequence, .S 1 ; T / .S1 ; T /. For k 0, set Ik WD ¹x 2 Sk j Œx; S1 D ¹0ºº. Clearly, I0 is an ideal of S0 . Suppose I0 ¤ ¹0º. Observe that ŒI0 ; S 1 D S 1 (as S 1 is a faithful irreducible S0 -module) and S0 D ŒS 1 ; S1 . It follows that I0 ŒI0 ; S0 D ŒŒI0 ; S 1 ; S1 D ŒS 1 ; S1 D S0 : Then ŒS0 ; S1 D ¹0º, hence ŒT; S1 D ¹0º. As S 1 is a faithful S0 -module, there is ı 2 .S 1 ; T / n ¹0º. Then ı 2 .S1 ; T /, a contradiction. Thus I0 D ¹0º. Since I 1 is a S0 -submodule of S 1 and S satisfies (g1), we also have that I 1 D ¹0º.

14.3

371


(a) Clearly M.S 0 / is a graded ideal of S 0 . Suppose M.S 0 / D ¹0º. Then Ik D ¹0º for all k 0. So it follows from Lemmas 14.3.9 and 14.3.10 that the graded Lie algebra S satisfies the conditions of Theorem 5.6.1. Then S is of classical, Cartan or Melikian type or is gl.n/=F with pjn.1 By assumption the first cases cannot occur, and the algebra gl.n/=F is not simple. Hence M.S 0 / ¤ ¹0º. (b) As I0 D I 1 D ¹0º and M.g/ D ¹0º, Theorem 5.6.1 is applicable to g. It says that g is either classical simple or gl.n/=F with pjn or X.mI n/.2/ g CX.mI n/, where X 2 ¹W; S; H; Kº, or a Melikian algebra M.n1 ; n2 /. In particular, g.1/ is simple of classical, Cartan, or Melikian type, and g.1/ g Der g.1/ . Observe that TR.g.1/ / P TR.S/ D 2. Since ŒS 1 ; S1 D S0 and S 1 \ CS .T / D ¹0º, we have that CS0 .T / D ¤0 ŒS 1; ; S1; . The definition of S4 shows that T is contained in the p-envelope of CS0 .T / in Der S. Consequently, X X S0 D ŒS 1; ; S1; C ŒCS0 .T /; S0; ¤0

¤0

and S˙1;˙ ŒS0 ; S˙1;˙ for all ¤ 0. Since S 1 ˚ S0 ˚ S1 Š g local Lie algebras, we therefore have that X .g 1 ˚ g0 ˚ g1; / g.1/ :

1

˚ g0 ˚ g1 as

¤0

This yields TR.g.1/ / D 2. (c) It is obvious by definition that g.1/ acts naturally on each factor space

i .M.S 0 //= i C1 .M.S 0 //. Let W be a composition factor of one of the g.1/ -modules

i .M.S 0 //= i C1 .M.S 0 //, and let G denote the restricted Lie algebra generated by g.1/ in gl.W /. By the above discussion, we can identify T with a 2-dimensional torus in .g.1/ /Œp , the semisimple p-envelope of g.1/ . There is a restricted epimorphism W G ! .g.1/ /Œp with ker D C.G/. By Schur’s lemma, C.G/ consists of scalar linear operators. If C.G/ ¤ ¹0º then 1 .T / is a 3-dimensional torus in G. However, this would imply that the semisimple p-envelope of S contains a 3-dimensional torus. Since the latter is false, C.G/ D ¹0º and G Š .g.1/ /Œp as restricted Lie algebras. It follows that W is a restricted .g.1/ /Œp -module. Suppose g.1/ is one of S.3I 1/.1/ , H.4I 1/.1/ , K.3I 1/, H.2I .1; 2//.2/ , M.1; 1/. Then annW T ¤ ¹0º (Theorem 10.7.2) which implies that annSk T ¤ ¹0º for some k < 0. This contradicts the inclusion CS .T / S.0/ . Suppose g.1/ is one of W .2I 1/; W .1I 2/; H.2I 1I ˆ.//.1/ ; H.2I 1I ˆ.1//. Then j.W; T /j p 2 1 (Theorem 10.7.3). This gives j.Sk ; T /j p 2 1 for some k < 0 and contradicts Lemma 14.3.7(1). Thus g.1/ is classical of type A2 , B2 or G2 . As a consequence, Der g.1/ D .1/ g forcing g D g.1/ . By Theorem 5.6.1, the grading of g must be natural. Then 1 see


372

14


g0 D g.ˇ/ Š gl.2/ for some root ˇ. Moreover, g0 is a restricted subalgebra of g Š Der g and its p-structure comes from the natural p-structure of gl.2/. Since S 1 ˚ S0 ˚ S1 Š g 1 ˚ g0 ˚ g1 as local Lie algebras, it follows that S0 Š gl.2/ as restricted Lie algebras. In order to finish the proof of our Classification Theorem it remains to show that g D S 0 =M.S 0 / cannot be classical simple of type A2 , B2 or G2 . Set t the image of T in Der g. Lemma 14.3.12. There holds g 6Š sl.3/. Proof. (a) Suppose the contrary and identify g with sl.3/. Since all 2–dimensional tori in g are conjugate under the adjoint action of G D SL.3; F / we shall assume that t is the Lie subalgebra of diagonal matrices in g. Then t D Lie T, where T is the group of diagonal matrices in G. As usual, we denote by i the rational character of T that sends a matrix in T to its i -th diagonal entry. Then the root system R of g (with respect to T) is the set of all i j with 1 i; j 3 and i ¤ j . We choose as simple roots ˛1 D 1 2 and ˛2 D 2 3 . The corresponding fundamental weights are then !1 D 1 13 .1 C 2 C 3 / and !2 D 1 C 2 32 .1 C 2 C 3 /. For ˛ D i j with i ¤ j , we choose as root vector e˛ the matrix Ei;j whose .i; j /-th entry equals 1 and all other entries are 0. Given x 2 g 1 ˚ g0 ˚ g1 let xQ denote the unique preimage of x in S 1 ˚ S0 ˚ S1 under the canonical epimorphism S 0 ! g. Since the grading of g is standard, we may assume (without loss of generality) that gDg

1

˚ g0 ˚ g1 ;

g0 D t ˚ F e

˛2

˚ F e ˛2 ;

g˙1 D F e˙˛1 ˚ F e˙.˛1 C˛2 / :

It follows that S 2 M.S 0 /. For each ˛ 2 R the tangent map d˛ W t ! F is a linear function on t D Lie T. In what follows we identify ˛ 2 R with d˛ 2 t , hence R with .g; t/. (b) Since M.S 0 / ¤ ¹0º (Lemma 14.3.11(1)) and S 1 Š g 1 is 2-dimensional, S 2 is a 1-dimensional subspace of M.S 0 / spanned by vQ 0 WD ŒeQ ˛1 ; eQ ˛1 ˛2 . In particular, .S 0 2 ; T / D ¹ 2˛1 ˛2 º: (14.3.2) Let h1 WD Œe˛1 ; e ˛1 D E1;1 E2;2 and h2 WD Œe˛2 ; e ˛2 D E2;2 E3;3 . Note that t D F h1 ˚ F h2 and !i .hj / D ıij , where 1 i; j 2. Note that S k D .ad S 1 /k 2 .S 2 / for k 3. Therefore the g-module V1 WD M.S 0 /= 2 .M.S 0 // is generated by v0 , the image of vQ 0 in V1 . As M.S 0 / \ S 1 D ¹0º, we must have Œe˛1 ; v0 D ŒeQ˛1 ; vQ 0 D 0: As dim S

2

D 1, Œe˛2 ; v0 D ŒeQ˛2 ; vQ 0 D 0:

14.3

373


Also, h1 v0 D 3v0 and h2 v0 D 0. Therefore, v0 is a primitive vector of weight p 3 .p 3/!1 in V1 . From this it is immediate that v1 WD e ˛1 v0 ¤ 0. This, in turn, yields that vQ 1 WD .ad eQ

p 3 .vQ 0 / ˛1 /

2S

pC1; ˛1 ˛2 n¹0º:

Since Œe˛2 ; e ˛1 D 0, we have that Œe˛2 ; v1 D 0 and Œh2 ; v1 D .p 3/v1 . Representation theory of sl.2/ now shows that .ad e ˛2 /i .v1 / ¤ 0 for i D 0; : : : ; p 3. Therefore, ¹˛1

˛2 ; : : : ; ˛1

.p

2/˛2 º

pC1 .S; T /:

(14.3.3)

Obviously, S pC1 ¤ ¹0º implies S pC3 ¤ ¹0º. Let 2 .S pC3 ; T / and u1 2 S pC3; n¹0º. As R .S; T / S.0/ , there is w1 2 Sp 3; such that Œu1 ; w1 ¤ 0. Note that Œu1 ; w1 2 CS0 .T / D T . Suppose S p 1 D ¹0º. Then Œu1 ; S pC1 S 2pC4 D ¹0º. Note that ad w1 maps S pC1; ı into S 2; ı . So it follows from (14.3.2) that ad .Œu1 ; w1 / annihilates S pC1; ˛1 Ck˛2 for all but at most one value of k 2 Fp . Due to (14.3.3) there are distinct m; n 2 Fp such that .˛1 m˛2 /.Œu1 ; w1 / D .˛1 n˛2 /.Œu1 ; w1 / D 0. But then Œu1 ; w1 D 0, a contradiction. (c) Thus S p 1 ¤ ¹0º. Therefore, S p ¤ ¹0º. The center of S0 acts trivially on S kp , hence S kp D S kp .˛2 /. As S0 .˛2 / Š gl.2/ is non-solvable, Lemma 14.3.7 implies that .S p ; T / D ¹˙j˛2 º for some j 2 Fp . But then hQ 2 has ex.1/

actly two eigenvalues on each composition factor of the S0 -module S p . Therefore, ˙j˛2 .h2 / D ˙1. As a consequence, .S p ; T / D ¹˙ 21 ˛2 º. Since S p 1 D ŒS 1 ; S p we have that .S p 1 ; T / .S 1 ; T / C .S p ; T /, whence .S

p 1; T /

°

˛1

3 ˛2 ; ˛1 2

1 1 ± ˛2 ; ˛1 C ˛2 : 2 2

(14.3.4)

The respective values of these linear functions at h2 are 2, 0, 2. Now it follows from representation theory of sl.2/ that 0 is an eigenvalue of ad hQ 2 on S p 1 . But then ˛1

1 ˛2 2 .S 2

p 1 ; T /:

(14.3.5)

.1/

Next we observe that S 3 D ŒS 2 ; S 1 Š S 1 as S0 -modules. It follows that .S 3 ; T / .S 1 ; T / C .S 2 ; T /, hence (see (14.3.2)) .S 3 ; T / D ¹ 3˛1

˛2 ; 3˛1

2˛2 º:

(14.3.6)

Note that S 4 ¤ ¹0º. Arguing as before we derive that .S 4 ; T / .S 1 ; T / C .S 3 ; T / D ¹ 4˛1 ˛2 ; 4˛1 2˛2 ; 4˛1 3˛2 º, that 0 is an .ad h2 /-eigenvalue on S 4 and that .4˛1 C 2˛2 / 2 .S 4 ; T /. Combining this with (14.3.2), (14.3.3)

374 and (14.3.5) we obtain that S D ˛1 C 12 ˛2 in Let WD ˛1

14


k; k.˛1 C 21 ˛2 /

¤ ¹0º for k 2 ¹2; 4; p 1; p C1º. Setting

Lemma 14.3.7 now yields p D 5. ˛2 . By (14.3.3) and (14.3.6), 2 .S 4 ; T / and 2 2 .S 3 ; T /. So Lemma 14.3.7 shows that is Hamiltonian and the grading of S./ is ruled by Lemma 14.3.3(4). Then 1; 2 2 ¹˙i0 ; 2i0 º, hence either 2i0 D 1 or 2i0 D 2. If 2i0 D 1, then i0 D 2 which gives a2 a1 D 4 and a2 D 3. If 2i0 D 2, then a1 D 4 and 3 D a2 a1 forcing a2 D 1. But then a2 < a1 a2 in both cases. This contradiction proves the lemma.

Lemma 14.3.13. There holds g 6Š sp.4/. Proof. (a) Suppose the contrary and identify g with sp.4/. Since all Witt bases of the symplectic linear space F 4 are conjugate under the natural action of G D Sp.4; F / on F 4 , all 2-dimensional tori in g D Lie G are conjugate under the adjoint action on G. Thus no generality is lost by assuming that t is the Lie subalgebra of diagonal matrices in g. Then t D Lie T, where T is the group of diagonal matrices in G. We are going to use Bourbaki’s notation [Bour]. The group of rational characters X.T/ will be embedded into an Euclidean space with orthonormal basis 1 ; 2 . The root system R of g (with respect to T) is the set ¹˙1 ˙ 2 ; ˙21 ; ˙22 º. We choose as simple roots ˛1 D 1 2 and ˛2 D 22 . Then ˛Q WD 2˛1 C ˛2 is the highest root. The corresponding fundamental weights are !1 D 1 and !2 D 1 C 2 , and X.T/ D Z!1 C Z!2 . The set of dominant weights N0 !1 C N0 !2 will be denoted by X C .T/. A dominant weight D a1 !1 Ca2 !2 is called p-restricted, if 0 ai p 1 for i D 1; 2. We identify the p-restricted weights with the corresponding tangent maps d W t ! F (this will cause no confusion since the kernel of the linear map d W X.T/ ! t equals pX.T/). For any ˛ 2 R choose h˛ 2 t such that .d/.h˛ / D h; ˛ _ i.mod p/ for all 2 X.T/. Choose e˛ 2 g˛ such that Œeˇ ; e ˇ D hˇ for all ˇ 2 R. Set hi WD h˛i , i D 1; 2, and QC WD N0 ˛1 C N0 ˛2 . (b) Since all roots having the same length are conjugate under the action of the Weyl group of R, all Levi subalgebras of g containing t and isomorphic to gl.2/ fall into two conjugacy classes under the adjoint action of NG .T/. Thus we may assume that either g0 D t˚F e˛1 ˚F e ˛1 or g0 D t˚F e˛2 ˚F e ˛2 . Given x 2 g 1 ˚g0 ˚g1 we let xQ 2 S 1 ˚ S0 ˚ S1 have the same meaning as in the proof of Lemma 14.3.12. We first suppose that g0 D t ˚ F e˛2 ˚ F e ˛2 . Since the grading of g is standard we may assume further that g˙1 D F e˙˛1 ˚ F e˙.˛1 C˛2 / . Then S 1 DPF eQ ˛1 ˚ F eQ .˛1 C˛2 / , S 2 D F eQ ˛Q where eQ ˛Q D ŒeQ ˛1 ; eQ .˛1 C˛2 / , and M.S 0 / D k 3 Sk . Since M.S 0 / ¤ ¹0º (Lemma 14.3.11(1)), we must have S 3 ¤ ¹0º. Since S 2 .1/ .1/ is a trivial S0 -module, the S0 -module S 3 D ŒS 2 ; S 1 is a homomorphic im.1/ age of S 1 . Hence S 3 Š S 1 as S0 -modules. It follows that .S 3 ; T / D ¹ 3˛1 ˛2 ; 3˛1 2˛2 º: It also follows that vQ 0 WD ŒeQ ˛1 ; eQ ˛Q 2 S 3 is nonzero and ŒeQ˛2 ; vQ 0 D 0. As S k D .ad S 1 /k .S 3 / for k 4 and S 3 Š S 1 is S0 -irreducible, the g-module V1 D M.S 0 /= 2 .M.G 0 // is generated by v0 , the image of vQ 0 in V1 .

14.3

As M.S 0 / \ S

2

375


D ¹0º, we must have ŒeQ˛1 ; vQ 0 D 0. Also,

ŒhQ 1 ; vQ 0 D h 3˛1

˛2 ; ˛1_ ivQ 0 D

6

2

.22 j1 2 / vQ 0 D .p .1 2 j1 2 /

4/vQ 0

and ŒhQ 2 ; vQ 0 D h 3˛1

˛2 ; ˛2_ ivQ 0 D

6

.1 2 j22 / .22 j22 /

Therefore, v0 is a primitive vector of weight D .p h; ˛Q _ i D 2

2 vQ 0 D vQ 0 :

4/!1 C !2 in V1 . Since

.j21 / ..p 3/1 C 2 j21 / D2 Dp .21 j21 / .21 j21 /

3;

p 3

Q Since the vector v1 WD e ˛Q v0 2 V1 is nonzero and has weight .p 3/˛. ˛Q ˛2 62 R we have that e˛2 v1 D 0. This implies that e ˛2 v1 ¤ 0 (because h .p 3/˛; Q ˛2_ i D h; ˛2_ i D 1). Observe that .p 3/˛Q C˛2 D 2.p 3/1 C22 D 2. It follows that .ad eQ

p 3 .vQ 0 / ˛2 /.ad eQ ˛Q /

2S

2pC3; n¹0º

(for vQ 0 2 S 3 , e ˛Q 2 S 2 and e ˛2 2 S0 ). As a consequence, S 3; ¤ ¹0º and S 2pC3; ¤ ¹0º. Lemma 14.3.7 now shows that 2 .S 3 ; T / is a Hamiltonian root of S. Since S0 ./ D t is solvable, the grading of S./ is ruled by Lemma 14.3.3(4), with a1 D 2p 3 and a2 D 3. But then a1 a2 > a2 . So this case is impossible. (c) Now suppose g0 D t ˚ F e˛1 ˚ F e ˛1 . Since the grading of g is standard we may assume that g˙1 D F e˙˛2 ˚ F e˙.˛1 C˛2 / ˚ F e˙.2˛1 C˛2 / and g˙k D ¹0º for k 2. Since S it is easy to see that S .1/

1

D F eQ

˛2

1 ˚S0 ˚S1

˚ F eQ

˛1 ˛2

Šg

1 ˚g0 ˚g1

˚ F eQ

˛Q

as local Lie algebras,

Š V .2/

as S0 -modules (recall that V .2/ stands for the 3-dimensional irreducible sl.2/-module). Since M.S 0 / ¤ ¹0º, S 2 D ŒS 1 ; S 1 is a nonzero S0 -submodule of M.S 0 /. V .1/ .1/ Now V .2/ Š V .2/ and 3 V .2/ Š F as S0 -modules. It follows that the S0 V2 modules V .2/ and V .2/ are isomorphic. As ŒS 1 ; S 1 is a homomorphic image V .1/ of 2 S 1 , we deduce that S 2 Š V .2/ as S0 -modules. From this it is immediate that S 2 D F ŒeQ ˛2 ; eQ ˛1 ˛2 ˚ F ŒeQ ˛2 ; eQ ˛Q ˚ F ŒeQ ˛1 ˛2 ; eQ ˛Q :

376

14


Moreover, vQ WD ŒeQ ˛2 ; eQ ˛1 ˛2 generates the S0 -module S 2 and has the property that ŒeQ˛1 ; v Q D 0. Since M.S 0 / \ S 1 D ¹0º, we must have ŒeQ˛2 ; v Q D 0. Also, ŒhQ 1 ; v Q D h ˛1

2˛2 ; ˛1_ ivQ D

2

4

.22 j1 2 / vQ D 2vQ .1 2 j1 2 /

and ŒhQ 2 ; v Q D h ˛1

2˛2 ; ˛2_ ivQ D

2

.1 2 j22 / .22 j22 /

4 vQ D .p

3/v: Q

As before, the g-module V1 D M.S 0 /= 2 .M.S 0 // is generated by v, the image of vQ in V1 . Let V10 be a maximal submodule of the g-module V1 , and VN1 WD V1 =V10 . Since v 62 V10 , the g-module VN1 is generated by v, N the image of v in VN1 . Let L./ denote the irreducible rational G-module with highest weight D 2!1 C .p 3/!2 2 X.T/. Let W G ! GL.L.// denote the corresponding representation of G, and d W g D Lie G ! gl.L.// the tangent map at 1 2 G. Since 2 X C .T/ is p-restricted, d is an irreducible restricted representation of g in L./ (see, e.g., [Bor 70]). By Lemma 14.3.11(4), VN1 is a restricted g-module. By construction, the g-module VN1 is irreducible and generated by a primitive vector of weight (or rather d 2 t ). Applying Curtis’s theorem we now obtain that VN1 Š L./ as g-modules (see, e.g., [Bor 70]). By the main result of [Pre 88], 2 X C .T/ is a T -weight of L./ if and only if 2 QC . Since 0 D ˛Q C.p 3/.˛1 C˛2 / 2 QC , 0 is a T-weight of L./. But then t D Lie T kills a nonzero vector of L./. This implies that annVN1 t ¤ ¹0º forcing annSk T ¤ ¹0º for some k < 0. This contradicts the inclusion CS .T / S.0/ and proves the lemma. Lemma 14.3.14. There holds g 6Š G2 . Proof. Suppose the contrary and let G be a simple algebraic group of type G2 such that g D Lie G (we may assume that g D Der O and then take G D Aut O). Let T be a maximal algebraic torus in G, and t0 D Lie T. Then t0 is a 2–dimensional torus in g. Let R be the root system of g with respect to T, B D ¹˛1 ; ˛2 º a basis of simple roots in R, ¹!1 ; !2 º the system of fundamental weights associated with B, and X C .T/ D N0 !1 C N0 !2 the set of dominant weights. We denote by L./ the irreducible rational G-module with highest weight 2 X C .T/. The Lie algebra g D Lie G acts on L./ via the differential at 1 2 G of the linear representation G ! GL.L.//. This gives L./ a canonical restricted g-module structure. By Lemma 14.3.11(1), M.S 0 / ¤ ¹0º. Then M.S 0 /= 2 .M.S 0 // is a nonzero gmodule. Let W be a composition factor of the g-module M.S 0 /= 2 .M.S 0 //. By Lemma 14.3.11(4), W is a restricted g-module. By Curtis’s theorem ([Bor 70]), there is D a1 !1 C a2 !2 2 X C .T/ with 0 a1 ; a2 < p such that W Š L./ as g-modules.

14.3


377

Let QC D N0 ˛1 C N0 ˛2 . A special feature of the present case is the inclusion ¹!1 ; !2 º QC which yields X C .T/ QC (see [Bour]). Hence 0 2 QC . But then zero is a T-weight of L./ (see [Pre 88]), hence annW t0 ¤ ¹0º. By Theorem 10.7.1, annW T ¤ ¹0º. This contradicts the inclusion CS .T / S.0/ . Thus g 6Š G2 . The conclusion of all the deliberations of this volume is Theorem 14.3.15. Every finite dimensional simple Lie algebra L of absolute toral rank 2 over an algebraically closed field of characteristic p > 3 is of classical, Cartan, or Melikian type.

Chapter 15

Supplements to Volume 1

It is very much appreciated if readers would inform me about improvements, suggestions, misprints, etc. There will be the opportunity to include all these into an extra chapter of the next volume. I would like to point out six more substantial changes in Volume 1. (1) p. 52, l. 8/ 7: h 2 L is called regular, if L0 .ad h/ has minimal dimension (and not “if L0 .ad h/ is minimal”). (2) p. 55, l. 16: all fibers of pk are finite (and not “of constant dimension nk ”). (3) There exists a stronger version of Theorem 4.1.4: Theorem. Suppose p > 3. Let L be a simple Lie algebra generated as a Lie algebra by a set S, such that every element x of S satisfies .ad x/p D 0 and Œ.ad x/i .u/; .ad x/j .v/ D 0 8 u; v 2 L; i C j p: Then L is of classical type. Pp 1 Proof. The assumptions are such that exp .ad x/ D i D0 iŠ1 .ad x/i is defined and, moreover, is a automorphism of L. Now the original proof works without any changes at all. (4) The final statement in Theorem 5.3.1 is wrong. Namely, the statement “If ˛ 2 GF .p/ (in particular if M is a restricted module), then ˛ D k D dim M 1” cannot be true. Whenever dim M D p, e p D 0, f p ¤ 0, ˛ 2 GF .p/, the module described in Theorem 5.3.1(2) is simple and satisfies e v0 D e v˛C1 D 0, h v0 D ˛v0 , hv˛C1 D .˛ C2/v˛C1 . The respective modules for ˛ and .˛ C2/ are isomorphic. The statement is true if M is a restricted module. (5) In Theorem 5.6.1 one can do a little bit better: the algebras gl.n/ and sl.kp/ in part (2) of the assertion do in fact not occur, because these have a nontrivial center and therefore do not satisfy assumption (c) or (d) ([B-G-P]). (6) Theorem 8.1.6(2) can be extended: (2) Suppose dim U D 1 and C ¤ ¹0º. Then 2 .G/ is semisimple and has the unique minimal ideal \n1 .C /n . Now add: Moreover, m 2.

379 The additional statement m 2 follows immediately from the first paragraph of part (c) in the original proof. There are some more minor suggestions. (1) p. 175, Theorem 3.5.8: In statement (1) of that theorem one has even more m ¤ 0 (not m 0), because A 1 D L 1 ¤ ¹0º and the grading is given by the grading of O.mI n/. (2) p. 282, l. 3: change to 8 ˆ I C F t; I Š sl.G 1 / if G0 Š gl.G ˆ ˆ ˆ

Simple Lie algebras over fields of positive characteristic. II. Classifying the absolute toral rank two case

Simple Lie Algebras over Fields of Positive Characteristic: II. Classifying the Absolute Toral Rank Two Case (De Gruyter Expositions in Mathematics)

Lie Groups and Lie Algebras II

Analysis in positive characteristic

Analysis in Positive Characteristic

Lie Algebras

Lie Algebras

Lie Algebras

Finite-dimensional division algebras over fields

Finite-Dimensional Division Algebras over Fields

Lie Algebras

Lie Algebras

Lie Algebras and Lie Groups

Lie groups and Lie algebras

Lie Groups and Lie Algebras

Simple groups of finite Morley rank

Introduction to Lie Algebras

Infinite-Dimensional Lie Algebras

Infinite-dimensional Lie algebras

Infinite-dimensional Lie algebras

Zeta Functions of Simple Algebras

Lie Algebras Madison 1987

Lie Algebras and Applications

Simple groups of Lie type

Introduction to Lie Algebras

Modular Lie Algebras

Complex semisimple Lie algebras

Complex Semisimple Lie Algebras

Infinite-Dimensional Lie Algebras

Notes on Lie Algebras

Free Lie algebras

Simple Lie algebras over fields of positive characteristic. II. Classifying the absolute toral rank two case