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UNIVERSITY OF AARHUS D EP A R T ME N T O F M AT H E MA T I C S ISSN: 1397-4076 SUBREGULAR NILPOTENT REPRESENTATIONS OF LIE ALGEBRAS IN PRIME CHARACTERISTIC By Jens Carsten Jantzen Preprint Series No.: 11 Ny Munkegade, Bldg. 530 8000 Aarhus C, Denmark June 1999 http://www.imf.au.dk institut@imf.au.dk Subregular nilpotent representations of Lie algebras in prime characteristic Jens Carsten Jantzen Introduction Let G be a reductive algebraic group over an algebraically closed eld K of prime characteristic p > 0. This paper deals with certain representations of the Lie algebra g of G. For the purpose of this introduction assume that G is semi-simple and simply connected, that the root system R of G is irreducible, and that p is larger than the Coxeter number h of R. (If R is of type E8 or F4, assume that p > h + 1; this restriction should be unnecessary, but my proofs require it.) Each simple g{module has a p{character; that is a linear form  on g such that all xp , x[p] , (x)p 1 with x 2 g annihilate the module. Here xp is the p{th power of x in the universal enveloping algebra U (g) and x 7! x[p] is the p{th power map on the Lie p{algebra g. A general result due to Kac and Weisfeiler reduces the problem of describing all simple modules basically to the case where the p{ character  is nilpotent; this means that  vanishes on some Borel subalgebra of g. Due to work by Curtis, Friedlander, Parshall, and Panov one then has a classi cation of the simple modules in case  has a certain special form (\standard Levi form"). For  not of this form so far no classi cation of the corresponding simple modules has been known. We look in this paper at the case where  is subregular nilpotent. Here \subregular" means that the orbit of  under the coadjoint action of G has dimension 2(N , 1) where 2N = jRj. A subregular nilpotent  has standard Levi form if and only if R has type An or Bn. In those two cases I have given a detailed description of the corresponding simple modules in [10]. For the other types the results in this paper are new. In order to describe them I rst need some notation. Let T be a maximal torus in G, let X be the character group of T and h the Lie algebra of T . Choose a basis  for the root system R  X . Set  equal to half the sum of the positive roots, and let 0 denote the unique short root that is a dominant weight. (If all roots have the same length, then all roots are short, and none is long.) Set e =  [ f, 0g. Consider the algebra U (g)G of G{invariants in U (g). Each  2 X de nes a \central character" cen : U (g)G ! K such that U (g)G acts via cen on a highest weight module with highest weight . Fix a subregular nilpotent . Denote for each  2 X the category of all nite dimensional g{modules M that are annihilated by all xp , x[p] , (x)p 1 with x 2 g and such that U (g)G acts via cen on all composition factors of M . Assume rst that all roots in R have the same length. (This is the case where our results are most complete.) Let  2 X such that 0 < h + ; _i < p for all 2 positive roots . (We denote by _ the coroot corresponding to .) Then there are up to isomorphism je j simple modules in C . We can denote them by L with 2 e such that dim L =  h + ; _ipN ,1; if 2 , _ N , 1 (p , h + ; 0 i)p ; if = , 0. (1) P De ne integers m for all 2 e by m, 0 = 1 and _0 = 2 m _. Let Q denote the projective cover of L in C . Then [Q : L] = jW j m m for all ; 2 e . (2) Here W is the Weyl group of R and we write [M : L] for the multiplicity of a simple module L as a composition factor of a module M . If  vanishes on the \standard" Borel subalgebra (corresponding to the positive roots), then one can de ne \baby Verma modules" Z (). One gets then [Z() : L ] = m for all 2 e . (3) The extension group (in C) of two non-isomorphic simple modules is given by Ext1(L ; L ) '  K; if ( ; ) < 0, 0; if ( ; ) = 0, (4) unless R is of type A1 where one has to replace K by K 2 . I do not know how big the Ext group is in case = ; it will be non-zero in most cases. The result on the number of simple modules in C as well as the formula in (2) for the Cartan matrix con rm conjectures by Lusztig. If we consider more generally  2 X such that 0  h + ; _ i < p for all positive roots , then the results in (1), (2), and (3) extend if we drop all L with h + ; _i = 0 and if we replace jW j in (2) by jW j. (We use here the `dot action' where w = w( + ) , .) In types An and Dn we get thus a description of all possible C (that depends only on W  + pX ), while there remain some open problems in type En. Consider next R of type Bn, Cn, or F4 . Choose again a weight  2 X such that 0 < h + ; _i < p for all positive roots . In this case there is one simple module L corresponding to each short root 2 e , and one expects that there are two simple module L;1 and L;2 corresponding to each long root 2 e . This expectation is known to hold in type Bn; in types Cn and F4 I cannot exclude that we get only one simple module corresponding to a given long simple root. The formulas in (1){(4) extend to the present situation as long as all roots occurring are short. The dimension formula (1) holds also for the simple modules corresponding to long roots. If our expectation holds, then one gets L;2 from L;1 by twisting with the adjoint action of a suitable element in the centraliser of  in G. In (2){(4) one has to make then minor changes in case long roots occur. 3 For example, if is long, then one should divide the right hand side in (3) by 2. If , are long with ( ; ) < 0, then one can choose the numbering of the simple modules such that Ext1(L  ;i ; L ;j ) '  K; if i = j , 0; otherwise. There are again similar results for all  2 X such that 0  h + ; _i < p for all positive roots . In types Bn and Cn these  suce to understand all possible C . In type G2 our results are much weaker. Let me brie y describe the main steps in the proof of the results. We rst construct a chain of submodules in a baby Verma module. Using the Kac-Weisfeiler conjecture (Premet's theorem) one can show that this is a composition series. (In the case of two root lengths the argument is actually slightly more complicated.) We then show that all composition factors can be described as induced modules from minimal parabolic subalgebras of g. (These results are contained in Section D and build on earlier work on translation functors and submodules of baby Verma modules in Sections B and C.) The main problem then is to decide when two factors in our composition series are isomorphic to each other. In order to show that two modules are not isomorphic to each other, we apply suitable translation functors and show that the modules behave di erently under them. In order to show that two modules are isomorphic to each other, we use a deformation argument to construct a non-zero homomorphism. This is done in Section F after some preparations in Sections A and E. This is followed by an investigation of the action of the centraliser in G of  on the simple modules. The results on the Cartan matrix and the Ext groups are contained in Sections G and H. Here we use \translation through the walls" as a main technique. Much of this work was done while I was visiting the SPIC Mathematical Institute (now: Chennai Mathematical Institute) and the Tata Institute of Fundamental Research. I would like to thank all those who made those stays possible and enjoyable, in particular C. S. Seshadri and K. N. Raghavan, V. B. Mehta and A. J. Parameswaran. I also would like to thank Jim Humphreys for several useful suggestions. 4 A A.1. Let K be an algebraically closed eld of characteristic p > 0. Let G be a reductive algebraic group over K and denote by g the Lie algebra of G. This is a restricted Lie algebra as the Lie algebra of an algebraic group; we denote the p{th-power map by x 7! x[p] . Let T be a maximal torus in G and set h = Lie(T ). Let X = X (T ) the (additive) group of all characters of T and let R  X be the root system of G. For each 2 R let g denote the corresponding root subspace of g. We choose a system R+ of positive roots. Set n+ equal to the sum of all g with > 0 and n, equal to the sum of all g with < 0. We have then the triangular decomposition g = n+  h  n, of g. Set b+ = h  n+ . All of these Lie subalgebras (h, n+ , n, , b+) of g are Lie algebras of closed subgroups of G and hence restricted subalgebras of g. For each 2 R let _ denote the corresponding coroot. Denote then by s ;n (for all n 2 Z) the (ane) re ection given by s ;n() =  , (h; _ i , n) for all  2 X . Write W for the Weyl group (generated by all s = s ;0), and write Wp for the ane Weyl group generated by all s ;mp with m 2 Z and 2 R. Then Wp contains all translations by p with 2 R, and Wp is generated by W and these translations. Each  2 X de nes by taking the derivative a linear form d on h. The map  7! d yields an embedding of X=pX into h. If 1, 2 ; : : : ; s is a basis of X over Z, then the di are a basis of h over K . There exists for each 2 R an element h 2 h such that d (h ) is the reduction modulo p of h; _ i for each  2 X . We choose for each root a basis vector x for g such that always [x ; x, ] = h . We have x[p] = 0 and h[p] = h for all 2 R. Let  2 X Z Q be a weight with h; _ i = 1 for all simple roots . We shall use the `dot action' of W or Wp on X given by w = w( + ) , . (It is independent of the speci c choice of .) We denote the universal enveloping algebra of g by U (g). For each  2  g set U (g) equal to the factor algebra of U (g) by the ideal generated by all xp , x[p] , (x)p with x 2 g. We use analogous notations for restricted subalgebras of g. A.2. Each f 2 h de nes a one dimensional h{module Kf [with each h 2 h acting as multiplication by f (h)]. There exists (f ) 2 h such that Kf is a U(f )(h){module, i.e., with (f )(h)p = f (h)p , f (h[p] ) for all h 2 h. So we have in particular (f )(h )p = f (h )p , f (h ) for all 2 R and f (h ) 2 Fp () (f )(h ) = 0: Any f1; f2 2 h satisfy obviously (f1 + f2 ) = (f1 ) + (f2 ): (1) (2) (3) 5 We have clearly () = 0 for all  2 X , hence (f + ) = (f ) for all f 2 h,  2 X . (4) Here (as usual) we write  instead of d by abuse of notation. For each  2 g set  = f f 2 h j jh = (f ) g: (5) Given f 2 h we can regard Kf as a b+{module with n+ acting trivially. This is then a U(b+){module for all  2 g satisfying (n+ ) = 0 and f 2 . For all these  we get then an induced U (g){module (a \baby Verma module") Z (f ) = U (g) U (b+ ) Kf : (6) Denote its \standard generator" 1 1 by vf . As before we usually write Z() instead of Z (d) for  2 X ; we should keep in mind that Z() depends only on d, hence on the coset  + pX . If f 0 2 h satis es f 0 (h ) = 0 for all 2 R, then we can extend Kf 0 to a g{module with all x acting as 0. This is then a U(f 0 ) (g){module where we extend (f 0 ) to a linear form on g such that (f 0 )(n+ + n, ) = 0. A trivial form of the tensor identity yields then Z(f ) Kf 0 ' Z+(f 0)(f + f 0 ) (7) for all f 2 h and  2 g with f 2  and (n+ ) = 0. A.3. Fix a simple root for the rest of Section A. Denote by p = b+ + g, the corresponding parabolic subalgebra. For all f 2 h and  2 g with (n+ ) = 0 and f 2  consider the induced U(p ){module Z; (f ) = U (p ) U (b+ ) Kf : (1) Denote the \standard generator" 1 1 now by vf0 . It satis es x vf0 = 0 for all 2 R+ and hvf0 = f (h)vf0 for all h 2 h; the xi, vf0 with 0  i < p are a basis of Z; (f ). We have x (xi, vf0 ) = i(f (h ) + 1 , i)xi,,1 vf0 for 0 < i < p and x (xi, vf0 ) = 0 for all 2 R+ with 6= . Suppose now that f (h ) 2 Fp: (2) Let d be the integer with 0 < d  p and f (h ) + 1 = d 1 in Fp. Then xd, v0 is annihilated by all x with > 0. It follows that we have a homomorphism of U (p ){modules 'f : Z; (f , d ) ,! Z; (f ) (3) 6 given by '(vf0 ,d ) = xd, vf0 . [Note that Z; (f , d ) makes sense because (f ) = (f , d ) by A.2(4).] If (x, ) 6= 0, then 'f is an isomorphism since xp, acts as the scalar (x, )p on these modules. If (x, ) = 0, then the image of 'f is spanned over K by all xi, vf0 with d  i < p. In this case the cokernel of 'f [the factor module of Z; (f ) by that image] has dimension d; let us denote it by L; (f ). Note that L; (f ) = Z; (f ) in case d = p. If d < p, then the image of 'f is clearly isomorphic to L; (f , d ). If f 0 2 h satis es f 0 (h ) = 0, then we can extend Kf 0 from a b+{module to a p {module such that x, acts as 0. This is then a U(f 0 )(p ){module where we extend (f 0 ) to a linear form on p such that (f 0 )(n+ + g, ) = 0. We get then [as in A.2(7)] rst Z; (f ) Kf 0 ' Z+(f 0); (f + f 0 ) and then L; (f ) Kf 0 ' L+(f 0); (f + f 0 ) (4) whenever L; (f ) is de ned. A.4. Let again f 2 h and  2 g with (n+) = 0 and f 2 . We have by transitivity of induction an isomorphism of U(g){modules  U (g) Z (f ) ,! (1)  U(p ) Z; (f ) that maps vf to 1 vf0 . If f (h ) 2 Fp, then we get from A.3 an induced homomorphism of U(g){modules 'f : Z(f , d ) ,! Z(f ) (2) where d is the integer with 0 < d  p and f (h ) + 1 = d 1 in Fp. If (x, ) 6= 0, then 'f is again an isomorphism. If (x, ) = 0, then we denote the cokernel of 'f by Z(f; ). So we have then an exact sequence f ' Z(f , d ) ,! Z(f ) ,! Z (f; ) ! 0: (3) Since induction is exact in our situation, we can identify Z(f; ) with the module induced from the cokernel of 'f : Z (f; ) ' U (g) U(p ) L; (f ): (4) We have dim Z (f; ) = dpN ,1 where N = jR+j. If d = p, then Z (f; ) = Z (f ). If d < p, then the image of 'f is isomorphic to Z(f , d ; ). We shall use the notation vf also for the image in Z (f; ) of the standard generator of Z(f ). If f 0 2 h satis es f 0 (h ) = 0 for all 2 R, then we get (as at the end of A.2) a U(f 0 )(g){module Kf 0 where we extend (f 0 ) to a linear form on g such that (f 0 )(n+ + n, ) = 0. One checks easily that A.2(7) induces an isomorphism Z(f; ) Kf 0 ' Z+(f 0)(f + f 0 ; ) (5) whenever Z(f; ) is de ned. 7 A.5. Let R0 be a root subsystem of R. Then g0 = h  L 2R0 g the Lie algebra of a reductive closed subgroup of G. We can then apply A.2 also to g0 working with b+ \ g0 instead of b+. We shall write then Z(f ; g0 ) = U (g0 ) U(b+\g0) Kf (1) for the analogue of Z(f ). In case 2 R0 we can also extend A.3/4 and get Z(f; ; g0 ) = U (g0 ) U (p \g0 ) L; (f ) (2) as an analogue of Z(f; ). We do not introduce extra notations for the analogues of Z; (f ) and L; (f ) because the analogous U (p \ g0 ){modules are just the restrictions of the corresponding U (p ){modules. A.6. In some cases we shall have to consider for a given f 2 h at the same time all  such that Z(f; ) is de ned. In that situation it will be convenient to work with a slightly modi ed notation. Set X = f f 2 h j f (h ) 2 Fp g: (1) This is a union of p ane subspaces of h of codimension 1 unless h = 0 (in which case X = h ). Set  = f  2 g j (p ) = 0 g: (2) Extend each (f ) to g such that (f )(n+ + n, ) = 0. Then Z(f )+(f; ) is de ned for all f 2 X and  2 . We write Z (f; ; ) = Z(f )+(f; ): (3) Note that then also all Z (f + ; ; ) with  2 X are de ned because f +  2 X. A.7. Proposition: Let 1 ; 2 2 X . For each integer m the set f(f; ) 2 X   j dim Homg (Z (f + 1; ; ); Z (f + 2; ; ))  m g is closed in X  . (1) Proof : The point is to show that the Hom space in (1) can be described as a space of solutions to a system of linear equations where the matrix of the system has size independent of (f; ) and entries that are polynomial functions of f and . If we have, say, l unknowns, then the Hom space has dimension  m if and only if the rank of the matrix is  l , m if and only if all (l , m + 1)  (l , m + 1) minors of the matrix are 0. This condition clearly de nes a closed subset of X  . In order to show that we are in a situation as described above, let me introduce some notation. Let R be the set of all R+ {tuples of non-negative integers. To each r = (r( )) 2 R we associate elements x,r = Y >0 xr,( ) and x+r = Y >0 xr( ) 8 in U (g) where these products are to be carried out in some xed order. That order is arbitrary except that xr,( ) should occur in x,r as the factor most to the right. Choose a basis h1, h2; : : : ; hn of h. Let S be the set of all n{tuples of nonnegative integers. Associate to each t = (t(i))i 2 S the element n Y ht = i=1 hti(i) in U (h)  U (g). So the x,r htx+s with r; s 2 R, t 2 S are a PBW basis of U (g). Let d1 and d2 be the integers with 0 < di  p and (f + i )(h ) + 1 = di1 in Fp. Set Ri equal to the set of all r 2 R with r( ) < di and r( ) < p for all other positive roots. Then the x,r vf +i with r 2 Ri are a basis of Z (f + i ; ; ). Claim: Let a 2 g. There exist cisr (a; f; ) 2 K [X  ] such that a x,r vf +i = X s2Ri cisr (a; f; ) x,s vf +i (2) for all r 2 Ri . Well, to start with there are cs;t;s0;r (a) 2 K (independent of f and , almost all equal to 0) such that a x,r = in U (g). It follows that a x,r vf +i = XX X s2R t2S s0 2R XX s2R t2S cs;t;s0;r (a)x,s htx+s0 cs;t;0;r(a)(f + i )(ht )x,s vf +i : (3) Here 0 denotes the R+{tuple with all components equal to 0. Given s 2 R there exist unique s0 ; s00 2 R with s( ) = s0 ( ) + ps00 ( ) and 0  s0 ( ) < p for all . We denote then s0 by sred. Then we have x,s vf +i = bs ()x,sred vf +i with bs () = Y s00 ( )>0 (x, )ps00( ): If sred( )  di, then x,sred vf +i = 0. So it is enough to consider in (3) only s with sred 2 Ri . We get now a formula as in (2) with cisr (a; f; ) = X X s0 2R;s=s0red t2S bs0 ()cs0 ;t;0;r (a)(f + i)(ht ): 9 This sum is a regular function on X  , since each bs0 () is polynomial in  and each n Y (f + i)(ht ) = (f + i)(hi )t(i) i=1 is polynomial in f . This nishes the proof of the claim. We now return to the proof of the proposition. The space of all linear maps from Z (f + 1; ; ) to Z (f + 2; ; ) has basis Esr with r 2 R1, s 2 R2 such that for all r0 2 R1  , 0 , Esr (xr0 vf +1 ) = xs vf +2 ; if r = r; 0; otherwise. Given a 2 g we can now use (2) to evaluate each (a
Esr )(x,r0 vf +1 ) = a (Esr (x,r0 vf +1 )) , Esr (ax,r0 vf +1 ) and get a
Esr = X t2R2 c2t;s(a; f; )Etr , X u2R1 c1r;u(a; f; )Esu : So we have a formula of the form a
Esr = X X r0 2R1 s0 2R2 bs0 ;r0 ;s;r (a; f; )Es0 r0 where the bs0 ;r0 ;s;r (a; f; ) are regular functions of (f; ) 2 X  . Now Homg (Z (f + 1 ; ; ); Z (f + 2 ; ; )) identi P es with the space of all (R2  R1 ){tuples (xsr )r;s of elements in K with a
r;s xsr Esr = 0 for all a 2 g. It actually suces to take for a all elements in a basis (or a generating system) of g. This shows that we indeed get the Hom space as a solution space of a linear system of equations as described at the beginning of the proof. The proposition follows. 10 B Keep the assumptions and notations from Section A. We shall have to make in this section (and in most sections to come) two restrictive assumptions on g: We assume that (B1) The derived subgroup DG of G is simply connected and (B2) The group X=ZR has no p{torsion. These assumptions are needed to introduce translation functors with \nice" properties, see B.1 and B.3 below. If G is semi-simple our assumptions mean that X is equal to the weight lattice of R and that p does not divide the index of connection, i.e., the index of the root lattice ZR in the weight lattice. If you want to compare with the assumptions in [11], 6.3: Our (B1) is called (H1) there, our (B2) follows from (H1) and (H3) there, see [11], 11.2. B.1. Fix  2 g with (b+) = 0. The set  from A.2(5) consists then of all d with  2 X . A simple U(g)-module is therefore the homomorphic image of some Z () with  2 X , cf. [10], 1.4 or [11], 6.7. The subalgebra U (g)G (cf. [11], 9.1) of the centre of U (g) acts on each Z() via a character. Let C denote the category of all nite dimensional U(g){modules M such that U (g)G acts on each composition factor of M via the same central character as on Z(). Our assumption (B1) implies that C = C ()  2 W  + pX; (1) cf. [11], 9.4. The simple modules in C are the simple homomorphic images of the Z(w) with w 2 W . Since all these Z(w) de ne the same class in the Grothendieck group (cf. [10], 1.5) we get also that the simple modules in C are the composition factors of Z (). The category C of all nite dimensional U (g){modules is the direct sum of all C with  running over representatives for the orbits on X of the semi-direct product of W and pX [acting via (w; p )
 = w + p ]. Each (closed) alcove with respect to Wp contains a representative of each orbit, cf. [11], 11.19 or [12], 4.1. In general, it will contain more than one representative. If M is in C , let pr(M ) denote the largest submodule of M that belongs to C . Then M is the direct sum of pr(M ) and other pr (M ). We get thus an exact functor pr : C ! C. B.2. Given ;  in a xed (closed) alcove C with respect to Wp, we de ne a translation functor T : C ! C (as usual) by T (V ) = pr(E V ) (1) 11 where E is the simple G{module with highest weight in W ( , ). (This makes sense: see the argument in [12], 4.7 where there is a more restrictive assumption on , which however does not a ect the argument here.) We have clearly T = Tww++pp for all w 2 W and  2 X . Note that we get (in general) more than one functor C ! C for xed  and 0 : if  and 0 are two distinct weights in C with 0 2 W  + pX , then T and T will be two (in general:) distinct functors from C to C = C0 . We have as usual that each T is exact and that T and T are adjoint to  Hom (M; E  N ) each other: The natural isomorphism Homg (E M; N ) ,! g induces an isomorphism  Hom (M; T  (N )); Homg(T (M ); N ) ,! (2) g  cf. [12], 4.7. B.3. For the next result we have to use the assumption (B2) made at the beginning of this section. Proposition: Let ;  2 X and w 2 Wp. Suppose that there exists a closed  alcove with respect to Wp containing  and . Then T Z (w) has a ltration with factors Z(ww1 ) with w1 2 Wp, w1 = . There is one factor for each weight of the form ww1. Proof : Consider E as in the de nition of T . The tensor identity (see [12], 1.12(1)) implies that E Z(w) has a ltration with factors Z(w +  ) with  running over the weights of E (counted with multiplicities). Then T Z (w) has a ltration with factors Z (w +  ) where  runs over all weights of E with w + p 2 W  + pX . Suppose that  has this property; so there are w1 2 W and 1 2 X with w +  = w1 + p1. We have w 2  + ZR and  2 ( , ) + ZR and w1  2  + ZR, hence p1 2 ZR. In other words, we have p(1 + ZR) = 0 in X=ZR. Assumption (B2) yields now 1 2 ZR, hence w1 + p1 2 Wp. So TZ (w) has a ltration with factors Z(w +  ) where  runs over those weights of E with w + p 2 Wp + pX . Now the claim follows from standard results, cf. [9], II.7.13. B.4. Note that Proposition B.3 implies in particular: If  is in the closure of the facet of  with respect to Wp, then we have T Z (w) ' Z (w) (1) for all w 2 Wp. Lemma: If  is in the closure of the facet of  with respect to Wp, then we have TL 6= 0 for all simple modules L in C . Proof : There exists w 2 W with Homg (Z (w); L) 6= 0. Now (1) and the adjointness property B.2(2) imply that Homg(Z (w); T L) ' Homg (Z(w); L) 6= 0; hence the claim. Remark : The claim extends of course to all non-zero modules in C . 12 B.5. Proposition: Suppose that  and  belong to the same facet with respect  to Wp. Then T L is simple for each simple module L in C. The functor T induces a bijection from the isomorphism classes of simple modules in C to the isomorphism classes of simple modules in C ; the inverse is induced by T. Proof : Consider a composition series Z() = Mr  Mr,1 


 M1  M0 = 0 of Z(). The exactness of T and B.4(1) yield a chain of submodules Z () ' T Mr  T Mr,1 


 T M1  T M0 = 0: Each T Mi=T Mi,1 ' T (Mi =Mi,1 ) is non-zero by Lemma B.4. It follows that the length of Z () is greater than or equal to the length of Z (). By symmetry we get also the reversed inequality. Therefore both modules have the same length. This implies that all T (Mi =Mi+1 ) are simple, hence the rst claim of the proposition. We get by symmetry: If L0 is simple in C , then also TL0 is simple. It follows that TT L is simple for all L simple in C . We have by adjointness Homg(T T L; L) ' Homg (T L; T L) 6= 0; hence T TL ' L (both modules being simple). We get by symmetry: TTL0 ' L0 for all L0 simple in C . The second claim follows. B.6. Let I be a subset of the set of simple roots. Let PI be the corresponding parabolic subgroup of G and pI its Lie algebra. So pI is the direct sum of b+ and the g with < 0 and 2 RI = R \ ZL I . Let GI  T be the standard Levi factor of PI and gI its Lie algebra, gI = h  2RI g . For each  2 X set Z;I () = U(pI ) U (b+ ) K : (1) Note that this generalises the case jI j = 1 considered in A.3(1) except for that we consider here more restrictive . We have [as in A.4(1)] by transitivity of induction an isomorphism of U(g){ modules  U (g) Z () ,! (2)  U (pI ) Z;I (): If we write v;I for the standard generator 1 1 of Z;I (), then the isomorphism in (2) takes v to 1 v;I . The nilradical nI of pI (the sum of the g with > 0 and 2= RI ) annihilates each Z;I (). Considered as a module over gI ' pI =nI , each Z;I () identi es with the corresponding baby Verma module for U(gI ). The Weyl group WI of RI identi es with the subgroup of W generated by the s with 2 I ; similarly for the corresponding ane Weyl group WI;p. The group GI again satis es (B1) and (B2); for (B2) note that ZRI = ZI is a direct summand of ZR. We de ne categories C (I ) of nite dimensional U(gI ){modules 13 analogous to the C: A nite dimensional U (gI ){module M belongs to C (I ) if and only if all its composition factors are composition factors of some Z;I (w) with w 2 WI . We can extend each U (gI ){module M to a U(pI ){module letting nI act trivially. We get then an induced U(g){module that we shall denote by indI M : indI M = U (g) U(pI ) M: We have by (2) and the remarks following it indI Z;I () ' Z(): (3) (4) Clearly indI is an exact functor. Lemma: Let  2 X . If M is in C (I ), then indI (M ) is in C. Proof : This is clear for M = Z;I (w) with w 2 WI by (4). It then follows (by the exactness of indI ) rst for all simple modules in C (I ) and then for all M in C (I ) . B.7. Keep the notations and assumptions from B.6. The category of all nite dimensional U(gI ){modules is the direct sum of all C (I ) with  running over suitable representatives. We denote the corresponding projection functors by pr(I ) . If  and  are weights in the same (closed) alcove with respect to WI;p then we can de ne a translation functor T (I ) working with the simple GI {module E 0 with highest weight in WI ( , ). If  and  are weights in the same (closed) alcove with respect to Wp then  and  belong also to the same (closed) alcove with respect to WI;p. So both T and T (I ) are de ned and we want to compare them. Choose w1, w2; : : : ; wr 2 Wp with wi  =  such that each w with w 2 StabWp  is conjugate to exactly one wi under the stabiliser of  in WI;p. Proposition: There exists for weach V in C (I ) a ltration of T (indI V ) with   factors isomorphic to indI (T (I ) i V ), 1  i  r. Proof : [Note that we do not claim that the factors in the ltration occur in the same order as the indices.] Let E be the simple G{module with highest weight in W ( , ). The tensor identity yields an isomorphism  U (g) E indI V = E (U U(g)(pI) V ) ,!  U (pI ) (E V ) that takes any e (1 v) with e 2 E and v 2 V to 1 (e v). We get thus a functorial isomorphism  pr U (g) T (indI V ) ,! (1) U(pI ) (E V ):   Consider a composition series of E as a PI {module. The unipotent radical of PI acts trivially on the factors (hence so does its Lie algebra nI ) and these factors 14 are simple GI {modules. It follows that T(indI V ) has a ltration with factors pr indI (L V ) with L running over the factors in a composition series of E as a PI {module. Each L V is the direct sum of all pr(I )0 (L V ) with 0 running over representatives for the orbits of WI on X=pX . So T (indI V ) has a ltration with factors pr indI pr(I )0 (L V ) (2) with L and 0 as above. Since  ,  is a weight of E with multiplicity 1, so is each wi ,  = wi , wi 2 W ( , ). Therefore our composition series of E as a PI {module contains exactly one factor, say Ei, such that wi ,  is a weight of Ei . Now wi ,  is an extremal weight of the GI {module Ei since it is an extremal weight of the G{module E . (The last statement means for all 2 R that either no wi ,  + n with n > 0 is a weight of E or that no wi ,  , n with n > 0 is a weight of E or both.) Therefore wi  ,  is conjugate to the highest weight of Ei under W and we can take Ei as the simple module E 0 used in the de nition I of T (I )w i  . This means that T (I )w i (V ) ' pr(I )wi  (Ei V ): So indI (T (I )w i wV )is one of the factors as in (2). [Note that indI (T (I )w i  V ) = pr indI (T (I ) i V ) by Lemma B.6.] Let us check next that distinct i lead here to distinct factors. Otherwise we have i 6= j with (Ei ; pr(I )wi  ) = (Ej ; pr(I )wj  ). This implies wi ,  2 WI (wj  , ) and wi 2 WI (wj ) + pX . The rst property yields wi  2 wj  + ZI ; since both X=ZR and ZR=ZI have no p{torsion, the second property implies that wi 2 WI (wj ) + pZI = WI;p(wj ). So  = wi ,  is a weight of the simple GI {module with extremal weight wj  ,  such that  +  2 WI (wj ). Now [9], II.7.7 implies that there exists w 2 WI;p with w =  and w(wj ) =  +  = wi. This is a contradiction to the choice of the wi. We have so far shown that all indI (T (I )w i  V ) occur in the ltration of  T (indI V ) with factors as in (2). It remains to be shown that all remaining factors as in (2) are 0. Using the exactness of pr , of all pr(I )0 , and of the induction one reduces rst to the case where V is simple and then to the case where V = Z;I (w) for some w 2 WI . That all other factors as in (2) are 0 in this case will follow if we can show that r X  dim T (indI V ) = dim indI (T (I )w i  V ) i=1 for V = Z;I (w). In this case indI V ' Z(w); so T (indI V ) has by B.3 a ltration with factors Z (w0 ) with w0  running over wStabWp (). Similarly, each T (I )w i  V has a ltration with factors Z;I (w0 wi ) with w0 wi running over wStabWI;p ()wi . Then indI (T (I )w i  V ) has a ltration with factors Z (w0 wi) and w0 as before. Now the claim follows because StabWp () is the disjoint union of the StabWI;p ()wi . 15 Remark : The following situation is particularly easy: Suppose that  is contained in the closure of the facet of  with respect to Wp. Then there is only one factor in the ltration (since w =  for all w in the stabiliser of ). So we get in this case an isomorphism  ind (T (I ) V ): T (indI V ) ,! (3) I  This map is functorial: We have for each morphism ' : V ! V 0 in C (I ) a commutative diagram  ind (T (I ) V ) T (indI V ) ,! I  ? T (indI ')? y T (indI V 0 ) ?? yindI (T (I )('))  ind (T (I ) V 0 ) ,! I  where the horizontal maps are isomorphisms as in (3). This follows from the functoriality of the isomorphism in (1) and the fact that idE ' induces morphisms pr(I )0 (L V ) ! pr(I )0 (L V 0 ) for all L and 0 as in (2). B.8. We want to apply B.6/7 in the case where jI j = 1. We rst have to prove  some results on T (I ) in that case. In order to simplify the notations we assume in this and the following subsection that gI = g. So assume now that G has semi-simple rank equal to 1. There is then exactly one positive root; denote it by . In this case our baby Verma module Z() coincides with the module Z; () in the notation from A.3. We can use the explicit description of a basis for this module there, and we have the homomorphism ' : Z ( , d ) ! Z () as in A.3(3) given by '(v,d ) = xd, v. Here d is the integer with 0 < d  p and h + ; _i  d (mod p), hence (h ) + 1 = d1 in Fp. We claim now that Homg (Z ( , d ); Z ()) = K' if d < p. (1) Indeed, our assumption (B2) implies that ,i 6 ,j (mod pX ) whenever 0  i; j < p and i 6= j . So the xi, v with 0  i < p belong to distinct eigenspaces with respect to h, hence span these eigenspaces. Each homomorphism from Z( , d ) to Z() has to take '(v,d ) to an eigenvector for the same eigenvalue, hence to a multiple of xd, v. So (1) follows. The same argument shows that Homg(Z ( , p ); Z ()) = Homg(Z (); Z ()) = K id : (2) (Note that Z ( , p ) = Z ().) If d = p, then ' is equal to multiplication by (x, )p since x[,p] = 0. Therefore (1) extends to the case d = p if (x, ) 6= 0. For (x, ) = 0, however, we have ' = 0 in this case. B.9. Keep the assumptions from B.8. Consider now two weights  and  such that  is in the closure of the facet of  with respect to Wp. There is a unique integer n with np < h + ; _i  (n + 1)p: 16 The assumption on the facet implies that np  h + ; _i  (n + 1)p: Set d = h + ; _i , np and d0 = h + ; _ i , np. Then d is the same number as considered in B.8. If d0 > 0, then it the analogue of d working with  instead of ; if d0 = 0, then that analogue is equal to p, however. Note that s ;np =  , d and s ;np =  , d0 . We get from B.4(1) isomorphisms T Z() ' Z () and T Z( , d ) ' Z ( , d0 ). We claim that, modulo these isomorphisms: Lemma: If d0 > 0, then T (') is a non-zero multiple of '. If d0 = 0, then T (') is a non-zero multiple of the identity. Proof : If d = p, i.e., if h + ; _i = (n + 1)p, then the assumption on the facets implies that also h + ; _i = (n + 1)p and d0 = p. In this case both ' and '  p are equal to (x, ) times the identity. Since the functor T takes a multiple of the identity to the corresponding multiple of the identity, the claim follows in this case. Assume from now on that d < p. Case I: We have h + ; _i  h + ; _ i. Then the module E involved in the construction of T has highest weight s ( , ) =  ,  + (d , d0 ) . If e0 is a highest weight vector of E , then v = e0 v,d is a highest weight vector of weight  , d0 in E Z( , d ) and generates the summand TZ ( , d ) ' Z( , d ). The map T (') is the restriction of idE ' to T Z( , d ). It maps v to e0 xd, v, hence is non-zero (since d < p). Now the claim follows from B.8(1) in case d0 > 0, or from B.8(2) in case d0 = 0. Case II: We have h + ; _i > h + ; _i. Now E has highest weight  , . It has a basis ei = xi, e0 with 0  i < d0 , d = h , ; _i where e0 is a weight vector of weight  ,  and satis es x e0 = 0. We have x ei = i(d0 , d + 1 , i)ei,1 for all i > 0. A straightforward calculation shows that T Z( , d ) ' Z ( , d0 ) is generated by 0 ,d  0  dX d e xd0 ,d,i v : v= i ,d , i=0 i (Note that v 6= 0 since the coecient of e0 now 0 ,d  0  dX d e T (')v = i i i=0 xd,0 ,dv,d is non-zero.) We have xd,0 ,i v,d : If d0 < p, then a look at the \leading term" e0 xd,0 v,d shows that T (') 6= 0 and the claim follows from B.8(1). If d0 = p, then T(' )v = e0 xp, v,d = (x, )pe0 v,d : We see thus that T (') 6= 0 if and only if (x, ) 6= 0 if and only if ' 6= 0. Now the claim is obvious in case (x, ) = 0 and follows otherwise from B.8(2). 17 B.10. Let G again be arbitrary. Choose a simple root and set I = f g. We get from A.4(2) for each  2 X a homomorphism ' : Z( , d ) ! Z() with ' (v,d ) = xd, v . Here d is again the integer with 0 < d  p and h + ; _i  d (mod p). We can identify Z() with indI Z; () [see A.4(1)]; then ' identi es with indI ('). Proposition: Let ;  2 X such that  is contained in the closure of the facet of  with respect to Wp. If h + ; _ i  0 (mod p) and h + ; _ i < h + ; _i, then T (' ) identi es with a non-zero multiple of the identity on Z (). Otherwise T (' ) identi es with a non-zero multiple of ' . Proof : As stated above, we can identify ' with indI ('). The functoriality in B.7(3) implies that we can identify T (' ) with indI (T (I ) '). Now the claim follows from Lemma B.9 applied to gI . (Note that the rst case in the present proposition corresponds to the case d0 = 0 in B.9.) B.11. Let be a simple root with (x, ) = 0. In this situation we have de ned in A.4 the module Z(; ) as the cokernel of ' . Corollary: Let ;  2 X such that  is contained in the closure of the facet of  with respect to Wp. If h + ; _i  0 (mod p) and h + ; _i < h + ; _i, then T Z(; ) = 0. Otherwise T Z(; ) ' Z(; ). Proof : This follows from Proposition B.10 when we apply the exact functor T to the exact sequence A.4(3). Remark : Let I be a subset of the set of simple roots such that the restriction of  to gI has standard Levi form (see [11], 10.1 or [12], 2.2). We have then for each  2 X a unique simple quotient L;I () of Z;I () and can de ne induced modules Z (; I ) = indI L;I (). We can apply B.7(3) and describe T Z(; I ) (for  in the closure of the facet of ) using [12], 4.11(4). The corollary above is a special case of that more general result. B.12. We shall always write [M : L] to denote the multiplicity of a simple module L as a composition factor of a module M . For each simple U (g){module L let QL denote the projective cover of L in the category of all U(g){modules. The proof of Lemma 10.9 in [11] shows that X dim QL = pN [Z() : L] (1)  where N = jR+j and where  runs over a system of representatives for X=pX . (In [11], 10.9 one has a more restrictive assumption on  that actually does not enter the argument.) Since QL and hence L belongs to C, only  2 W  + pX can contribute a positive multiplicity [Z() : L]. Since all Z() with  2 W  + pX de ne the same class in the Grothendieck group (cf. [10], 1.5) we get from (4) that dim QL = pN jW ( + pX )j [Z () : L]: (2) 18 Remark : Our assumptions (B1) and (B2) are not needed to prove (1). For (2) we just need (B1). Similarly, one can check that the rst seven paragraphs in B.13 below do not require the assumptions while (B1) suces for the remainder of B.13 and for B.14. B.13. In this and the next subsection we drop our assumption that (b+) = 0 and consider arbitrary  2 g . (But we shall soon assume  to be nilpotent.) Let C () denote the category of all nite dimensional U (g){modules; as before we just write C when it is clear which  we consider. Let g 2 G. We write Ad(g) for the adjoint action of g on g and for the induced action on U (g). If M is a g{module, then we write g M for M \twisted by g", i.e., we take g M = M as a vector space and let any x 2 g (or in U (g)) act on g M as Ad(g,1 )(x) does on M . (See also the more general discussion at the beginning of [10], 1.13.) Clearly M 7! g M is an exact functor that takes simple modules to simple modules. If M is a G{module considered as a g{module via the derived action, then  g we have an isomorphism M ,! M given by m 7! gm. If M is a U(g){module, then g M is a Ug (g){module where g is the image of  under the coadjoint action of g given by (g)(x) = (Ad(g,1)x). The functor M 7! g M is an equivalence of categories from C () to C (g). It takes simple modules to simple modules. Assume now that  is nilpotent. This means that  vanishes on a Borel subalgebra of g, or, equivalently, that there exists g 2 G with (g)(b+ ) = 0, see [13]. If M is a U(g){module, then each u 2 U (g)G acts on each g M as it does on M . Suppose for the moment that (g)(b+) = 0; if M is simple, then U (g)G acts on g M (and hence also on M ) as on some Zg () with  2 X . Set C () equal to the full subcategory of C () consisting of those N in C () such that U (g)G acts on each composition factor of N as on Zg (). Then C () is the direct sum of all C () with  in a suitable set of representatives. Note: If we have already (b+) = 0, then the de nition of C () given above coincides with that one that we get by applying B.1 directly, because each u 2 U (g)G acts on Z() and on Zg () by the same scalar (for each  2 X ), cf. [10], 1.7. This observation implies for arbitrary  that the de nition of C () is independent of the choice of g with (g)(b+ ) = 0. We have (for each nilpotent ) projection functors pr : C () ! C () and translation functors T : C () ! C () de ned as in B.1/2. Let g 2 G. If M is a U (g){module in C () , then g M belongs to C (g). The functor M 7! g M restricts to an equivalence of categories from C () to C (g) . We get for arbitrary M in C () that pr( g M ) = g (pr M ): (1) This implies for M in C () that g (T  M ) ' T  ( g M ):   (2) 19 Indeed, we use the same G{module E when we de ne T on C () and on C (g) and get therefore g (T  M ) = g (pr (E M )) = pr g (E     g ' pr (E M ) = T ( g M ): M ) = pr ( g E gM) Lemma: Suppose that  2 g and g 2 G with (b+) = (g)(b+) = 0. Let  2 X and let L be a simple U (g){module. Then [Zg () : g L] = [Z() : L]: (3) Proof : If QL is the projective cover of L in the category of all U (g){modules, then g (QL ) is the projective cover of L in the category of all Ug (g){modules. Applying B.12(2) to g instead of , we get dim g (QL ) = pN jW ( + pX )j [Zg () : g L]: (4) Since QL and g (QL ) have the same dimension, a comparison of (4) with B.12(1) yields (3). B.14. Let again g 2 G. Given a Lie subalgebra q of g and a q{module M , then we get an Ad(g)q{module g M by an obvious generalisation of the de nition in B.13. If q is a restricted Lie subalgebra of g and if M is a U (q){module, then g M is a Ug (Ad(g)q){module. It is then easy to check that we get for all  2 g an isomorphism for the induced modules g (U (g)  U (q) M ) ,! Ug (g) Ug (Ad(g)q) gM (1) induced by u m 7! Ad(g)(u) m. Let B+ = P; (cf. B.6) be the Borel subgroup of G with Lie algebra b+. If  2 g with (b+) = 0, then we get applying (1) with q = b+ g Z () ' Zg () for all  2 X and g 2 B+ (2) since Ad(g)(b+ ) = b+ and since Ad(g) acts trivially on b+=n+. Let be a simple root and let P  B+ be the standard parabolic subgroup with Lie algebra p = b+ + g, . Suppose that (p ) = 0. Then we claim that g Z (; ) ' Zg (; ) for all  2 X and g 2 P . (3) We want to apply (1) using A.4(4). We can replace  by a weight in  + pX and assume that 0  h; _ i < p. Then we get L; () by taking the derived action of p on the simple P {module with highest weight . This then implies g L; () ' L; (), hence via (1) the claim. 20 C Keep all assumptions and notations from Section B, in particular (B1) and (B2). However, one may check that (B1) and (B2) are used only for C.1{4 and C.9{10. We introduced in A.1 an element  2 X Z Q. If (B1) holds, then we can choose  2 X . We assume in future that we have made such a choice whenever (B1) holds. (If G is not semi-simple, then  is not necessarily half the sum of the positive roots.) C.1. Set C0 = f  2 X j 0  h + ; _i  p for all 2 R+ g (1) and C00 = f  2 X j 0  h + ; _i < p for all 2 R+ g: (2) So C0 is the usual \ rst dominant alcove" C0 as in [11], 11.19(1) or [12], 4.1(1). We begin now (for certain ) an investigation of the subcategories C with  2 C00 . If R has no components of exceptional type, then one can show (see H.1 below) that there exists for each  2 X a weight  2 C00 with  2 W  + pX , hence with C = C . So in those case we do not lose anything by the restriction to weights in C00 . Lemma: Let  2 C00 . If w 2 W and  2 X with w + p = , then  = 0. Proof : If w + p = , then p 2 ZR since w 2  + ZR. So our assumption (B2) implies that  2 ZR. This means that the map x 7! wx + p belongs to the stabiliser of  in the ane Weyl group Wp. Since Wp is a re ection group, that stabiliser is generated by all s ;rp with 2 R+ and h + ; _i = rp. The de nition of C00 implies that then necessarily rp = 0. Therefore the stabiliser of  in Wp is contained in W ; this yields  = 0. C.2. Proposition: Let  2 C00 . There exists a simple module L in C with projective cover isomorphic to T, Z(,); it satis es [Z() : L] = 1. Proof : Let us abbreviate Q = T, Z(,). Proposition B.3 implies that Q has a ltration with factors Z () with  2 W , one factor for each such . We get in particular that dim Q = jW j pN (1) with N = jR+j. One knows that Z(,) is projective, cf. [7], Thm. 4.1. Any translation functor T takes projective modules to projective modules since the adjoint functor T is exact. Therefore Q is projective. Let L be a simple quotient of Q. Then the projective cover QL of L has to be a direct summand of Q. We have by B.12(2) dim QL = pN jW ( + pX )j [Z () : L]: (2) 21 Now Lemma C.1 implies that the stabiliser of  + pX in W is equal to the stabiliser of  in W . Using (1) and the fact that QL is a summand of Q, we get that dim Q = pN jW j  dim QL = pN jW j[Z() : L]  pN jW j: So we have to have equality everywhere, hence QL = Q and [Z() : L] = 1. So the claim follows. C.3. Proposition C.2 yields a representation theoretic proof of the following special case of a recent theorem of Brown and Gordon in [2], 3.18: Corollary: If  2 C00 , then the subcategory C is a block of the category of all nite dimensional U (g){modules. Proof : If C is not a block, then it is a direct sum of two non-trivial subcategories that are closed under taking subquotients. The indecomposable module Q considered in C.2 would have to belong to one of them. Then so would be all subquotients Z(w) with w 2 W of Q, hence all simple modules in C . Then the other subcategory will contain no simple modules at all, hence be trivial. Remark : The theorem in [2] says that all C with  2 X are blocks. That proves the conjecture by Humphreys in [8], Section 18. (For arbitrary type such a result had previously been known only for  in standard Levi form, cf. [8].) The corollary here together with H.1 yields that conjecture in case R has no components of exceptional type (and p 6= 2 if it has components not of type A). C.4. Let w0 2 W denote the unique element with w0(R+ ) = ,R+. Proposition: Let  2 C00 , let L denote the simple module in C with projective cover isomorphic to T,Z (,). a) Up to isomorphism L is the only simple module in C with T,L = 6 0. b) The socle of Z () and the head of Z(w0 ) are isomorphic to L. We have dim Homg (Z (w0); Z ()) = 1: (1) Each non-zero homomorphism from Z(w0 ) to Z() has image equal to the socle of Z (). Proof : a) It is known that Z(,) is simple, cf. [7], Thm. 4.2. (The assumption in that theorem that p is good for R is not needed for this particular result.) Therefore Z(,) is the only simple module in C, (up to isomorphism). If M is a module in C with T,M 6= 0, then we get 0 6= Homg(Z (,); T, M ) ' Homg (T, Z(,); M ): If we assume additionally that M is simple, then we get that M is a homomorphic image of the projective cover of L, hence isomorphic to L. 22 b) Let E be a G{module; suppose that 1, 2; : : : ; r are the weights of E (counted with multiplicities) enumerated such that i > j implies i < j . Then any E Z () has a ltration with factors Z ( + i) such that Z ( + 1) occurs at the bottom, then Z( + 2 ) as the next higher factor, and so on. Each pr (E Z ()) has then a similar ltration where only the Z( + i) with  + i 2 W  + pX occur. In particular, if i1 (resp. i2 ) is minimal (resp. maximal) among the i with  + i 2 W  + pX , then Z( + i1 ) is a submodule of pr (E Z()), and Z ( + i2 ) is a homomorphic image of pr (E Z ()). Applied to Q = T,Z (,) this shows that Z() is a submodule of Q and that Z(w0 ) is a factor module. Since Q (being the projective cover of L) has simple head isomorphic to L, so has Z(w0 ). Since U (g) is a symmetric algebra (see [6], Prop. 1.2), the simple module L is also isomorphic to the socle of Q, hence to that of Z(). It is now clear that we get a non-zero homomorphism from Z(w0) to Z() by projecting rst onto L and then embedding L into Z (). Conversely, let ' be a non-zero homomorphism from Z (w0) to Z(). The image of ' has simple head and simple socle, both isomorphic to L. Since [Z() : L] = 1, the image of ' has to be isomorphic to L. Therefore the image is equal to the socle of Z(), and the kernel of ' has to be equal to the radical of Z(w0 ). Now (1) follows from Schur's Lemma. Remark : The entries in the row (or column) of the Cartan matrix of C corresponding to L are equal to jW ( + pX )j[Z () : L0 ] with L0 running over the simple modules in C . In particular, the diagonal entry is equal to jW ( + pX )j and the other entries are non-zero multiples of that number. This implies that Lusztig's original conjecture in [15], 14.5 for the Cartan matrix could not work in type D4 (for example). C.5. Fix  2 C00 . If w 2 W and if is a simple root with w,1 > 0, then 0  hw( + ); _ i < p and we have a homomorphism 'w;s w : Z(s w) ,! Z(w) (1) given by _ 'w;s w (vs w) = xh,w(+); ivw: (2) If hw( + ); _ i = 0, then 'w;s w is the identity map on Z(). If hw( + ); _ i > 0, then 'w;s w is up to a non-zero scalar multiple of the map 'w from A.4(2). We get therefore from A.4: Lemma: If (x, ) 6= 0, then 'w;s w is an isomorphism. If (x, ) = 0, then and where N = jR+j. im 'w;s w ' Z(s w; ) (3) dim im('w;s w ) = (p , hw( + ); _ i)pN ,1 (4) 23 C.6. We now want to de ne for each w 2 W a homomorphism 'w : Z (w) ,! Z() (1) as follows: If w = s1s2 : : : sr is a reduced decomposition (with si = s i for some simple root i ) then we want to have 'w = '1;sr  'sr ;sr,1sr 


 's2:::sr,1sr ;s1s2:::sr,1sr (2) where the single factors are de ned in C.5. One has to check the independence of the right hand side in (2) of the chosen reduced decomposition. It is (as usual) enough to check the \braid relations" for each pair ; , ( 6= ) of simple roots. For example, if s s has order 3, then we have to check for each x 2 W with x,1 ; x,1 > 0 that 'x;s x  's x;s     s x  's s x;s s s x = 'x;s x  's x;s s x  's s x;s s s x : This follows from the Verma relations, see [4], 7.8.8; similarly in the cases where s s has order 2, 4, or 6. Now that the 'w are well-de ned, we get for all w 2 W and all simple roots with w,1 > 0 that 's w = 'w  'w;s w (3) since we get a reduced decomposition of s w when we multiply a reduced decomposition of W on the left by s . C.7. Set for all w 2 W sbm(w; ) = im('w )  Z () (1) the subm odule corresponding to w. Lemma: Let be a simple root and w an element in W with w,1 > 0. a) We have sbm(s w; )  sbm(w; ). b) If (x, ) 6= 0, then sbm(s w; ) = sbm(w; ). c) If (x, ) = 0, then sbm(s w; ) is a homomorphic image of Z (s w; ), and sbm(w; )= sbm(s w; ) is a homomorphic image of Z (w; ). Proof : The identity 's w = 'w  'w;s w from C.6(3) implies a). If (x, ) 6= 0, then 'w;s w is an isomorphism; this yields b). Suppose now that (x, ) = 0. Denote the image of 'w;s w by M . Then M is isomorphic to Z (s w; ), and Z(w)=M is isomorphic to Z(w; ). Now the claim in c) follows from sbm(s w; ) = 'w (M ) and sbm(w; ) = 'w (Z (w)). Remark : It is sometimes more convenient to restate the last part of the lemma as follows: Let w 2 W and 2 R+ such that w is a simple root with (x,w ) = 0. Then sbm(ws ; ) is a homomorphic image of Z(ws ; w ), and sbm(w; )= sbm (ws ; ) is a homomorphic image of Z(w; ). [Note that sw w = ws .] 24 C.8. We now want to generalise Lemma C.7.a to positive roots that are not necessarily simple: Proposition: Let w 2 W and 2 R+ with w,1 > 0. Then sbm(s w; )  sbm(w; ). Proof : This will follow in the same way as in C.7 if we can nd a homomorphism 'w;s w : Z(s w) ! Z(w) 's w = 'w  'w;s w : with (1) Suppose rst that G is semi-simple and simply connected; drop the assumption that (B2) should hold. In order to construct the map in (1) we make a detour to characteristic 0. Let gC be a complex semisimple Lie algebra of the same type as g. Fix a triangular decomposition gC = n,C  hC  n+C and a Chevalley basis of gC . Denote by gZ the span over Z of our Chevalley basis. This is a Lie algebra over Z with a triangular decomposition gZ = n,Z  hZ  n+Z induced by that of gC . We can and shall assume that g = gZ Z K , similarly for n and h. We have then also U (g) = U (gZ ) Z K and similarly for n and h. We denote (by abuse of notation) the vectors in our Chevalley basis of gC by x and h ( ; 2 R, simple). We assume that we have chosen the x 1 as our root vectors in g (which we usually denote by x ). The group X can be identi ed with the lattice of integral weights of hC . We have for each  2 X a Verma module M ( )C for gC with highest weight  ; we denote its standard generator by z . We de ne for each w 2 W a homomorphism fw : M (w)C ! M ()C in the same way as we de ned in C.6(2) the 'w . There is a unique element uw 2 U (n,C ) with fw (zw) = uw z: (2) The construction shows that uw is a product of powers of the x, with simple, hence contained in U (n,Z ), and that 'w (vw) = (uw 1)v: (3) Let 1 , 2 ; : : : ; n denote the simple roots. If we write u(w; ) as a linear combination of the usual PBW basis of U (n,C ), then uw = n Y i=1 xr,i i + lower order terms if  , w = n X i=1 ri i (4) where \lower order terms" refers to the canonical ltration of an enveloping algebra as in [4], 2.3.1. Now consider w and as in the proposition. The theory of Verma modules shows that there exists a unique homomorphism  fw;s w : M (s w)C ! M (w)C   with fw  fw;s w = fs w ; (5) 25 see [4], 7.6.6 and 7.6.23. There is a unique element   uw;s w 2 U (n,C ) with fw;s w (zs w ) = uw;s w zw : A comparison with (2) shows that us w = uw;s w uw: A look at the terms of highest order shows that uw;s w = n Y i=1 xr,i i + lower order terms if w , s w = n X i=1 (6) (7) ri i : (8) P If we write d = hw( + ); _ i, then we have above d = ni=1 ri i . This shows that uw;s w is equal to the element denoted by S ;d(w) in [5], Section 3. (Note that there is a misprint in the last displayed equation on p. 66 of [5]: One should replace r by dr.) The results in [5] say that uw;s w 2 U (n,Z ): (9) (See the remarks on the top of page 67 in [5].) We now want to use this element to de ne 'w;s w by 'w;s w (vs w ) = (uw;s w 1) vw: (10) If this is possible, then (7) and (3) yield the equality 's w = 'w  'w;s w , hence the proposition. The right hand side in (10) has weight s w; it therefore suces to show that this term is annihilated by all x 1 with > 0. We have to start with in U (gZ ) X x uw;s w = F c + terms in U (gZ )n+Z  where the F are (as in [5]) the elements in a PBW basis of U (n,Z ) and where all c 2 U (hZ). We have then X 0 = x uw;s w zw = (w)(c )F zw:  Since the F zw are linearly independent, we get (w)(c ) = 0 for all . Since X (x 1)(uw;s w 1) vw = (w)(c )(F 1)vw = 0  the claim follows (for G semi-simple and simply connected). The extension to the case where G is a direct product of a semi-simple and simply connected group with a torus is immediate. In general, G is a quotient of such a group, say Ge, by a central subgroup. The corresponding homomorphism  from eg = Lie(Ge) to g satis es g = h + (eg). If we consider a baby Verma module for g as a eg{module via , we get a baby Verma module for eg. We can use the construction above to get a map 'w;s w as in (1) that is a homomorphism of eg{modules. It then suces to show that this map also commutes with h. That, however, follows from the fact that the element uw;s w has weight ,d also for h. 26 Remark : Let  denote the usual Chevalley-Bruhat order on W (with smallest element 1). The proposition implies for all w1; w2 2 W : w1  w2 =) sbm(w1; )  sbm(w2 ; ): (11) C.9. Lemma: Let  be a weight in the closure of the facet of  with  2 C00 . For all w 2 W and 2 R simple and w,1 > 0 we can identify T ('w;s w ) with a non-zero multiple of 'w;s w . We can identify T ('w ) with a non-zero multiple of 'w and have T sbm(w; ) ' sbm(w; ) (1) for all w 2 W . Proof : In order to prove the claim for the 'w;s w we distinguish two cases: If hw( + ); _i = 0, then the assumption on  implies that also hw( + _ ); i = 0. Then both 'w;s w and 'w;s w are the identity. So is T('w;s w ); the claim follows in this case. If hw( + ); _ i > 0, then the claim follows easily from Proposition B.10 because 'w;s w is a non-zero multiple of the map 'w considered there. The rest of the lemma follows now from the de nition of 'w , the functor property of T and its exactness. C.10. We can apply Lemma C.9 to  = ,. Since all ',w are the identity, hence non-zero, we get that 'w = 6 0 for all w 2 W . (1) This holds in particular for w = w0. Proposition C.4.b implies therefore that sbm(w0; ) = soc Z (): (2) C.11. Set I = f 2 R j simple; (x, ) 6= 0 g: (1) Let WI denote the subgroup of W generated by all s with 2 I . Lemma: a) We have sbm(w1 ; ) = sbm(w2; ) for all w1; w2 2 W with WI w1 = WI w2. b) Let w 2 WI and be a simple root with 2= I . Then sbm(s w; ) ' Z(s w; ) (2) and dim sbm(s w; ) = (p , hw( + ); _i)pN ,1 (3) where N = jR+j. Proof : The claim in a) follows easily from Lemma C.7.b. In b) we have w,1 > 0 since w 2 WI and 2= I . It follows that 's w = 'w  'w;s w . Furthermore 'w is a composition of certain 'w0;s w0 with 2 I , hence an isomorphism. Therefore the image sbm(s w; ) of 's w is isomorphic to the image of 'w;s w . Now the claims in b) follow from Lemma C.5. 27 C.12. Here and in the next two subsections we x a simple root with (x, ) = 0 and consider I as in C.11(1). Note that WI  ( + ZI ) \ R  R+: (1) Pick for each 2 WI an element w 2 WI with w,1 = and set M = sbm(s w ; ): (2) dim M = (p , h + ; _i)pN ,1 (3) Lemma C.11.b implies that and that M ' Z(s w ; ): (4) Claim: The submodule M of Z() depends only on , not the choice of w . Proof : If w0 is a second element in WI with (w0 ),1 = , then w = w0 w,1 2 WI satis es w = , hence ws = s w. It follows that s w0 = ws w 2 WI s w ; now apply Lemma C.11.a. C.13. Lemma: Let be a simple root, let w 2 W with w 2 WI . Then sbm(w; )= sbm (ws ; ) is a homomorphic image of Z (ww w; ). Proof : The element x = ww w satis es x = , hence xs = s x. Since ww 2 WI , Lemma C.11.a implies that sbm(w; ) = sbm(x; ) and sbm(ws ; ) = sbm(xs ; ) = sbm(s x; ): Now the claim follows from Lemma C.7.c. C.14. Proposition: Let 2 WI and 2 I with h ; _ i < 0. a) We have and M  Ms (1) dim(M =M s ) = jh ; _ij h + ; _ipN ,1 : (2) b) Assume that h ; _i = h ; _ i = ,1. Then 0 = w ( + ) is a root and belongs to WI ; the element x = w 0 s w 2 W satis es x,1 = and M =M + ' Z(x; ) if h + ; _i > 0. (3) 28 Proof : a) We have clearly s 2 WI and could choose ws = w s . So we have M s = sbm(s w s ; ): On the other hand s = , h ; _i is a positive root not in ZI . Therefore also 0 = w s is positive. Using s = w,1 s w , we get 0 = s w , hence s 0 s w = (s w s w,1s )s w = s w s : Since (s w ),1 0 = > 0, we get from Proposition C.8 that sbm(s 0 s w ; )  sbm(s w ; ); hence (1). Now (2) follows from C.12(3) and (s )_ = _ , h ; _i _ . b) Now our assumptions imply that s = + = s . It follows that + 2 WI and that 0 = w s = w ( + ) 2 WI . We observed above that 0 = s w ; so we can apply Lemma C.13 with w = s w . Now x = w 0 s w is the same x as in the proof of Lemma C.13 and satis es = x . We have M = sbm(w; ) since w s is a possible ws surjective homomorphism =w and + M + = sbm(ws ; ) . Now Lemma C.13 says that we have a Z(x; ) ,! M =M + : (4) The left hand side has dimension equal to hx( + ); _ ipN ,1 = h + ; _ipN ,1 as long as h + ; _i > 0; otherwise its dimension is equal to pN . In the rst case, the surjection in (4) has to be an isomorphism by dimension comparison; this implies (3). Remark : In the situation from b) one can deduce the inclusion in (1) directly from Lemma C.7.a (without the more complicated argument in C.8) using x,1 = > 0 and the equalities M = sbm(x; ) and M + = sbm(s x; ): 29 D We keep the assumptions from Sections B and C. We assume in addition: (D1) The prime p is good for R. and (D2) G is almost simple. The rst assumption is crucial so that we can apply the Kac-Weisfeiler conjecture proved by Premet. The second assumption is mainly meant to simplify the statement of the results. D.1. We call a linear form  2 g subregular if its orbit under the coadjoint action has dimension equal to 2(N , 1) where N = jR+j. If so, then the Kac- Weisfeiler conjecture as proved by Premet says: If M is a U (g){module, then dim(M ) is divisible by pN ,1. Recall that we set p = b+ + g, for each simple root ; we denote (as in B.14) by P the corresponding parabolic subgroup of G. The following result is a translation of well known results on orbits in g: Lemma: There exists a unique subregular nilpotent orbit O in g . If is a simple root then O intersects the set of all  2 g with (p ) = 0 in an open and dense subset. This intersection is one orbit under P . Proof : We can (under our assumptions on p) identify g and g as G{modules. The classi cation of the nilpotent orbits in g is the same as for the Lie algebra over C of the same type (since p is good). In particular, there is exactly one subregular nilpotent orbit in g; this yields the rst claim. The elements  2 g with (b+) = 0 and L (x, ) = 0 correspond (under  g ' g ) to the elements in the nilradical n = >0; 6= g of the parabolic subalgebra p . The theory of the Richardson orbits (cf. [3], 5.2.3) says that there exists exactly one nilpotent orbit for G that intersects n in an open and dense subset. That intersection is one orbit under P ; furthermore the dimension of the orbit (under G) is equal to the codimension in g of a Levi factor of p . Since that codimension is equal to 2(N , 1), we get the remaining claims. Remark : For each simple root 6= the set of all  with (p ) = 0 and (x, ) 6= 0 is an open and dense subset of the set of all  with (p ) = 0. It follows: The set of all subregular  2 g with (p ) = 0 and (x, ) 6= 0 for all simple roots 6= is an open and dense subset of the set of all  with (p ) = 0. D.2. Each subset J of the set of all simple roots de nes a facet F (J ) contained in C00 as follows: A weight  2 C00 belongs to F (J ) if and only if h + ; _i > 0 for all 2 J and h + ; _i = 0 for all simple roots 2= J . Write $ for the fundamental weight corresponding to a simple root P. Then  2 X belongs to F (J ) if there are integers m > 0 such that  +  = 2J m $ and if h + ; _i < p for all 2 R+ . Each facet (with respect to Wp) that is contained in C00 has the form F (J ). 30 Lemma: Let J be as above and let 2 J . Suppose that we have for all  2 F (J ) a module N () in C with dim N () = h + ; _ipN ,1 such that T N () ' N () for all ;  2 F (J ). If  is subregular, then each N () with  2 F (J ) is simple. Proof : We may assume that F (J ) = 6 ;. Then the weight  = , + P 2J $ belongs to F (J ) since h + ; _ i  h + ; _i < p for all  2 F (J ) and all 2 R+. We have now dim N () = pN ,1. Therefore Premet's theorem implies that N () is simple. Proposition B.5 yields now simplicity in general. Remark : Let m be a positive integer. Suppose that we have in the lemma dim N () = m h+; _ipN ,1 , while all the remaining assumptions are unchanged. Then the same proof as above yields that the length of N () is at most equal to m. D.3. Here are two types of situations where we shall apply Lemma D.2. Let be a simple root and let  be subregular nilpotent with (p ) = 0. Lemma: a) Let w 2 W such that w,1 is a simple root. Then Z (w; ) is simple for all  2 C00 with hw( + ); _ i > 0. b) Let be a simple root, let w1; w2 2 W such that sbm(w1; )  sbm(w2 ; ) and dim sbm(w2; )= sbm(w1; ) = h + ; _ipN ,1 for all  2 C00 . Then sbm(w2; )= sbm(w1; ) is simple for all  2 C00 with h + ; _i > 0. Proof : a) Let J be the set of simple roots with  2 F (J ). Then hw( + ); _ i > 0 is equivalent to w,1 2 J . If so, then dim Z(w; ) = hw( + ); _ ipN ,1 = h + ; w,1 _ipN ,1: Now the claim follows from Corollary B.11 and Lemma D.2. b) This follows from C.9(1) and Lemma D.2. (Note that C.14(2) provides us with cases where our assumption is satis ed.) D.4. Let 0 denote the unique short root that is a dominant weight. Set h = h; _0 i + 1 equal to the Coxeter number of R. Proposition: If R is of type E8, F4, or G2 suppose that p > h + 1. Let be a simple root and let  be subregular nilpotent with (p ) = 0. Let  2 C00 . a) Let w 2