subregular nilpotent representations of lie algebras in prime ...

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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) su ces 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 .(Butwe 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 adjointactionofg 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) doesonM. (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 we have an isomorphism g M ,! M given by m 7! gm. IfM 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( )toC(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;ifM 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 thatU(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 (foreach nilpotent ) projection functors pr : C( ) !C( ) and translation functors T : C( ) !C( ) de ned as in B.1/2. Let g 2 G. IfM 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 This implies for M in C( ) that pr ( g M)= g (pr M): (1) g (T M) ' T ( g M): (2)
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 ' pr (E g M)=T ( g M): g (E M) =pr ( g E g M) 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 :IfQL is the projectivecover 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) =p N 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 anobvious 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) g M (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 aweight in + pX and assume that 0 h ; _ i
Page 1: U N I V E R S I T Y O F A A R H U S
Page 4 and 5: 2 positive roots . (We denote by _
Page 6 and 7: 4 A A.1. Let K be an algebraically
Page 8 and 9: 6 given by '(v 0 f ,d )=xd , v 0 f
Page 10 and 11: 8 in U(g) where these products are
Page 12 and 13: 10 B Keep the assumptions and notat
Page 14 and 15: 12 B.5. Proposition: Suppose that a
Page 16 and 17: 14 are simple GI{modules. It follow
Page 18 and 19: 16 The assumption on the facet impl
Page 22 and 23: 20 C Keep all assumptions and notat
Page 24 and 25: 22 b) Let E be a G{module; suppose
Page 26 and 27: 24 C.8. We nowwant to generalise Le
Page 28 and 29: 26 Remark: Let denote the usual Che
Page 30 and 31: 28 Proof :a)Wehave clearly s 2 WI a
Page 32 and 33: 30 Lemma: Let J be asabove and let
Page 34 and 35: 32 0 + (b ) = 0 and 0 (x, 0)=0. Sin
Page 36 and 37: 34 Theorem: Assume that is subregul
Page 38 and 39: 36 D.9. Consider now R of type Bn o
Page 40 and 41: 38 and, if 1 2 J( ), dim T $1, (M3=
Page 42 and 43: 40 Claim: The element w1 = s wIs 0
Page 44 and 45: 42 hence that L is isomorphic to Z
Page 46 and 47: 44 E We return to the more general
Page 48 and 49: 46 for all 2 R1 and m 2 M. Proof :
Page 50 and 51: 48 E.6. We now drop the assumptions
Page 52 and 53: 50 F.3. Lemma: Proposition F.1 hold
Page 54 and 55: 52 factors shows that Z (s ; ) has
Page 56 and 57: 54 So the Hom space in (2) is isomo
Page 58 and 59: 56 Consider nally R of type F4. Wem
Page 60 and 61: 58 with j as above and j 0 = h 1; i
Page 62 and 63: 60 if h + ; _ 2 i > 0. Proof : Note
Page 64 and 65: 62 Remark: Let L be a simple module
Page 66 and 67: 64 Remarks: 1) This result con rms
Page 68 and 69: 66 H.2. Let 2 g be subregular nilpo
Page 70 and 71:
68 Assume that R is not of type A1.
Page 72 and 73:
70 (Here L runs over representative
Page 74 and 75:
72 Let 0 2 J( ) with 0 6= .Ifwewrit
Page 76 and 77:
74 and Homg(M 0 ; M) ' Homg(T ( ) M
Page 78 and 79:
76 Remarks: 1) The lemma can be ext
Page 80:
78 References [1] N. Bourbaki, Grou
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