subregular nilpotent representations of lie algebras in prime ...

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2 positive roots . (We denote by _ the coroot corresponding to .) Then there are up to isomorphism jej simple modules in C .We can denote them by L with 2 e suchthat dim L = h + ; _ ip N ,1 ; if 2 , (p ,h + ; _ 0 i)p N ,1 ; if = , 0. De ne integers m for all 2 e by m, 0 = 1 and _ 0 = P 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 ofa 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 )oftwo non-isomorphic simple modules is given by Ext 1 (L ;L ) ' K; if ( ; ) < 0, 0; if ( ; )=0, 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
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 Ext 1 (L ;i ;L ;j ) ' K; if i = j, 0; otherwise. There are again similar results for all 2 X such 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 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 20 and 21: 18 Remark: Our assumptions (B1) and
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
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52 factors shows that Z (s ; ) has
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54 So the Hom space in (2) is isomo
Page 58 and 59:
56 Consider nally R of type F4. Wem
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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
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68 Assume that R is not of type A1.
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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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