subregular nilpotent representations of lie algebras in prime ...

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16 The assumption on the facet implies that np h + ; _ i (n +1)p: Set d = h + ; _ i,np and d 0 = h + ; _ i,np. Then d is the same number as considered in B.8. If d 0 > 0, then it the analogue of d working with instead of ; if d 0 = 0, then that analogue is equal to p, however. Note that s ;np = , d and s ;np = , d 0 . We get from B.4(1) isomorphisms T Z ( ) ' Z ( )andT Z ( , d ) ' Z ( , d 0 ). We claim that, modulo these isomorphisms: Lemma: If d 0 > 0, then T (' ) is a non-zero multiple of ' . If d 0 =0, then T (' ) is a non-zero multiple of the identity. Proof :Ifd = p, i.e., if h + ; _ i =(n +1)p, then the assumption on the facets implies that also h + ; _ i =(n +1)p and d 0 = p. In this case both ' and ' are equal to (x, ) p 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
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 )=x d , v . Here d is again the integer with 0
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 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
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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