subregular nilpotent representations of lie algebras in prime ...

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72 Let 0 2 J( ) with 0 6= .Ifwewrite(ww 0 ) ,1 _ as a linear combination of the _ with simple, then 0_ occurs with coe cient cm 0 with m 0 as in D.6(1). Since _ 0 is the largest root of R _ this implies jcm 0j m 0, hence jcj 1. If c = 1, then _ occurs above with coe cient1+m , a contradiction. So we get c =0orc = ,1. If c = 0, then a look at the coe cientof _ implies (ww 0 ) ,1 >0; if c = ,1, then a look at the coe cientof 0_ [with 0 as before] implies (ww 0 ) ,1
Arguing as in H.4 one checks that L has a ltration with factors Z (w ; ) ' L and Z (ws have ; ). We want to apply Lemma H.6 to the second factor. We (ws ) ,1 _ = s _ = _ ,h ; _ _ i : Now the claim in a) follows immediately from H.6. In most cases Lemma H.4 implies that 0L has a ltration with factors Z (w ; ) ' L and Z (ws0 ; ) ' Z (ws 0 ; ). The only exception occurs when hw( (0) + ); _ i = h (0) + ; _ i is congruent to 0 modulo p. The choice of (0) implies that this can happen only if = 0, hence only if R has type A1. Set that case aside for the moment. We want to apply Lemma H.6 to Z (ws 0 ; ); we have (ws 0) ,1 _ = s 0 _ = _ ,h 0; _ i _ 0 : Since _ 0 is the largest root in R_ and since R has not type A1, wehave h 0; _ i2 f0; 1g. It follows that (ws 0) ,1
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U N I V E R S I T Y O F A A R H U S
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2 positive roots . (We denote by _
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4 A A.1. Let K be an algebraically
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6 given by '(v 0 f ,d )=xd , v 0 f
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8 in U(g) where these products are
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10 B Keep the assumptions and notat
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12 B.5. Proposition: Suppose that a
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14 are simple GI{modules. It follow
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16 The assumption on the facet impl
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18 Remark: Our assumptions (B1) and
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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 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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