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74 and Homg(M 0 ; M) ' Homg(T ( ) M 0 ;T ( ) M) (5) for all modules M, M 0 in C . It follows that each L with ; 2 [f0g has simple head and simple socle isomorphic to L . (These are standard arguments due to Vogan.) We have soc L rad L . In case ; 2 we get now from (1) rad L = soc L ' L ; if ( ; ) < 0, 0; if ( ; )=0. One gets similar formulas from (2) and (3). Note that (6) implies in case ( ; ) = 0 that we have a non-split extension (6) 0 ! L ,! L ,! L ! 0: (7) Let me write Ext i for Ext groups in the category of U (g){modules. So (7) says that Ext 1 (L ;L ) 6= 0 for all 2 such that there exists a simple root with ( ; ) = 0. Using (2) one get similarly Ext 1 (L ;L ) 6= 0 in case ( 0; ) = 0, and from (3) one gets Ext 1 (L 0 ;L 0 ) 6= 0 if there exists a simple root with ( 0; )=0. This shows that Ext 1 (L; L) 6= 0 for all simple modules in C unless R has type A2 (compatibly with Remark 1 in [10], 2.19) or type D4 where L = L 2 is the only possible exception. I have no idea how large these non-vanishing Ext groups are. Proposition: Suppose that all roots in R have the same length and that R is not of type A1. If and are simple roots with 6= , then If is a simple root, then Ext 1 (L ;L ) ' K; if ( ; ) < 0, 0; if ( ; )=0. Ext 1 (L ;L 0) ' Ext 1 (L 0;L ) ' K; if ( 0; ) > 0, 0; if ( 0; )=0. Proof : This follows again from standard arguments due to Vogan. When dealing with (8), one applies the functor Homg( ;L ) to the short exact sequence 0 ! rad L ,! L ,! L ! 0: The adjointness of T ( ) and T ( ) implies that is self-adjoint. This implies for all i 0, see G,13(2). It follows that Ext i ( L ;L ) ' Ext i (L ; L )=0 Ext 1 (L ;L ) ' Homg(rad L ;L ): Now apply (6). The proof of (9) is similar, working with (2) and (3) instead of (1). (8) (9)
Remark: IfR has type A1 and if we denote the unique simple root by , then one has [ 0L ]=[ L 0]=2[L ]+2[L 0] It is well known that in this case (Note that = 0 and look at [19].) Ext 1 (L ;L 0) ' Ext 1 (L 0;L ) ' K 2 : H.10. We nowwant to extend some results from H.9 to the cases with two root lengths. Recall that we exclude type G2. Note rst that H.9(4) and H.9(5) hold without restriction. We next look at H.9(7). Lemma: Let be a simple root and let L be a simple module in C with (L) = f g. If is a simple root with ( ; )=0or if =0and ( 0; )=0,thenwe have a short exact sequence that does not split. 0 ! L ,! L ,! L ! 0 (1) Proof : Lemma H.8 shows that L has length 2 with both composition factors, say L1 and L2, satisfying (L1) = (L2) =f g. Wewant toshow that L1 ' L2 ' L. If so, then we get a short exact sequence as in (1). If that sequence splits, then we get dim Homg(L; L) = 2 contradicting H.9(5). If is short, then L1 ' L2 ' L follows from Theorem F.5. So consider the case where is long. We may assume that there exists a simple root with (p )=0andw 2 W with w = and L ' Z (w ; ). The proof of Lemma H.8 shows that our claim will follow ifwe can show that Z (ws ; ) is isomorphic to L where s = s in case 2 and s = s 0 in case =0. We distinguish two cases: The type ofR is Bn. We may assume that we are in the situation of [10], Section 3. In particular, we have = 1 and = j with 1 j1). So we get indeed s"j = "j and the claim follows in type Bn. The type of R is Cn or F4. We may assume that we are in the situation of F.6(1),(2) or F.6(3){(5). We have to show that ws 2 W2w in the notations from F.6(1) or F.6(3). Since ws = sw 0w with 0 = or 0 = 0 this means that sw 0 2 W2. Now( 0 ; ) = 0 and w = imply that w 0 is orthogonal to . So it su ces to show that w 0 is a short root. (Recall the de nition of W2.) That is obvious in case = 0. Otherwise note that in type Cn or F4 all simple roots orthogonal to a long simple root are short. 75
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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
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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 78 and 79: 76 Remarks: 1) The lemma can be ext
Page 80: 78 References [1] N. Bourbaki, Grou
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