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76 Remarks: 1) The lemma can be extended to the case where L = L 0 and where is a simple root with ( ; 0) =0. 2) We see as in H.9 that Ext 1 (L; L) 6= 0 in almost all cases. The only additional exceptions can occur in types C2 = B2 and C3. H.11. Lemma: Let 2 [f0g, letL be a simple module in C with L 6= 0. a) The head and the socle of L are both isomorphic to L. b) If L 0 is a simple module in C with T ( ) L 0 ' T ( ) L, then L 0 ' L. Proof :IfL 0 is a simple module in C , then H.9(4),(5) imply that L 0 occurs with multiplicity 1 both in head and in the socle of L if T ( ) L 0 ' T ( ) L. Otherwise it does not occur at all. In particular L itself occurs with multiplicity 1. This shows that a) will follow from b). Suppose that L 0 is a simple module in C with T ( ) L 0 ' T ( ) L. Lemma H.7 implies that (L) = (L 0 ). If L 0 is not isomorphic to L, then there has to be a long simple root with (L) =f g. Lemma H.8 shows that both L and L 0 occur with multiplicity 1 as composition factors in L. Since they occur both in the head and in the socle they have to be direct summands. The discussion above of the socle implies that there cannot be any other contributions to the socle, hence that L ' L L 0 . Then Lemma H.8 implies that ( ; ) = 0 in case 2 , resp. ( 0; ) = 0 in case =0. Now Lemma H.10 yields a contradiction. H.12. Proposition: a) If and are simple roots with ( ; )=0, then we have Ext 1 (L; L 0 )=0for all simple modules L and L 0 in C with (L) =f g and (L 0 )=f g. b) Let , be simple roots with ( ; ) < 0, letL be asimplemodule L in C with (L) =f g. Then there exists a simple module L 0 in C with (L 0 )=f g and Ext 1 (L; L 0 ) ' K ' Ext 1 (L 0 ;L): (1) If and have the same length, then the condition Ext 1 (L; L 0 ) 6= 0determines L 0 up to isomorphism. If and have di erent lengths, then (1) holds for all possible L and L 0 . c) If be a simple root. If R is not of type A1, then Ext 1 (L; L 0 ) ' Ext 1 (L 0 ;L) ' K; if ( 0; ) > 0, 0; if ( 0; )=0 for all simple modules L in C with (L) =f g. Proof : a & b) Consider simple modules L and L 0 in C with (L) =f g and (L 0 )=f g. Arguing as in H.9 one gets Ext 1 (L; L 0 ) ' Homg(rad L; L 0 ): (3) (2)
If we use the short exact sequence 0 ! L ,! L ,! L= soc L ! 0 and apply the functor Homg(L 0 ; ), we get similarly Ext 1 (L 0 ;L) ' Homg(L 0 ; L= soc L): (4) If ( ; ) = 0, then H.8 shows that L 0 is not a composition factor of L; this implies now the claim in a). Suppose that ( ; ) < 0. If and have the same length, then h ; _ i = ,1. Then H.8 and H.11 imply: Given L as above there exists a simple module L 0 1 with (L 0 1)=f g and rad L= soc L ' L 0 1 : Then (3) and (4) show that L 0 1 satis es (1) while these Ext groups are 0 for any simple module L 0 2 with (L 0 2)=f g and L 0 2 6' L 0 1. Consider now the case where and have di erent lengths. The claim in the proposition is symmetric in and .Sowemay assume that is short and is long. So there is up to isomorphism only one choice for L 0 while there are one or two possibilities for L. Wehave h ; _ i = ,1 and h ; _ i = ,2. Lemma H.8 yields therefore rad L= soc L ' L 0 : This implies that (1) holds for all possible L and L 0 . c) The proof is similar and left to the reader. Remark: Let , be simple roots with ( ; ) < 0suchthat is short and is long. We knowthat there exists up to isomorphism only one simple module L in C with invariant f g. Consider now M = rad L = soc L . Lemma H.8 tells us that M has length 2, with both factors having invariant f g. Part b) of the proposition implies using (3) and (4) that each simple module L in C invariant f g occurs with multiplicity 1 in the socle and in the head of M. with There are now two possibilities: If there are (as expected) two isomorphism classes (with representatives L ;1 and L ;2 ) of simple modules in C with invariant f g, then we can apply the discussion to L = L ;1 and to L = L ;2. It follows that M ' L ;1 L ;2 in this case. If, however, there is only one such class, then M has to be a non-split extension of L by itself. So, as in other situations, it is desirable to show that the middle M is completely reducible. 77
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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
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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 80: 78 References [1] N. Bourbaki, Grou
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