subregular nilpotent representations of lie algebras in prime ...

More documents

Recommendations

Info

36 D.9. Consider now R of type Bn or Cn with n 2. We choose such that = 1 (in the notations from [1]) and get then (in those notations) WI = f"1 "i j 2 i ng. The sequence from D.7(1) is now equal to "1 , "2;"1 , "3;:::;"1 , "n;"1 + "n;:::;"1 + "3;"1 + "2: We have i = "i+1 , "i+2 = i+1 for 1 i < n , 1, and n,1 = n and n+i = "n,i,1 , "n,i = n,i,1 for 0 i 0; otherwise this factor module is equal to 0. If R is of type Bn, then and hence all i are long. We get therefore h i; _ i i = ,1 for 1 i
simple module in C of the same dimension with (L 0 )=f ng by B.13(2) and dim T L 0 = p N ,1 . Since L 0 is in C it has to be a composition factor of Z ( ). All Mi=Mi+1 with i 6= n,1 and M2n,2 are simple with invariant di erent fromf ng. Therefore L 0 has to be a composition factor of Mn,1=Mn. It follows that this factor module has length equal to 2 and that the second composition factor, say L 00 , has to have the same dimension. Furthermore, the exactness of the translation functors implies that T (Mn,1=Mn) has a ltration with factors T L 0 and T L 00 . By comparison of dimensions we get that also T L 00 has dimension equal to p N ,1 . This shows that n 2 (L 00 ). On the other hand, since L 00 is a composition factor of Mn,1=Mn we have (L 00 ) (Mn,1=Mn) =f ng: So we get equality: (L 00 )=f ng. This completes the proof of the claim concerning the composition factors of Mn,1=Mn, hence that of the theorem for type Cn. D.10. Consider R of type F4. Wechoose such that = 4 in the notations from [1]. So is short. We can choose the chain in D.7(1) such that the corresponding simple roots are (in this order) 3; 2; 1; 3; 2; 3: We get r = 7 and 7 = 1 +2 2 +3 3 + 4, hence _ 7 =2 _ 1 +4 _ 2 +3 _ 3 + _ 4 = _ 0 , _ 4 : Set M8 = sbm(w0; ). We get from D.7(4){(6) dim(M0=M1) = dim(M7=M8) =h + ; _ 4 ipN ,1 ; dim(M1=M2) = dim(M4=M5) = dim(M6=M7) =h + ; _ 3 ip N ,1 ; dim(M2=M3) = dim(M5=M6) =2h + ; _ 2 ip N ,1 ; dim(M3=M4) =2h + ; _ 1 ip N ,1 : Lemma D.3.b implies that M0=M1, M7=M8, M1=M2, M4=M5, andM6=M7 are simple or 0; if non-zero, then the rst two modules have invariant f 4g, and the remaining three modules have invariant f 3g. Each factor module M2=M3, M5=M6, M3=M4 has length at most equal to 2 by the Remark D.2. We have (M2=M3) = (M5=M6) =f 2g and (M3=M4) =f 1g: More precisely, wehave, if 2 2 J( ), dim T $2, (M2=M3) = dim T $2, N ,1 (M5=M6) =2p 37
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 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 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

subregular nilpotent representations of lie algebras in prime ...

Create successful ePaper yourself

Delete template?

Save as template?