subregular nilpotent representations of lie algebras in prime ...

More documents

Recommendations

Info

34 Theorem: Assume that is subregular nilpotent with (b + )=0. Exclude the case where R is of type G2; ifR is of type E8 or F4, assume that p>h+1. Let m . We can denote the factors in a 2 C 0 0 . Then Z ( ) has length 1+P 2J( ) composition series of Z ( ) by L0 and L ;i with that 2 J( ) and 1 i m such dim L ;i = h + ; _ N ,1 ip and (L ;i) =f g (2) for all and i, while dim L0 =(p ,h + ; _ N ,1 0 i)p and (L0) =J( ) [f0g: (3) Remarks: 1) The restrictions on the type are hoped to be unnecessary. IfRis of type E8 or F4 and p h + 1, then the theorem will hold for all 2 C 0 0 where the socle of Z ( ) has dimension equal to (p ,h + ; _ 0 i)pN ,1 . The same remark applies to the results in D.12/13. 2) The theorem does not say whether di erent factors in the composition series are isomorphic to each other. If L and L 0 are composition factors with (L) 6= (L 0 ), then clearly L 6' L 0 . So the question is whether for xed 2 J( )them factors L ;i are isomorphic to each other. We shall see in Section F that such isomorphisms exist in certain cases. 3) The theorem con rms in part my speculations in [11], 11.15 (where I look only at those 2 C 0 0 that are p{regular, i.e., satisfy h + ; _ i > 0 for all 2 R + ). The part of those speculations not con rmed is that the factors in a composition series should be pairwise non-isomorphic. As mentioned in the preceding remark, that turns out to be wrong. D.7. We now begin to prove the theorem. Lemma B.13 and D.5(3) show: If Theorem D.6 holds for one subregular , then it holds for all subregular . We assume from now on that is a simple root with (p )=0and (x, ) 6= 0 for all simple roots 6= . Remark D.1 shows that we can nd with this property for each . Later on we shall make speci c choices for . [Usually this is done such that the right hand side in (6) below is as small as possible.] Set L0 equal to the simple socle of Z ( ). It satis es D.6(3) by Proposition D.4.b and D.5(2). By our choice of the set I as in C.11(1) consists of all simple roots 6= .We shall use the construction from C.12{14 in order to nd the remaining composition factors of Z ( ). We can construct inductively a chain 1 = ; 2;:::; r (1) of roots in WI such that h r; _ i 0 for all 2 I and such that there exists for each i
The root r is uniquely determined as the only weightinWI that is dominant with respect to the root system RI and its basis I. Set Mi = M i for 1 i r (2) and M0 = Z ( ). We have by C.14(1) a descending chain of submodules in Z ( ) M0 = Z ( ) M1 M2 Mr sbm(w0; ) 0 (3) where the inclusion Mr sbm(w0; )follows from C.10(2). Furthermore, C.14(2) yields dim(Mi=Mi+1) =jh i; _ i ij h + ; _ N ,1 i ip 35 for 1 i
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 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

subregular nilpotent representations of lie algebras in prime ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?