subregular nilpotent representations of lie algebras in prime ...

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58 with j as above and j 0 = h 1; i +2(n , 1) , j: We have b2n = h 1; i +2n , 1 and bi + b2n+1,i = b2n for 1 i 2n, see [10], 3.2(1) and 3.6(8). If j runs from bi to bi+1 , 1 (and if i
Claim: We have for all 2 C0 with h + ; _ ni > 0. g Z ( ; n) ' Z (s2"n,1 ; n) (1) Proof : Set J = f n,1; ng and use the abbreviation Z ;J( ; n) =U (pJ) U (p n ) L ; n( ) for all 2 X. Wehave by transitivity of induction Z ( ; n) ' U (g) U (pJ ) Z ;J( ; n) for all . Note that g belongs to the standard Levi factor GJ of PJ since we can nd there at least one representative ofs"n,1+"n [coming from the root subgroups U ("n,1+"n)]. Therefore we have for all g Z ( ; n) ' U (g) U (pJ) g Z ;J( ; n); cf. B.14(1). So our claim will follow ifwe can show that g Z ;J( ; n) ' Z ;J(s2"n,1 ; n): (2) The nilradical of pJ acts as 0 on both sides in (2). So we just have to nd an isomorphism of gJ{modules. The centre of gJ [equal to the intersection in h of ker( n) and ker( n,1)] acts on both sides via the restriction of .Sowejusthave to nd an isomorphism as modules over the derived Lie algebra DgJ of gJ. We want to apply F.9(2) to DgJ in order to get the isomorphism in (2) and thus the claim. The restriction of to DgJ has standard Levi form. We have GJ = Z(GJ) 0 DGJ [where Z(GJ) 0 T is the connected centre] and can thus write g = zg 0 with z 2 Z(GJ) 0 . Since z trivially xes on DgJ, sodoesg 0 . Since g 0 is still a representative fors"n,1+"n, we can now apply F.9(2) and get the claim. F.11. Suppose now that R is of type F4. Set x = x 2 + x 3 + x 4 + x 1+ 2+ 3+ 4: (1) Let be the linear form corresponding to x. Then is subregular (see the end of the proof of D.13) with (p 1) = 0. We can nd a representative g 2 NG(T ) for the re ection s"2 = s 1+ 2+ 3 such that g = . Proposition F.7 follows in this case from F.6(4),(5) and: Claim: Let 2 C0. We have if h + ; _ 1 i > 0, and g Z ( ; 1) ' Z (s"2+"3 ; 1) (2) g Z (s2s1 ; 1) ' Z (s"2+"3s2s1 ; 1) (3) 59
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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 _
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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
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 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 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
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