subregular nilpotent representations of lie algebras in prime ...

More documents

Recommendations

Info

40 Claim: The element w1 = s wIs 0 satis es and and sbm(w1; ) sbm(w1s 0; ) sbm(w1s 0s ; ) sbm(w1s + sbm(w1; ) sbm(w1s ; ) sbm(w1s s 0; ) sbm(w1s + 0; ) (4) 0; ) (5) sbm(w1s 0; )= sbm(w1s 0s ; ) ' L ;2: (6) Proof (of Claim): We have s 0 = 0 , and s 0 0 = 0, hence and w1 = s wI( + 0 )=s ( 0 , , 0 )= 0 , , 0 w1 0 = s wI(, 0 )=s 0 = + 0 : Since both w1 and w1 0 are positive, we have in the Bruhat-Chevalley order and w1
D.13. We want tohave an analogue to Lemma D.12 for the simple modules whose invariant is a long simple root. Suppose that R is of type Bn, Cn, orF4; if R is of type F4 assume that p>h+1. Set = 1 if R is of type Bn or F4, set = n if R is of type Cn. So is a long simple root. Let be subregular with (p )=0. We claim under these assumptions: Lemma: Let 2 C 0 0 . Let be a long simple root with 2 J( ). IfLis a simple module in C with (L) =f g, then there exists an element x 2 W with x = and L ' Z (x ; ). Proof : As in D.12 it su ces to prove the claim for one special subregular with (p )=0. Consider rst R of type Bn. Then we may assume that has the standard Levi form considered in [10], Section 3. Now the claim follows from [10], 3.13. Consider next R of type Cn. Wemay assume that (x, i) 6= 0 for all 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 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 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?