subregular nilpotent representations of lie algebras in prime ...

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22 b) Let E be a G{module; suppose that 1, 2;:::; r are the weights of E (counted with multiplicities) enumerated such that i > j implies i0, then 0 hw( + ); _ i
C.6. We nowwant to de ne for each w 2 W a homomorphism ' w : Z (w ) ,! Z ( ) (1) as follows: If w = s1s2 :::sr is a reduced decomposition (with si = s i for some simple root i) then we want tohave ' w = ' 1;sr ' sr;sr,1sr ' s2:::sr,1sr;s1s2:::sr,1sr (2) where the single factors are de ned in C.5. One has to check the independence of the right hand side in (2) of the chosen reduced decomposition. It is (as usual) enough to check the \braid relations" for each pair ; ,( 6= ) of simple roots. For example, if s s has order 3, then we have tocheck for each x 2 W with x ,1 ; x ,1 >0 that ' x;s x ' s x;s s x ' s s x;s s s x = ' x;s x ' s x;s s x ' s s x;s s s x : This follows from the Verma relations, see [4], 7.8.8; similarly in the cases where s s has order 2, 4, or 6. Now that the ' w are well-de ned, we get for all w 2 W and all simple roots with w ,1 >0 that ' s w = ' w ' w;s w (3) since we get a reduced decomposition of s w when we multiply a reduced decomposition of W on the left by s . C.7. Set for all w 2 W the submodule corresponding to w. sbm(w; )=im(' w ) Z ( ) (1) Lemma: Let be a simple root and w an element in W with w ,1 >0. a) We have sbm(s w; ) sbm(w; ). b) If (x, ) 6= 0, then sbm(s w; )=sbm(w; ). c) If (x, )=0, then sbm(s w; ) is a homomorphic image of Z (s w ; ), and sbm(w; )= sbm(s w; ) is a homomorphic image of Z (w ; ). Proof : The identity ' s w = ' w ' w;s w from C.6(3) implies a). If (x, ) 6= 0, then ' w;s w is an isomorphism; this yields b). Suppose now that (x, ) = 0. Denote the image of ' w;s w by M. ThenM is isomorphic to Z (s w ; ), and Z (w )=M is isomorphic to Z (w ; ). Now the claim in c) follows from sbm(s w; )=' w(M) and sbm(w; )=' w(Z (w )). Remark: It is sometimes more convenient to restate the last part of the lemma as follows: Let w 2 W and 2 R + such that w is a simple root with (x,w )=0. Then sbm(ws ; ) is a homomorphic image of Z (ws ; w ), and sbm(w; )= sbm (ws ; ) is a homomorphic image of Z (w ; ). [Note that sw w = ws .] 23
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 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 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
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