subregular nilpotent representations of lie algebras in prime ...

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8 in U(g) where these products are to be carried out in some xed order. That order r( ) is arbitrary except that x, should occur in x , r as the factor most to the right. Choose a basis h1, h2;:::;hn of h. Let S be the set of all n{tuples of nonnegative integers. Associate to each t =(t(i))i 2 S the element in U(h) U(g). So the x , r htx + s ht = nY i=1 h t(i) i with r;s2 R, t 2 S are a PBW basis of U(g). Let d1 and d2 be the integers with 0
This sum is a regular function on X each , since each bs0( ) is polynomial in and is polynomial in f. (f + i)(ht) = nY i=1 (f + i)(hi) t(i) This nishes the proof of the claim. We now return to the proof of the proposition. The space of all linear maps from Z(f + 1; ; )toZ(f + 2; ; ) has basis Esr with r 2 R1, s 2 R2 such that for all r 0 2 R1 Esr(x , r 0vf+ 1) = x, s vf+ 2; if r 0 = r; 0; otherwise. Given a 2 g we can now use (2) to evaluate each and get (a Esr)(x , r 0 vf+ 1) =a (Esr(x , r 0vf+ 1)) , Esr(ax , r 0vf+ 1) a Esr = X t2R2 So we have aformula of the form a Esr = X c 2 t;s (a; f; )Etr , X X r02R1 s02R2 u2R1 c 1 r;u (a; f; )Esu: bs 0 ;r 0 ;s;r(a; f; )Es 0 r 0 where the bs 0 ;r 0 ;s;r(a; f; ) are regular functions of (f; ) 2 X . Now Homg(Z(f + 1; ; );Z(f + 2; ; )) identi es with the space of all (R2 R1){tuples (xsr)r;s of elements in K with a P r;s xsrEsr =0foralla 2 g. It actually su ces to take fora all elements in a basis (or a generating system) of g. This shows that we indeed get the Hom space as a solution space of a linear system of equations as described at the beginning of the proof. The proposition follows. 9
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 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 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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