subregular nilpotent representations of lie algebras in prime ...

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64 Remarks: 1) This result con rms for p{regular (i.e., for 2 C 0 0 with jW j = W ) the revised conjecture by Lusztig, as in [17], 2.6. (The formulation there looks somewhat di erent, but can be checked to yield the same numbers.) 2) Suppose that R is of type Bn, Cn, orF4. Thenwecan apply G.3(5) for L = L0 and for L = L with short. Therefore (1) holds, (2) holds if is short, and (3) holds if both and are short. If is a long root in J( ) and if there (as expected) exist two isomorphism classes of simple modules with invariant f g in C ,thenweget [Q0 : L ;i] =jW j m 2 (5) and [Q : L ;i] =jW m m j 2 (6) for all short 2 J( ) [and for i =1; 2] in both cases. By the symmetry of the Cartan matrix we get also [Q ;i : L0] and [Q ;i : L ]. 3) For R of type Bn we can choose to have standard Levi form. Then QL has a Z{ ltration for each simple L in C, cf. [11], 10.11. This implies that G.3(5) holds for all L even though QL =2 P 0 in general (see the nal remark in G.4). We getin this case that [Q ;i : L ;j ]=jW j m m 4 for all long ; 2 J( ). One may speculate whether (8) also holds in types Cn and F4. We keep the assumptions from Section D. H H.1. Recall from D.6(1) that we write _ 0 = P 2 m _ where is the set of simple roots. The fundamental weight $ corresponding to a simple root is minuscule (in the sense of [1], Ch. VI, x1, exerc. 24) if and only if m =1. For all with this property sety = y 0 w0 2 W where w0 is the longest element inW and where y 0 is the longest element in the subgroup of W generated by all s with 6= . Now Prop. 6 in [1], Ch. VI, x2, (applied to R _ instead of R) shows that y C0 + p$ = C0. More precisely, the map x 7! y x + p$ maps the `real' alcove ofallx 2 X Z R with 0 hx + ; _ i p for all 2 R + to itself. It therefore permutes the \walls" of this alcove, i.e., the hyperplanes with equations hx + ; _ i = 0 with 2 and the hyperplane with equation hx + ; _ 0 i = p. So y permutes [f, 0g. In fact, one checks easily that y (, 0) = and that hx + ; _ 0 i = p () hy (x + )+p$ ; _ i =0: (1) The simple root ,w0 satis es y (,w0 )=,y 0 ( ) < 0 since y 0 ( ) > 0. It follows that y (,w0 )=, 0, hence that hx + ; (,w0 ) _ i =0 () hy (x + )+p$ ; _ 0 i = p: (2) (8)
If we apply y to _ 0 , then we get easily that m,w0 =1 and my = m for all 2 ; 6= ,w0 . (3) Proposition: Suppose that R is not of exceptional type. Then there exists for each 2 X aweight2C0 0 with 2 W + pX. Proof : Since C0 is a fundamental domain for Wp, wemayassume that 2 C0. If h + ; _ 0 i 1 65 m h + ; _ i: (4) So far we did not use any assumption on R. Wedo that now. If R is of type An, then we havem = 1 for all 2 ; so the right hand side in (4) is equal to 0: a contradiction. If R is of type Bn or Cn (with n 2), or Dn (with n 4), then m 2f1; 2g for all 2 . Then (4) turns into p =2 X h + ; _ i: m =2 Since p 6= 2 in these cases by (B2), we get a contradiction. Remarks: 1) If we drop our assumptions (B1) { (D2), then we can extend the proposition to all cases where R has no components of exceptional type provided that p 6= 2ifR has a component not of type A. The proposition above gives the result in case G is semi-simple and R indecomposable. If G is semi-simple and R arbitrary, then X is the direct sum of the weight lattices of the irreducible components of R. These components are stable under W ;we get the result for X immediately from that for each component. For arbitrary reductive G set X0 equal to the subgroup of all 2 X with h ; _ i =0 for all 2 R. Then X=X0 identi es with the weight lattice of R, and the canonical map X ! X=X0 commutes with the action of W . Therefore the claim for X=X0 implies the claim for X. 2) It is clear that the proposition cannot extend to the types E8, F4, andG2. In these cases W + pX = Wp ; since C0 is a fundamental domain for Wp, we cannot move any 2 C0 with h + ; _ 0 i = p to an element inC 0 0 . For R of type E6 and E7 the proof of the proposition shows that we can handle all 2 C0 with h + ; _ 0 i = p for which there exists a simple root with m = 1 and h + ; _ i > 0. The simple roots with this property are 1 and 6 for E6, resp. 7 for E7 (in the numbering from [1]). However there will now exist other weights that we cannot handle since there exist simple roots ; with m = 2 and m =3.
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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 _
Page 6 and 7:
4 A A.1. Let K be an algebraically
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6 given by '(v 0 f ,d )=xd , v 0 f
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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 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 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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