subregular nilpotent representations of lie algebras in prime ...

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28 Proof :a)Wehave clearly s 2 WI and could choose ws = w s .Sowehave M s = sbm(s w s ; ): On the other hand s = ,h ; _ i is a positive rootnotinZI. Therefore also 0 = w s is positive. Using s = w ,1 s w ,we get 0 = s w , hence s 0s w =(s w s w ,1 s )s w = s w s : Since (s w ) ,1 0 = >0, we get from Proposition C.8 that sbm(s 0s w ; ) sbm(s w ; ); hence (1). Now (2) follows from C.12(3) and (s ) _ = _ ,h ; _ i _ . b) Now our assumptions imply that s = + = s . It follows that + 2 WI and that 0 = w s = w ( + ) 2 WI .Weobserved above that 0 = s w ; so we can apply Lemma C.13 with w = s w .Now x = w 0s w is the same x as in the proof of Lemma C.13 and satis es = x .Wehave M = sbm(w; ) and M + =sbm(ws ; ) since w s is a possible ws = w + . Now Lemma C.13 says that we have a surjective homomorphism The left hand side has dimension equal to Z (x ; ) ,! M =M + : (4) hx( + ); _ ip N ,1 = h + ; _ N ,1 ip as long as h + ; _ i > 0; otherwise its dimension is equal to p N . In the rst case, the surjection in (4) has to be an isomorphism by dimension comparison; this implies (3). Remark: In the situation from b) one can deduce the inclusion in (1) directly from Lemma C.7.a (without the more complicated argument in C.8) using x ,1 = >0 and the equalities M = sbm(x; ) and M + = sbm(s x; ):
We keep the assumptions from Sections B and C. We assume in addition: (D1) The prime p is good for R. and (D2) G is almost simple. D The rst assumption is crucial so that we can apply the Kac-Weisfeiler conjecture proved by Premet. The second assumption is mainly meant to simplify the statement of the results. D.1. We call a linear form 2 g subregular if its orbit under the coadjoint action has dimension equal to 2(N , 1) where N = jR + j. If so, then the Kac- Weisfeiler conjecture as proved by Premet says: If M is a U (g){module, then dim(M) is divisible by p N ,1 . Recall that we setp = b + + g, for each simple root ;we denote (as in B.14) by P the corresponding parabolic subgroup of G. The following result is a translation of well known results on orbits in g: Lemma: There exists a unique subregular nilpotent orbit O in g . If is a simple root then O intersects the set of all 2 g with (p )=0in an open and dense subset. This intersection is one orbit under P . Proof :We can (under our assumptions on p) identify g and g as G{modules. The classi cation of the nilpotent orbits in g is the same as for the Lie algebra over C of the same type (since p is good). In particular, there is exactly one subregular nilpotent orbit in g; this yields the rst claim. The elements 2 g with (b + ) = 0 and (x, ) = 0 correspond (under g ' g ) to the elements in the nilradical n = L >0; 6= g of the parabolic subalgebra p . The theory of the Richardson orbits (cf. [3], 5.2.3) says that there exists exactly one nilpotent orbit for G that intersects n in an open and dense subset. That intersection is one orbit under P ; furthermore the dimension of the orbit (under G) is equal to the codimension in g of a Levi factor of p . Since that codimension is equal to 2(N , 1), we get the remaining claims. Remark: For each simple root 6= the set of all with (p ) = 0 and (x, ) 6= 0 is an open and dense subset of the set of all with (p ) = 0. It follows: The set of all subregular 2 g with (p )=0and (x, ) 6= 0for all simple roots 6= is an open and dense subset of the set of all with (p )=0. D.2. Each subset J of the set of all simple roots de nes a facet F (J) contained in C 0 0 as follows: A weight 2 C 0 0 belongs to F (J) if and only if h + ; _i > 0 for all 2 J and h + ; _i = 0 for all simple roots =2 J. Write $ for the fundamental weight corresponding to a simple root . Then 2 X belongs to F (J) if there are integers m > 0 such that + = P 2J m $ and if h + ; _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 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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