subregular nilpotent representations of lie algebras in prime ...

simple module in C of the same dimension with (L 0 )=f ng by B.13(2) and
dim T L 0 = p N ,1 .
Since L 0 is in C it has to be a composition factor of Z ( ). All Mi=Mi+1 with
i 6= n,1 and M2n,2 are simple with invariant di erent fromf ng. Therefore L 0
has to be a composition factor of Mn,1=Mn. It follows that this factor module has
length equal to 2 and that the second composition factor, say L 00 , has to have the
same dimension. Furthermore, the exactness of the translation functors implies
that T (Mn,1=Mn) has a ltration with factors T L 0 and T L 00 . By comparison
of dimensions we get that also T L 00 has dimension equal to p N ,1 . This shows
that n 2 (L 00 ). On the other hand, since L 00 is a composition factor of Mn,1=Mn
we have
(L 00 ) (Mn,1=Mn) =f ng:
So we get equality: (L 00 )=f ng. This completes the proof of the claim concerning
the composition factors of Mn,1=Mn, hence that of the theorem for type
Cn.
D.10. Consider R of type F4. Wechoose such that = 4 in the notations from
[1]. So is short. We can choose the chain in D.7(1) such that the corresponding
simple roots are (in this order)
3; 2; 1; 3; 2; 3:
We get r = 7 and 7 = 1 +2 2 +3 3 + 4, hence
_
7 =2 _
1 +4 _
2 +3 _
3 + _
4 = _
0 , _
4 :
Set M8 = sbm(w0; ). We get from D.7(4){(6)
dim(M0=M1) = dim(M7=M8) =h + ; _
4 ipN ,1 ;
dim(M1=M2) = dim(M4=M5) = dim(M6=M7) =h + ; _
3 ip N ,1 ;
dim(M2=M3) = dim(M5=M6) =2h + ; _
2 ip N ,1 ;
dim(M3=M4) =2h + ; _
1 ip N ,1 :
Lemma D.3.b implies that M0=M1, M7=M8, M1=M2, M4=M5, andM6=M7 are
simple or 0; if non-zero, then the rst two modules have invariant f 4g, and the
remaining three modules have invariant f 3g.
Each factor module M2=M3, M5=M6, M3=M4 has length at most equal to 2
by the Remark D.2. We have
(M2=M3) = (M5=M6) =f 2g and (M3=M4) =f 1g:
More precisely, wehave, if 2 2 J( ),
dim T $2,
(M2=M3) = dim T $2, N ,1
(M5=M6) =2p
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