subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
subregular nilpotent representations of lie algebras in prime ...
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34<br />
Theorem: Assume that is <strong>subregular</strong> <strong>nilpotent</strong> with (b + )=0. Exclude the<br />
case where R is <strong>of</strong> type G2; ifR is <strong>of</strong> type E8 or F4, assume that p>h+1. Let<br />
m . We can denote the factors <strong>in</strong> a<br />
2 C 0 0 . Then Z ( ) has length 1+P 2J( )<br />
composition series <strong>of</strong> Z ( ) by L0 and L ;i with<br />
that<br />
2 J( ) and 1 i m such<br />
dim L ;i = h + ; _ N ,1<br />
ip and (L ;i) =f g (2)<br />
for all and i, while<br />
dim L0 =(p ,h + ; _ N ,1<br />
0 i)p<br />
and (L0) =J( ) [f0g: (3)<br />
Remarks: 1) The restrictions on the type are hoped to be unnecessary. IfRis <strong>of</strong><br />
type E8 or F4 and p h + 1, then the theorem will hold for all 2 C 0<br />
0 where the<br />
socle <strong>of</strong> Z ( ) has dimension equal to (p ,h + ; _<br />
0 i)pN ,1 . The same remark<br />
app<strong>lie</strong>s to the results <strong>in</strong> D.12/13.<br />
2) The theorem does not say whether di erent factors <strong>in</strong> the composition series are<br />
isomorphic to each other. If L and L 0 are composition factors with (L) 6= (L 0 ),<br />
then clearly L 6' L 0 . So the question is whether for xed 2 J( )them factors<br />
L ;i are isomorphic to each other. We shall see <strong>in</strong> Section F that such isomorphisms<br />
exist <strong>in</strong> certa<strong>in</strong> cases.<br />
3) The theorem con rms <strong>in</strong> part my speculations <strong>in</strong> [11], 11.15 (where I look only<br />
at those 2 C 0 0 that are p{regular, i.e., satisfy h + ; _ i > 0 for all 2 R + ).<br />
The part <strong>of</strong> those speculations not con rmed is that the factors <strong>in</strong> a composition<br />
series should be pairwise non-isomorphic. As mentioned <strong>in</strong> the preced<strong>in</strong>g remark,<br />
that turns out to be wrong.<br />
D.7. We now beg<strong>in</strong> to prove the theorem. Lemma B.13 and D.5(3) show: If<br />
Theorem D.6 holds for one <strong>subregular</strong> , then it holds for all <strong>subregular</strong> . We<br />
assume from now on that is a simple root with (p )=0and (x, ) 6= 0 for<br />
all simple roots 6= . Remark D.1 shows that we can nd with this property<br />
for each . Later on we shall make speci c choices for . [Usually this is done<br />
such that the right hand side <strong>in</strong> (6) below is as small as possible.]<br />
Set L0 equal to the simple socle <strong>of</strong> Z ( ). It satis es D.6(3) by Proposition<br />
D.4.b and D.5(2).<br />
By our choice <strong>of</strong> the set I as <strong>in</strong> C.11(1) consists <strong>of</strong> all simple roots 6= .We<br />
shall use the construction from C.12{14 <strong>in</strong> order to nd the rema<strong>in</strong><strong>in</strong>g composition<br />
factors <strong>of</strong> Z ( ).<br />
We can construct <strong>in</strong>ductively a cha<strong>in</strong><br />
1 = ; 2;:::; r (1)<br />
<strong>of</strong> roots <strong>in</strong> WI such that h r; _ i 0 for all 2 I and such that there exists for<br />
each i