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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16<br />
The assumption on the facet imp<strong>lie</strong>s that<br />
np h + ; _ i (n +1)p:<br />
Set d = h + ; _ i,np and d 0 = h + ; _ i,np. Then d is the same number as<br />
considered <strong>in</strong> B.8. If d 0 > 0, then it the analogue <strong>of</strong> d work<strong>in</strong>g with <strong>in</strong>stead <strong>of</strong> ;<br />
if d 0 = 0, then that analogue is equal to p, however. Note that s ;np = , d<br />
and s ;np = , d 0 .<br />
We get from B.4(1) isomorphisms T Z ( ) ' Z ( )andT Z ( , d ) '<br />
Z ( , d 0 ). We claim that, modulo these isomorphisms:<br />
Lemma: If d 0 > 0, then T (' ) is a non-zero multiple <strong>of</strong> ' . If d 0 =0, then<br />
T (' ) is a non-zero multiple <strong>of</strong> the identity.<br />
Pro<strong>of</strong> :Ifd = p, i.e., if h + ; _ i =(n +1)p, then the assumption on the facets<br />
imp<strong>lie</strong>s that also h + ; _ i =(n +1)p and d 0 = p. In this case both ' and '<br />
are equal to (x, ) p times the identity. S<strong>in</strong>ce the functor T takes a multiple <strong>of</strong><br />
the identity to the correspond<strong>in</strong>g multiple <strong>of</strong> the identity, the claim follows <strong>in</strong> this<br />
case.<br />
Assume from now on that d