subregular nilpotent representations of lie algebras in prime ...

Claim: We have
for all 2 C0 with h + ; _ ni > 0.
g Z ( ; n) ' Z (s2"n,1 ; n) (1)
Proof : Set J = f n,1; ng and use the abbreviation
Z ;J( ; n) =U (pJ) U (p n ) L ; n( )
for all 2 X. Wehave by transitivity of induction
Z ( ; n) ' U (g) U (pJ ) Z ;J( ; n)
for all . Note that g belongs to the standard Levi factor GJ of PJ since we can
nd there at least one representative ofs"n,1+"n [coming from the root subgroups
U ("n,1+"n)]. Therefore we have for all
g Z ( ; n) ' U (g) U (pJ) g Z ;J( ; n);
cf. B.14(1). So our claim will follow ifwe can show that
g Z ;J( ; n) ' Z ;J(s2"n,1 ; n): (2)
The nilradical of pJ acts as 0 on both sides in (2). So we just have to nd an
isomorphism of gJ{modules. The centre of gJ [equal to the intersection in h of
ker( n) and ker( n,1)] acts on both sides via the restriction of .Sowejusthave
to nd an isomorphism as modules over the derived Lie algebra DgJ of gJ.
We want to apply F.9(2) to DgJ in order to get the isomorphism in (2) and
thus the claim. The restriction of to DgJ has standard Levi form. We have
GJ = Z(GJ) 0 DGJ [where Z(GJ) 0 T is the connected centre] and can thus
write g = zg 0 with z 2 Z(GJ) 0 . Since z trivially xes on DgJ, sodoesg 0 . Since
g 0 is still a representative fors"n,1+"n, we can now apply F.9(2) and get the claim.
F.11. Suppose now that R is of type F4. Set
x = x 2 + x 3 + x 4 + x 1+ 2+ 3+ 4: (1)
Let be the linear form corresponding to x. Then is subregular (see the end of
the proof of D.13) with (p 1) = 0. We can nd a representative g 2 NG(T ) for
the re ection s"2 = s 1+ 2+ 3 such that g = . Proposition F.7 follows in this
case from F.6(4),(5) and:
Claim: Let 2 C0. We have
if h + ; _ 1 i > 0, and
g Z ( ; 1) ' Z (s"2+"3 ; 1) (2)
g Z (s2s1 ; 1) ' Z (s"2+"3s2s1 ; 1) (3)
59