subregular nilpotent representations of lie algebras in prime ...

Lemma: If m runs through a basis of M, then the set of all
Y
>0; 2R1nZI
r( )
x, Y
2R +
2
x p,1
, m (2)
45
with 0 r( ) 0; 2R1nZI
r( )
x, Y
2R +
2
r( )
x, Y
2R +
3
r( )
x, m (3)
with m as above and all exponents runnning from 0 to p , 1 are a basis of V . The
elements in (2) are a subset of this basis, hence linearly independent.
We have dimu = jR +
2 j + jR+ 3 j. Therefore a comparison of (3) and (2) shows
that the elements in (2) span a subspace of dimension equal to dim(V )=p dim u .
This is by E.1(4) also the dimension of V u . So the claim will follow ifwe can show
that all elements in (2) belong to V u .
There exists a group homomorphism d : ZR ! Z with d( ) = 0 for all
2 R1 and d( ) > 0 for all 2 R with g u, i.e., for all 2 (,R +
2 ) [ R+ 3 .
(If belongs to the basis of R used to construct u in E.1, then set d( )=1if
=2 R1 and d( ) = 0 otherwise.) We get then a Z{grading on g such that each g
is homogeneous of degree d( ) and such that h is homogeneous of degree 0. Then
l is the homogeneous component of degree 0 and u is the sum of the homogeneous
components of positive degree. This induces a grading on U(g) and then also on
U (g). (For the last claim one uses that vanishes by E.1(2) on all x not in l.)
The subalgebra pI of g is graded (since it is spanned by h and certain x , hence
by homogeneous elements). If we giveMthe trivial grading (where everything has
degree 0), then it becomes a graded pI{module. (The homogeneous components
of pI of non-zero degree belong to the nilradical of pI and act as 0.) This leads
then to a grading on the induced U (g){module V .
Each basis element in (3) is homogeneous [of degree , P r( )d( )]. We get
elements of maximal degree if we choose r( )=p , 1 whenever d( ) < 0, and
r( ) = 0 whenever d( ) > 0. We haveto admit arbitrary m and arbitrary r( )
whenever d( )=0. If 2 R + ,thenwehaved( ) = 0 if and only if 2 R1, and
d( ) < 0 if and only if g, u if and only if 2 R +
2 . It follows that the elements
in (2) are precisely the elements in (3) that have maximal degree. So they are a
basis for the homogeneous component of maximal degree of V . Since each x 2 u
maps an element of some degree to an element of higher degree, all elements in
(2) are annihilated by u. The lemma follows.
E.3. Lemma: We have
x Y
2R +
2
x p,1
,
Y
m =
2R +
2
x p,1
, x m (1)