subregular nilpotent representations of lie algebras in prime ...

This sum is a regular function on X
each
, since each bs0( ) is polynomial in and
is polynomial in f.
(f + i)(ht) =
nY
i=1
(f + i)(hi) t(i)
This nishes the proof of the claim. We now return to the proof of the
proposition. The space of all linear maps from Z(f + 1; ; )toZ(f + 2; ; )
has basis Esr with r 2 R1, s 2 R2 such that for all r 0 2 R1
Esr(x ,
r 0vf+ 1) = x, s vf+ 2; if r 0 = r;
0; otherwise.
Given a 2 g we can now use (2) to evaluate each
and get
(a Esr)(x ,
r 0 vf+ 1) =a (Esr(x ,
r 0vf+ 1)) , Esr(ax ,
r 0vf+ 1)
a Esr = X
t2R2
So we have aformula of the form
a Esr = X
c 2
t;s (a; f; )Etr , X
X
r02R1 s02R2 u2R1
c 1
r;u (a; f; )Esu:
bs 0 ;r 0 ;s;r(a; f; )Es 0 r 0
where the bs 0 ;r 0 ;s;r(a; f; ) are regular functions of (f; ) 2 X .
Now Homg(Z(f + 1; ; );Z(f + 2; ; )) identi es with the space of all
(R2 R1){tuples (xsr)r;s of elements in K with a P r;s xsrEsr =0foralla 2 g.
It actually su ces to take fora all elements in a basis (or a generating system) of
g. This shows that we indeed get the Hom space as a solution space of a linear
system of equations as described at the beginning of the proof. The proposition
follows.
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