subregular nilpotent representations of lie algebras in prime ...

C.6. We nowwant to de ne for each w 2 W a homomorphism
' w : Z (w ) ,! Z ( ) (1)
as follows: If w = s1s2 :::sr is a reduced decomposition (with si = s i for some
simple root i) then we want tohave
' w = ' 1;sr ' sr;sr,1sr ' s2:::sr,1sr;s1s2:::sr,1sr (2)
where the single factors are de ned in C.5. One has to check the independence of
the right hand side in (2) of the chosen reduced decomposition. It is (as usual)
enough to check the \braid relations" for each pair ; ,( 6= ) of simple roots.
For example, if s s has order 3, then we have tocheck for each x 2 W with
x ,1 ; x ,1 >0 that
' x;s x ' s x;s s x ' s s x;s s s x = ' x;s x ' s x;s s x ' s s x;s s s x :
This follows from the Verma relations, see [4], 7.8.8; similarly in the cases where
s s has order 2, 4, or 6.
Now that the ' w are well-de ned, we get for all w 2 W and all simple roots
with w ,1 >0 that
' s w = ' w ' w;s w (3)
since we get a reduced decomposition of s w when we multiply a reduced decomposition
of W on the left by s .
C.7. Set for all w 2 W
the submodule corresponding to w.
sbm(w; )=im(' w ) Z ( ) (1)
Lemma: Let be a simple root and w an element in W with w ,1 >0.
a) We have sbm(s w; ) sbm(w; ).
b) If (x, ) 6= 0, then sbm(s w; )=sbm(w; ).
c) If (x, )=0, then sbm(s w; ) is a homomorphic image of Z (s w ; ),
and sbm(w; )= sbm(s w; ) is a homomorphic image of Z (w ; ).
Proof : The identity ' s w = ' w ' w;s w from C.6(3) implies a). If (x, ) 6= 0,
then ' w;s w is an isomorphism; this yields b).
Suppose now that (x, ) = 0. Denote the image of ' w;s w by M. ThenM
is isomorphic to Z (s w ; ), and Z (w )=M is isomorphic to Z (w ; ). Now
the claim in c) follows from sbm(s w; )=' w(M) and sbm(w; )=' w(Z (w )).
Remark: It is sometimes more convenient to restate the last part of the lemma as
follows: Let w 2 W and 2 R + such that w is a simple root with (x,w )=0.
Then sbm(ws ; ) is a homomorphic image of Z (ws ; w ), and sbm(w; )= sbm
(ws ; ) is a homomorphic image of Z (w ; ). [Note that sw w = ws .]
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