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