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Applications: Finding A Maximal Two-sided Ideal In A Given Left Ideal - Results
After the first iteration, we obtain the ideal
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==>
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B[1]=eh-ea+2e
B[2]=2ef+ha-a2-2h
B[3]=e2
B[4]=h2a-2ha2+a3+2ha-2a2
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Since a is a parameter, we replace B[4]
with B[4]=h2-2ha+a2+2h-2a. Continuing iterations with
A=B, we obtain
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==>
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B[1]=4ef+h2-a2-2h-2a
B[2]=h3-3h2a+3ha2-a3+6h2-12ha+6a2+8h-8a
B[3]=eh2-2eha+ea2+6eh-6ea+8e
B[4]=e2h-e2a+4e2
B[5]=e3
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The first candidate is B[1]=4ef+h2-a2-2h-2a, since it has lower degree than the other
elements. We can prove our hypothesis by performing some more iterations :
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==>
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B[1]=4ef+h2-a2-2h-2a
B[2]=h4-4h3a+6h2a2-4ha3+a4+12h3-36h2a+36ha2-12a3+44h2-88ha+44a2+48h-48a
B[3]=eh3-3eh2a+3eha2-ea3+12eh2-24eha+12ea2+44eh-44ea+48e
B[4]=e2h2-2e2ha+e2a2+10e2h-10e2a+24e2
B[5]=e3h-e3a+6e3
B[6]=e4
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And we see that we are right - the answer is the ideal, generated by the
4ef+h2-2h-a2-2a.
Moreover, the answer is nothing else but the central element
4ef+h2-2h of
U(sl2) minus
the central character a2+2a
of the ideal <e,h-a> .
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