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# Conversely suppose that the killing form of l has a

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Unformatted text preview: emisimple iﬀ its Killing form is nondegenerate. Proof. Suppose that the Killing form of L is nondegenerate. Then L must be semisimple since any abelian ideal is contained in the kernel of κ which is zero. Conversely, suppose that the Killing form of L has a nonzero kernel S . Let LC = L ⊗F C be the complexiﬁcation of L. Since L ⊆ LC , it is clear that L is abelian iﬀ LC is abelian. This implies the L is solvable iﬀ LC is solvable since D(LC ) = (DL)C . One can also see that the Killing form κC of LC is the complexiﬁcation of the Killing form κ of L. So, the kernel of κC is SC . We know that SC is solvable from the proof of Cartan’s Theorem. Therefore S is solvable and L is not semisimple. ￿ Example 5.2.6. Let L = sl(2, R). This has basis ￿ ￿ ￿ ￿ 01 00 x= , y= , 00 10 h= with respect to this basis we have 0 0 −2 0 00 ad x = 0 0 0 , ad y = 0 0 2 , 01 0 −1 0 0 Then ￿ ￿ 10 0 −1 200 ad h = 0 −2...
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## This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.

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