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I deleted the claim that nilpotency is equivalent to vanishing of the
Killing form as it is false. There are non-nilpotent Lie algebras whose Killing form vanishes. Hopefully it won't violate the no original research principle if I give a counterexample further down. I haven't found any reference
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Counterexample: let L be a complex vector space with a basis {a,b,c}. There is a unique alternating bilinear form such that = b, = i.c , = 0. The only non-trivial case of the Jacobi identity is ] + ] + ] = 0, which holds because each term vanishes individually. Hence we have a Lie algebra.
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For any x, y, is a linear combination of b, c, so ] = ] = 0. Hence (ad b)(ad x) = (ad c)(ad x) = 0, so the
Killing form * satisfies b*x = c*x = 0. Finally, a*a = tr((ad a)^2)) - tr((diag(0,1,i))^2) = tr(diag(0,1,-1) = 0. Thus * vanishes.
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only refers to the solvable case as "Cartan's criterion" (p.66), though it discusses the semisimple case too (p.68). The only reference to nilpotency I could find is
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Also, it seems that the standard usage of "Cartan's criterion" refers only to either the solvable or the semisimple case. Eg.
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Not sure why this claim is made in this article anyways, besides the fact that it is false: take e.g. the associative algebra
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so the common kernel of the adjoint action is trivial). Alternatively, observe that = <b,c: -->
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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in fact has a non-degenerate bilinear form but is not reductive... it has a single proper ideal
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Finally, L is not nilpotent because it has no center (since ker(ad a) = <a: -->
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explicitly discusses both cases under the heading of "Cartan's criterion", and
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is reductive if and only if it admits a nondegenerate invariant bilinear form.
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which is abelian, so the adjoint representation is not reducible.
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348:{\displaystyle {\mathfrak {g}}=sl_{2}(\mathbb {C} )\otimes A}
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