Talk:Cartan's criterion

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 to the fact that this equivalence is false.

Also, it seems that the standard usage of "Cartan's criterion" refers only to either the solvable or the semisimple case. Eg. this explicitly discusses both cases under the heading of "Cartan's criterion", and this 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 here, which correctly states that nilpotency implies that the Killing form vanishes (it is the converse that is false) but does not refer to this fact as Cartan's criterion.

Counterexample: let L be a complex vector space with a basis {a,b,c}. There is a unique alternating bilinear form [,] such that [ab] = b, [ac] = i.c , [bc] = 0. The only non-trivial case of the Jacobi identity is [a[bc]] + [b[ca]] + [c[ab]] = 0, which holds because each term vanishes individually. Hence we have a Lie algebra.

For any x, y, [xy] is a linear combination of b, c, so [b[xy]] = [c[xy]] = 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.

Finally, L is not nilpotent because it has no center (since ker(ad a) = <a>, ker(ad b) = ker(ad c) = <b,c> so the common kernel of the adjoint action is trivial). Alternatively, observe that [L,L] = <b,c> and [L,<b,c>] = <b,c>, so the upper central series stabilises at <b,c>.

Jeremy Henty (talk) 22:01, 20 June 2008 (UTC)

Non-degenerate invariant bilinear form =/=> reductive

More generally, a finite-dimensional Lie algebra ${\mathfrak {g}}$ is reductive if and only if it admits a nondegenerate invariant bilinear form.

Not sure why this claim is made in this article anyways, besides the fact that it is false: take e.g. the associative algebra $A=\mathbb {C} [z]/\langle z^{2}\rangle$ and consider the Lie algebra ${\mathfrak {g}}=sl_{2}(\mathbb {C} )\otimes A$ (usual definition with $[x\otimes f(z),y\otimes g(z)]=[x,y]\otimes f(z)g(z)$). Then ${\mathfrak {g}}$ in fact has a non-degenerate bilinear form but is not reductive... it has a single proper ideal $sl_{2}(\mathbb {C} )\otimes z$ which is abelian, so the adjoint representation is not reducible.

See also e.g. Favre & Santharoubane (1987).

Adagioing (talk) 00:19, 14 December 2010 (UTC)