# Can the Killing form induce an endomorphism in a Lie algebra?

Let $$\mathfrak{g}$$ be a Lie algebra over $$\mathbb{R}$$ of the finite dimensional Lie group $$G$$; let $$\langle \cdot , \cdot \rangle$$ be a left-invariant Riemannian metric on $$G$$. If $$B:\mathfrak{g}\times \mathfrak{g}\to \mathbb{R}$$ is the Cartan-Killing form $$(X,Y)\mapsto \text{Tr}( \text{ ad}_X \circ \text{ad}_Y)$$. Is it true that there is a symmetric endomorphism $$\phi$$ on $$\mathfrak{g}$$ such that for every $$X\in \mathfrak{g}$$ we have $$\langle X,X \rangle=B(\phi(X),X)$$?.

• I think you need the extra condition of the Lie algebra being semi-simple. Jul 8, 2020 at 17:14

You need the Killing form to be nondegenerate, which is equivalent to requiring that the Lie algebra is semisimple. Otherwise there exists some $$X \in \mathfrak{g}$$ such that $$B(\phi(X),X)=0$$ for any possible definition of $$\phi$$.
When $$\mathfrak{g}$$ is semisimple the statement holds. First of all there is a unique isomorphism $$\phi:\mathfrak{g} \to \mathfrak{g}$$ such that $$\langle X, Y \rangle = B(\phi(X),Y)$$ by nondegeneracy and then $$\langle \phi^{-1}(Z),Y \rangle = B(Z,Y) = B(Y,Z) = \langle \phi^{-1}(Y),Z \rangle.$$