Can the Killing form induce an endomorphism in a Lie algebra

Question

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 $\langle X,X \rangle=B(\phi(X),X)$?.

Answer 1

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 $\langle \phi^{-1}(Z),Y \rangle = B(Z,Y) = B(Y,Z) = \langle \phi^{-1}(Y),Z \rangle.$