Ado's theorem for metric Lie algebras?

Question

Background

Ado's Theorem states that every finite-dimensional Lie algebra over a field of zero characteristic admits a faithful representation.

More precisely, if $\mathfrak{g}$ is a finite-dimensional Lie algebra over a field $K$ of zero characteristic, then there is a Lie algebra monomorphism $\rho: \mathfrak{g} \to \operatorname{End}(K^N)$ for some $N$, where $\operatorname{End}(K^N)$ is the Lie algebra of endomorphisms of $K^N$ relative to the commutator.

Now let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $K$ and let $\langle-,-\rangle$ be an ad-invariant symmetric inner product; that is, a nondegenerate symmetric $K$-bilinear form such that for all $x,y,z \in \mathfrak{g}$, $$\langle [x,y], z\rangle = \langle x,[y,z]\rangle.$$ We call such a Lie algebra a metric Lie algebra.

For example, $\operatorname{End}(K^N)$ itself is a metric Lie algebra, relative to the inner product $$\langle X,Y \rangle := \operatorname{Tr}(XY),$$ for endomorphisms $X,Y$.

Semisimple and, more generally, reductive Lie algebras are metric, but there are others. The relevant structure theorem is due to Medina and Revoy (MathSciNet link).

A final definition, added after Victor's comment below, is the following. Given two metric Lie algebras, the orthogonal direct sum of the underlying vector spaces can again be given the structure of a metric Lie algebra, in which the original Lie algebras sit as orthogonal ideals. This gives rise to the notion of an indecomposable metric Lie algebra as one which is not isomorphic (as metric Lie algebra) to the direct product of orthogonal (nonzero) ideals.

The following question is motivated by trying to construct Chern-Simons forms for Lie groups admitting a bi-invariant metric. But this motivation aside, I think the question is natural.

Question

Does every (finite-dimensional) indecomposable metric Lie algebra admit a faithful representation $$\rho: \mathfrak{g} \to \operatorname{End}(K^N),$$ for some $N$, such that for all $x,y\in\mathfrak{g}$, $$\langle x,y \rangle = c \operatorname{Tr}(\rho(x)\rho(y)),$$ for some nonzero $c \in K$?

Answer 1

The answer is negative if $\mathfrak g$ is solvable and non commutative. It follows from the "Critère de Cartan" (Bourbaki, algèbres de Lie, chapitre 1, par. 5).

Answer 2

Given $\mathfrak{g}$, a necessary and sufficient condition for the existence of a single finite-dimensional representation whose trace form defines a non-degenerate bilinear form on the Lie algebra, is that $\mathfrak{g}$ is reductive.

And indeed there are many non-reductive Lie algebras with a symmetric invariant bilinear form. For instance, for every Lie algebra $\mathfrak{h}$, the semidirect product $\mathfrak{h}\ltimes\mathfrak{h}^*$ (with $\mathfrak{h}^*$ abelian, action being the coadjoint representation) admits such a form, for which both $\mathfrak{h}$ and $\mathfrak{h}^*$ are isotropic subspaces and the bilinear form is induced by the dual pairing.

Besides, for a general reductive Lie algebra $\mathfrak{g}$, not every non-degenerate bilinear form is a pull-back of the trace form of a representation.

For instance, if $\mathfrak{g}$ is semisimple and nonzero (and, to simplify, $K$ is uncountable) the number of possible representations (up to equivalence) is countable while the number of forms is uncountable (just using scalar multiplication). Even if one counts forms "up to scalar multiplication" (as in the OP's question), as soon as the space of invariant bilinear forms has dimension $\ge 2$ (which means that $\mathfrak{g}$ is nonzero and not absolutely simple), the number of forms "up to scalar multiplication" is too large.