### Background


[Ado's Theorem](https://en.wikipedia.org/wiki/Ado%27s_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](https://mathscinet.ams.org/mathscinet-getitem?mr=826103) (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 $c \in K$?
> 
> 
> 



Ado's theorem for metric Lie algebras?
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).


