Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


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.


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 $c \in K$?

share|cite|improve this question
Is there a particular class of metric Lie algebras that you are interested in? There is a silly counterexample for a metric Lie algebra of the form $g_1\oplus g_2,$ where the summands $g_i$ are split simple, the restriction of the metric to $g_i$ is the $c_i$-multiple of the Killing form on $g_i$, and the ratio $c_1/c_2$ is irrational. – Victor Protsak Aug 6 '10 at 4:12
Sorry, I don't understand what $L$ is. Is it $K$? Or is it just a formal symbol? – Pierre-Yves Gaillard Aug 6 '10 at 5:57
PYG: My guess is that, L being the first character in "Lie", it indicates that $\text{End}_L(K^N)$ is viewed as a Lie algebra. – Victor Protsak Aug 6 '10 at 6:03
@Pierre-Yves: As Victor pointed out, I was making a distinction between the Lie algebebra and the underlying associative algebra by attaching the $L$ to $\operatorname{End}$. Given my choice of notation for the field, I can see how this was confusing. I will edit. – José Figueroa-O'Farrill Aug 6 '10 at 10:19
@Victor: Sorry, I forgot to add that the indecomposability condition! Thanks for pointing this out. I will edit. – José Figueroa-O'Farrill Aug 6 '10 at 10:20

1 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).

share|cite|improve this answer
Bienvenue Michel! – Alain Valette Nov 12 '11 at 17:40
Bienvenue, Michel ! et merci d'avance d'y apporter ton expérience. – Duchamp Gérard H. E. Jun 10 at 11:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.