# Whitehead's lemma (Lie algebras) for reductive Lie algebras.

I move the question here.

Whitehead's lemma (Lie algebras) is: Let $\mathfrak{g}$ be a semisimple Lie algebra over a field of characteristic zero, $V$ a finite-dimensional module over it and $f$: $\mathfrak{g} \to V$ a linear map such that $f([x, y]) = xf(y) - yf(x)$. The lemma states that there exists a vector $v$ in $V$ such that $f(x) = xv$ for all $x$.

If we let $\mathfrak{g}$ be a reductive Lie algebra (for example, let $\mathfrak{g} = \mathfrak{gl}_n$), the conclusion of Whitehead's lemma is still true or not (or we need to add some other conditions)? Are there some references about this? Any help will be greatly appreciated!

Edit: I would like to know in particular the case of $\mathfrak{gl}_n$. Do we also have the following: all Lie bialgebra structure on $gl_n$ are coboundary (see also the post)?

• Are you asking specifically about $\mathfrak{gl}_n$ or in general? Because an example where it fails was given in the comments there. – Tobias Kildetoft Feb 16 '16 at 13:20
• @Tobias Kildetoft, thank you very much. I am asking specifically about $\mathfrak{gl}_n$. I will edit the post. – LJR Feb 16 '16 at 13:22
• On MO (mathoverflow.net/questions/231298) I gave you an example showing that the result is false in general for reductive Lie algebras (with $\mathfrak{g}$ 1-dimensional and $V$ 2-dimensional). – YCor Feb 16 '16 at 14:43

Since $\mathfrak{gl}(n)$ is not semisimple (it has a nontrivial center), Whitehead's first Lemma must fail for $\mathfrak{gl}(n)$ and some finite-dimenisonal module $V$ (and it does).