# Whitehead's lemma (Lie algebras) for reductive Lie algebras [closed]

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!

## closed as off-topic by Vladimir Dotsenko, Stefan Kohl, Wolfgang, Stefan Waldmann, YCorFeb 16 '16 at 17:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Stefan Kohl, Wolfgang, Stefan Waldmann, YCor
If this question can be reworded to fit the rules in the help center, please edit the question.

• Cohomology of $gl_n$ with coefficients in finite-dimensional modules is well known. You are asking a question about $H^1$. If figuring this out and locating the answer in the literature leads to substantial difficulties, you should be asking it on MSE, not MO. – Vladimir Dotsenko Feb 16 '16 at 12:17
• The conclusion fails in general. Take $\mathfrak{g}$ 1-dimensional abelian, $V$ the 2-dimensional module defined by $x(y,z)=(xz,0)$. Define $f(x)=(0,x)$, so $xf(y)=(xy,0)$ is symmetric in $x,y$ and thus $f([x,y]-xf(y)+yf(x)=0$ for all $x,y$. It does not have the form $f(x)=xv=x(v_1,v_2)=(xv_2,0)$. – YCor Feb 16 '16 at 12:20
• @Vladimir Dotsenko, thank you very much. I moved the question to MSE. – Jianrong Li Feb 16 '16 at 13:20
• You haven't moved the question, you have cross-posted it. – YCor Feb 16 '16 at 18:00

Now it suffices to say that $\mathfrak{gl}(n)$ is not semisimple.