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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!

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

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  • $\begingroup$ 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. $\endgroup$ – Vladimir Dotsenko Feb 16 '16 at 12:17
  • $\begingroup$ 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 $f(x)=xv=x(v_1,v_2)=(xv_2,0)$. $\endgroup$ – YCor Feb 16 '16 at 12:20
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    $\begingroup$ @Vladimir Dotsenko, thank you very much. I moved the question to MSE. $\endgroup$ – Jianrong Li Feb 16 '16 at 13:20
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    $\begingroup$ You haven't moved the question, you have cross-posted it. $\endgroup$ – YCor Feb 16 '16 at 18:00
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The following result is proved in Bourbaki's book on Lie algebras:

Theorem (A converse to the First Whitehead Lemma). Any finite-dimensional Lie algebra over the field of characteristic zero such that its first cohomology with coefficients in any finite-dimensional module vanishes, is semisimple.

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

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