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, 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
  • 2
    $\begingroup$ @Vladimir Dotsenko, thank you very much. I moved the question to MSE. $\endgroup$ – Jianrong Li Feb 16 '16 at 13:20
  • 1
    $\begingroup$ You haven't moved the question, you have cross-posted it. $\endgroup$ – YCor Feb 16 '16 at 18:00

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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