# “If $g$ is semisimple, It is not too hard to see that $H^2(g,a)=0$. With a little supplementary argument…”

This is a statement made in Knapp, Lie groups, Lie algebras, Cohomology Chpt 4 last paragraph of Sec 2.

$$H^i(g,a)$$ is the $$i-$$th cohomology group of complex $$Hom(\wedge^i g,a)$$ with $$a$$ abelian lie algebra.

"If $$g$$ is semisimple, it is not too hard to see that $$H^2(g,a)=0$$. With a little supplementary argument, it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra. The same theorem and proof apply over $$R$$ and a structure theorem for lie groups drops out. Results for $$H^1$$ then furnishes a uniqueness theorem"

$$\textbf{Q1:}$$ Maybe this is too obvious for others. I do not see why obviously $$H^2(g,a)=0$$. Fix $$\pi:g\to End(a)$$ representation. I am using Brian, Hall's Lie algebra's semisimple meaning reductive and trivial center. I could see all semi simples are decomposed into simples and this is unique upto reordering. So given an extension $$0\to a\to h\to g\to 0$$. Now $$h=a\oplus_\pi g$$ as semi-direct product but the lie bracket between $$a,g$$ will be twisted by representation of $$g\to End(a)$$. I do not see why all extensions are equivalent to $$0\to a\to a\oplus_\pi g\to g$$.

$$\textbf{Q2:}$$ Why it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra?

$$\textbf{Q3:}$$ Why the same theorem and proof apply over $$R$$ and a structure theorem for lie groups drops out? Over $$R$$, I no longer have unitary representation which may not allow me to have decomposition of $$g$$ into simple ones.

$$\textbf{Q4:}$$ What are results for $$H^1$$ furnishing a uniqueness theorem? What is the meaning of uniqueness theorem here?

• The result in Q1 is called "Whitehead's second lemma"; that might be a useful keyword. – Qiaochu Yuan Mar 10 at 20:46
• $\mathfrak{a}$ is not just an abelian Lie algebra (the question would make no sense, or would assume implicitly that the action is trivial), but is also endowed with a $\mathfrak{g}$-action. – YCor Mar 11 at 11:16
• Q2: This is Levi's theorem. There is a short proof of it using Whitehead's second Lemma from above (which can be found in several books, but is also a nice exercise). – Dietrich Burde Mar 11 at 16:17