I was recently trying to prove the following "well-known" theorem for myself, given that I could not find a proof in the literature that I could understand. In what that follows, I will implicitly assume that everything is finite-dimensional.


**Theorem.** If every abelian extension of the Lie algebra $\mathfrak g$ splits, then every extension of $\mathfrak g$ splits.


The proof considers a splitting $0\rightarrow \mathfrak h\rightarrow \mathfrak e\rightarrow \mathfrak g\rightarrow 0$ and proceeds by induction on the dimension of $\mathfrak h$. The base case uses the assumption about abelian extensions, specifically for the case of one-dimensional $\mathfrak h$. If $\mathfrak h$ contains a non-trivial, proper subideal $\mathfrak k$, then we can apply the induction hypothesis to get a splitting of $$0\rightarrow \mathfrak h/\mathfrak k\rightarrow \mathfrak e/\mathfrak k\rightarrow \mathfrak g\rightarrow 0$$
Then we can lift this splitting to a splitting of the original sequence (a lemma that also took me a while to understand). But suppose that no such $\mathfrak k$ exists. Then if $\mathfrak h^\perp$ is the perp-space with respect to the Killing form on $\mathfrak e$, we either have $\mathfrak h\cap \mathfrak h^\perp=0$ or $\mathfrak h\subseteq \mathfrak h^\perp$. In the former case, this perp-space $\mathfrak h^\perp$ yields a splitting. In the latter case, the Killing form vanishes on $\mathfrak h$ and so $\mathfrak h$ is solvable by Cartan's criterion. This implies that $[\mathfrak h,\mathfrak h]\neq \mathfrak h$ and thus $[\mathfrak h,\mathfrak h]=0$ (because we assumed that $\mathfrak h$ contains no proper, non-trivial subideals). But then $\mathfrak h$ is abelian and so we get a splitting by assumption.


In this proof, we twice used the assumption that every abelian extension of $\mathfrak g$ splits. But the first instance only used the seemingly weaker assumption that every one-dimensional extension splits, whereas the second instance used the full power of the assumption. This leads me to wonder:


**Question:** If every one-dimensional extension of $\mathfrak g$ splits, does every extension of $\mathfrak g$ split? Or phrased negatively, is there a Lie algebra $\mathfrak g$ for which every one-dimensional extension splits, but which also admits a non-splitting extension
$$0\rightarrow \mathfrak h\rightarrow \mathfrak e\rightarrow \mathfrak g\rightarrow 0\,?$$



Lie algebras for which all one-dimensional extensions split
Here is an example of a Lie algebra $\mathfrak g$ for which every one-dimensional extesion splits, but not every extension splits (which is based on @YCor's comment above). We can work over any field $k$ of characteristic zero. If $V$ is a $\mathfrak g$-module, then $H^2(\mathfrak g,V)=0$ if and only if the only extension of $\mathfrak g$ by $V$ is the semidirect product $\mathfrak g\ltimes V$. Hence, we are looking for a Lie algebra $\mathfrak g$ and a $\mathfrak g$-module $V$ such that:


* $H^2(\mathfrak g,V)\neq 0$;
* $H^2(\mathfrak g,k)=0$ for any $\mathfrak g$-module structure on $k$.


We will choose $\mathfrak g$ to be perfect (meaning that $[\mathfrak g,\mathfrak g]=\mathfrak g$), which implies that the only $\mathfrak g$-module structure on $k$ is the trivial one, making the second condition easier to verify. We use the following two lemmas:


**Lemma 1.** If $\mathfrak g$ is a perfect Lie algebra and $V$ is a non-trivial, irreducible $\mathfrak g$-module, then the semidirect product $\mathfrak g\ltimes V$ is also perfect. <https://mathoverflow.net/a/60500/147463>


**Lemma 2.** Suppose that $\mathfrak s$ is semisimple and $V$ is an $\mathfrak s$-module such that $\big(\bigwedge^{\!2}V^\*\big)^{\mathfrak s}=0$. Considering $k$ as a trivial module, we then have $H^2(\mathfrak s\ltimes V,k)=0$. <https://math.stackexchange.com/q/3478712/716726>


Since $\mathfrak{so}\_3$ is simple and $\mathfrak{so}\_3$-module structure on $k^3$ is irreducible, we can see that $\mathfrak g=\mathfrak{so}\_3\ltimes k^3$ is perfect. If $\omega$ is an $\mathfrak{so}\_3$-invariant 2-form on $k^3$, then $\ker \omega$ is a submodule. But $k^3$ is irreducible, so this implies that $\ker \omega=0$ or $\ker \omega=k^3$. Since $k^3$ is odd-dimensional, it does not admit any non-degenerate 2-forms. So we must have $\ker \omega=k^3$ and thus $\omega =0$. By Lemma 2, we then have $H^2(\mathfrak g,k)=0$ (the only $\mathfrak g$-module structure on $k$ is the trivial one, because $\mathfrak g$ is perfect).


Now take the $\mathfrak g$-module structure on $k^3$ induced by the quotient map $\mathfrak g\rightarrow \mathfrak{so}\_3$ and the previous structure of $k^3$ as an $\mathfrak{so}\_3$-module. The I claim that $H^2(\mathfrak g,k^3)\neq 0$. To see this, consider the alternating bilinear form $\varphi:\mathfrak g\times \mathfrak g\rightarrow k^3$ given by the cross-product: $$\varphi\big((A,u),(B,v)\big)=u\times v.$$
Recall that $\mathfrak{so}\_3$ acts on $k^3$ by derivations with respect to the cross-product. Thus, we have
$$\varphi\Big(\big[(A,u),(B,v)\big],(C,w)\Big)=\varphi\Big(\big([A,B],Av-Bu\big),(C,w)\Big)=Av\times w-Bu\times w,$$
$$-(B,v)\cdot \varphi\big((C,w),(A,u)\big)=-B(w\times u)=B(u\times w)=Bu\times w+u\times Bw.$$
Summing both of these over cyclic permutations of $(A,u)$, $(B,v)$ and $(C,w)$, all of the terms cancel and we get $d\varphi=0$. Thus $\varphi$ is a cocycle. If $\varphi=df$ for some $f:\mathfrak g\rightarrow k^3$, then
$$u\times v=\varphi\big((0,u),(0,v)\big)=f\Big(\big[(0,u),(0,v)\big]\Big)-0\cdot f(0,v)+0\cdot f(0,u)=0.$$ But the cross-product is not always zero, so this is a contradiction. Thus $\varphi$ represents a non-zero element of $H^2(\mathfrak g,k^3)$.


