# Commutator of radical and Levi factor in Lie algebra

Write a complex finite-dimensional Lie algebra as $L=S\ltimes R$ (Levi decomposition). Then the subalgebra $[S,R]$ (generated by brackets $[s,r]$, $s\in S$, $r\in R$) is an ideal, by a simple verification based on the Jacobi and the equality $[S,R]=[S,[S,R]]$.

Is there a classical reference for this probably well-known fact ($[S,R]$ is an ideal)?

Remark: I indeed had in mind the proof given by Robin Goodfellow and I'm pretty sure I once saw it written. Moreover $[S,R]$ can be defined with no reference to any $S$: this is the intersection $R\cap\bigcap_n L^{(n)}$, where $\bigcap_n L^{(n)}$ is the intersection of the derived series (and thus the largest perfect subalgebra in $L$, which is also the ideal generated by $S$).

• My guess would be 'Lie Algebras' by Jim Humpreys (who is also on mathoverflow, but I don't know if there is a way to 'tag' him in this comment). However I don't have the book lying around right now so I cannot check if it is really in there. – Vincent Nov 23 '16 at 13:31
• @Vincent: Actually, I didn't get into the Levi decomposition in general in my old textbook. But Bourbaki's Chapter I in their treatise on Lie groups and Lie algebras is still a standard reference. – Jim Humphreys Nov 25 '16 at 23:22
• @JimHumphreys Let me just quickly grab the opportunity to say that I learned a LOT from your Lie algebra book! – Vincent Nov 28 '16 at 9:07
• Concerning references, I don't see this written down in Bourbaki, Serre, etc. But an easy corollary of Levi's theorem is found on page 91 of Jacobson's 1962 book Lie Algebras (now in Dover reprint): in your notation, $[L,L] \cap R = [L,R]$, which is therefore an ideal of $L$. It's a quick argument not using $[S,S]=S$ or the Jacobi identity. The fact that $[S,R]$ is an ideal takes a little more argument but is also an elementary consequence of Levi's theorem. (Note that Mal'cev's conjugacy theorem for Levi subalgebras isn't used.) – Jim Humphreys Nov 28 '16 at 22:56
• @JimHumphreys That both $[L,L]\cap R$ and $[L,R]$ are ideals is immediate regardless of Levi's theorem. That $[S,R]$ is an ideal, I can hardly say that it relies on Levi's theorem since its bare definition relies on Levi's theorem. That the obvious inclusion $[L,R]\subset [L,L]\cap R$ is an equality is a consequence of the vanishing of $H_2(L/R)$ (which is equivalent to Levi' theorem for Lie algebras with central radical). – YCor Nov 29 '16 at 5:07

Of course, $[S,R]\leq R$, since $R$ is an ideal, so $[S,[S,R]]\leq[S,R]$. Going the other way, since $S$ is perfect, $[S,R]=[[S,S],R]\leq[[S,R],S]+[S,[S,R]]=[S,[S,R]]$, so $[S,R]=[S,[S,R]]$.
As for showing $[R,[S,R]]\leq[S,R]$, we simply need to use the Jacobi identity and the previous result: $$[R,[S,R]]=[R,[S,[S,R]]]\leq[[R,S],[S,R]]+[S,[R,[R,S]]]\leq[S,R].$$
• I guess "this fact" refers to the fact $[S,R]$ is an ideal, not just the equality $[S,R]=[S,[S,R]]$. – YCor Nov 25 '16 at 1:34
• In general it's just $\subset$. For instance, if $S$ is $SL_2$, $R$ is the 3-dimensional Heisenberg Lie algebra (with standard action), then $[S,R]=R$, so $[R,[S,R]]$ is the 1-dimensional center, but $[S,[R,R]]=0$. So the "Jacobi identity for subalgebras" here is a proper inclusion. – YCor Nov 25 '16 at 18:45