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Given a Lie Algebra $L$ and the adjoint homomorphism $ad \colon L \rightarrow gl (L)$, if $ad\, L$ is solvable, then so is $L$.

We know that Ker $ ad \,L= \{x \in L \mid [x, y] = 0 \,\,\forall y \in L\} = Z(L)$. In order to use this result, one could build the quotient algebra, $L/Z(L)$. In that case, if $L/Z(L)$ is solvable then $L$ is solvable, meaning that one has to prove that $L/Z(L)$ is solvable.

Can somebody provide some suggestion how to prove this ?

cgb5436
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user249018
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2 Answers2

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By assumption $L/Z(L)\cong {\rm ad}(L)$ is solvable. Also $Z(L)$ is abelian, hence solvable. Now $L$ is an extension of $Z(L)$ by $L/Z(L)$, and hence solvable itself. This follows from the general result (see your lecture notes, or the book by Erdmann and Wildon you are using), that extensions of solvable Lie algebras by solvable Lie algebras are solvable. In terms of exact sequences, $$ 0\rightarrow Z(L)\rightarrow L\rightarrow L/Z(L)\rightarrow 0. $$

Dietrich Burde
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  • Thanks. In the book by Erdmann one encounters in proof of the Cartan's criteria of solvability the relation: $ \overline {ad, d} = ad ,\overline d.$ I can not figure out how this works. Can you give me a suggestion there ? I tried the following: $\overline {ad, d} (z)= \overline {[d,z]} = \overline {dz-zd}=...=ad ,\overline {d} (\overline z). $ – user249018 Nov 17 '21 at 22:32
  • See Torsten's comment here. $d$ is the "diagonal" part. – Dietrich Burde Nov 18 '21 at 11:04
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You have shown that $L/Z(L)\cong \operatorname{ad} L$, so you just need to show that $L/Z(L)$ solvable means that $L$ is solvable as well.

See if you can show $$\left[L/Z(L),L/Z(L)\right]=[L,L]/Z(L),$$ in the sense that it is $\{x+Z(L)\colon x\in [L,L]\}$. Use this to relate the derived series of $L/Z(L)$ to the derived series of $L$. If $(L/Z(L))^{(n)}=0$, you should be able to conclude something about $L^{(n)}$. Can you use this to show $L$ is solvable?

Carl Schildkraut
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