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Let $\mathfrak{g}$ be a finite-dimensional Lie algebra, let $\mathfrak{g}^{(i)}$ with $i \geq 0$ denote the derived series of $\mathfrak{g}$ (with $\mathfrak{g}^{(0)} = \mathfrak{g}$) and let $\operatorname{rad}(\mathfrak{g})$ be the solvable radical of $\mathfrak{g}$, i.e. the unique maxmimal solvable ideal of $\mathfrak{g}$. There exist minimal natural numbers $n$ and $m$ with $$ \mathfrak{g}^{(n)} = \mathfrak{g}^{(n+1)} \, \qquad\text{and}\qquad \operatorname{rad}(\mathfrak{g})^{(m)} = 0 \,. $$

Question. Is there some relation between the numbers $n$ and $m$?

Some thoughts:

  • If $\mathfrak{g}$ is solvable, then $\operatorname{rad}(\mathfrak{g}) = \mathfrak{g}$ and $n = m$.
  • If $\mathfrak{g}$ is semisimple, then $n = 0 = m$.
  • There exist perfect, non-semisimple Lie algebras. It may therefore happen that $n = 0 < m$. But the only examples I know have $m = 1$.
Jendrik Stelzner
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2 Answers2

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I assume some field of characteristic zero is fixed. The answer is then: the possible $(n,m)$ are exactly those $(n,m)$ with $0\le n\le m$.

Proof:

  1. All $(n,m)$ with $0\le n\le m$ can be achieved.

Indeed, $(0,0)$ is clear (it means semisimple), so suppose $m\ge 1$ Let $\mathfrak{v}_m$ be the free $2^m$-step nilpotent Lie algebra on $m$-generators, and consider $\mathfrak{g}_m=\mathfrak{sl}_m\ltimes\mathfrak{v}_m$. It's perfect ans has $m(\mathfrak{g}_m)=m$. Hence for $\mathfrak{g}_{n,m}=\mathfrak{v}_n\times\mathfrak{g}_m$, for $n\le m$, achieves $(n,m)$.

(The dimension is clearly not optimal but I don't really pay attention. The smallest achieving $(0,1)$ is the perfect 5-dimensional Lie algebra $\mathfrak{sl}_2(K)\ltimes K^2$ and the smallest achieving $(0,2)$ is the 6-dimensional $\mathfrak{sl}_2(K)\ltimes \mathfrak{hei}_3(K)$.)

  1. Conversely, $n\le m$ is a necessary condition.

Indeed, let $\mathfrak{r}_2$ be the radical of $\mathfrak{g}$ and $\mathfrak{r}_1$ the quotient of $\mathfrak{g}$ by the intersection of its derived series (this is the largest solvable quotient of $\mathfrak{g}$). So by definition $n(\mathfrak{g})$ is the solvability length of $\mathfrak{r}_1$ and $m(\mathfrak{g})$ is the solvability length of $\mathfrak{r}_2$.

The inequality then follows from the observation that $\mathfrak{r}_1$ is quotient of $\mathfrak{r}_2$. Indeed, write $\mathfrak{r}=\mathfrak{s}\ltimes\mathfrak{r}_2$. Then the quotient map $\mathfrak{g}\to\mathfrak{r}_1$ is trivial on $\mathfrak{s}$, hence is surjective in restrictio to the radical $\mathfrak{r}_2$.

YCor
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Benayadi has constructed a perfect non-semisimple Lie algebra $L$ in dimension $25$, where $\operatorname{rad}(L)$ is $2$-step solvable, so with $m=2$. It is given by \begin{align*} L & =\mathfrak{sl}_2(\Bbb C)\ltimes (V(7)\oplus V(5)\oplus V(7)\oplus V(3)) \\ & =\mathfrak{sl}_2(\Bbb C)\ltimes (L_1\oplus L_2\oplus L_3\oplus L_4) \end{align*} where $V(n)$ is the $n$-dimensional irreducible representation of $\mathfrak{sl}_2(\Bbb C)$ and $$ [L_1,L_1]=L_3,\; [L_1,L_2]=L_4,\; [L_2,L_2]=L_4,\; [L_1,L_3]=L_4. $$ So the radical is not abelian. This can be generalized.

Benayadi had another motivation of course, he wanted to find a non-semisimple perfect Lie algebra of lowest dimension satisfying
$$ H^0(L,L)=H^1(L,L)=0. $$

Reference: S. Benayadi: Structure of perfect Lie algebras without center and outer derivations. Annales de la faculte des sciences de Toulouse $6^e$ serie, tiome 5, no.2, (1996), p. 203-231.

Dietrich Burde
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  • There's a much easier 6-dimensional example with these properties. I assume the motivation for such a complicated construction was this vanishing of adjoint cohomology (trivial center, every derivation is inner). The 6-dimensional example has both nontrivial center and nontrivial exterior derivation algebra. Examples with trivial center appear in dimension 8 if I'm correct: $\mathfrak{sl}_2\ltimes\mathfrak{f}$ where $\mathfrak{f}$ is free 3-step nilpotent on 2 generators, with irreducible action on the abelianization. – YCor Dec 17 '20 at 12:54
  • Yes, this is what I found, too. Of course there are easier examples. I have chosen the complicated example, because I just have done some computations with it, so that I had it ready (and I thought, it deserves some attention because of its adjoint cohomology - it doesn't satisfy $H^2(L,L)=0$ though). On the other hand, this was not really the question here. – Dietrich Burde Dec 17 '20 at 14:34