Solvable equivalent to nilpotency of first derived Lie algebra?

Question

The Wikipedia "Solvable Lie Algebra" page lists the following property as a notion equivalent to solvability: $\mathfrak{g}$ is solvable iff the first derived algebra $[\mathfrak{g},\,\mathfrak{g}]$ is nilpotent.

I can't see this at all and it does sound dodgy: solvability and nilpotency different by "only one rung" in the derived series. Is this correct (I suspect I'm missing something trivial- I can't seem to find anything relevant in Knapp "Lie Groups: Beyond an Introduction" (mine is reprint of the first edition))? If not universally correct, is it correct with further assumptions about the field which $\mathfrak{g}$ is a vector space over? (For example, a field of characteristic nought, algebraically closed, ...). Can someone either give a proof, a counterexample or a reference to one?

I'd also like to put a reference on the Wikipedia page quotation, or clearly have it stricken from the page if incorrect.

Answer 1

Wikipedia says, correctly, that this is equivalent if $\mathfrak{g}$ is finite-dimensional over a field of characteristic zero. I don't think it's true in general.

The hard direction is to show that this condition holds if $\mathfrak{g}$ is solvable. Here's a sketch. Since the image $\text{ad}[\mathfrak{g}, \mathfrak{g}]$ of $[\mathfrak{g}, \mathfrak{g}]$ in the adjoint representation of $\mathfrak{g}$ differs from it by a central extension, it suffices to show that $\text{ad}[\mathfrak{g}, \mathfrak{g}]$ is nilpotent. By Lie's theorem, over an algebraic closure $\bar{k}$ of the ground field the elements of $\text{ad}(\mathfrak{g})$ are simultaneously upper triangularizable. It follows that over $\bar{k}$ the elements of $\text{ad}[\mathfrak{g}, \mathfrak{g}]$ can be represented by strictly upper triangular matrices, and hence $\text{ad} [\mathfrak{g}, \mathfrak{g}]$ is nilpotent over $\bar{k}$. But nilpotence just means that certain words vanish identically, and whether this is true doesn't depend on whether we extend the ground field or not; hence $\text{ad}[\mathfrak{g}, \mathfrak{g}]$ is nilpotent.

Answer 2

Here's a simple proof that there exists, over any field, a solvable Lie algebra (of infinite dimension) whose derived subalgebra is non-nilpotent.

Consider the free complex class-3 solvable Lie algebra on countably many generators. Suppose by contradiction that its derived subalgebra is nilpotent, say $c$-nilpotent. Then from the universal property it follows that every countably generated (in part., every finite-dimensional) 3-solvable Lie algebra has a $c$-nilpotent derived subalgebra. We then get a contradiction as follows: for $n\ge 2$, let $\mathfrak{h}_n$ be the Lie algebra with basis $(e_1,f_1,\dots,f_{n-1})$ and nonzero brackets $[e_1,f_i]=e_{i+1}$ for $1\le i\le n-2$. Let $D$ be the derivation of this Lie algebra given by $e_1\mapsto e_1$, $f_i\mapsto if_i$, and consider the corresponding (2-solvable) semidirect product $\mathfrak{g}_n=\mathfrak{a}_1\ltimes_{D}\mathfrak{h}_n$, where $\mathfrak{a}_n$ is 1-dimensional abelian. Note that $[\mathfrak{h}_n,\mathfrak{h}_n]$ is abelian, so that $\mathfrak{g}_n$ is 3-solvable. Also $[\mathfrak{g}_n,\mathfrak{g}_n]=\mathfrak{h}_n$ (regardless of the characteristic), whose nilpotency class is exactly $n-1$. Since $n$ is unbounded, this yields contradiction and actually the free complex class-3 solvable Lie algebra has its derived subalgebra non-nilpotent.

(Alternatively, to avoid invoking free objects/ universal properties, the product $\prod_n\mathfrak{g}_n$ (or the restricted product if you like) is 3-solvable and its derived subalgebra is non-nilpotent).

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Edit: here's a variant, yielding a finitely generated algebra.

Consider the free metabelian Lie algebra $\mathfrak{k}$ on two generators $x,y$ (over any field). It is naturally graded in $\mathbf{Z}^2$ with $x$ of degree $(1,0)$ and $y$ of degree $(0,1)$. It is not nilpotent (because it admits the standard filiform $n$-dimensional algebra as a a quotient for every $n$, and the latter has a nontrivial $(n-1)$-th term of the central series). Consider the commuting derivations $D,E$ of $\mathfrak{f}$ where $D$ (resp. $E$) acts by multiplication by $i$ (resp. $j$) on $\mathfrak{f}_{(i,j)}$

The semidirect product of $\mathfrak{f}$ by a 2-dimensional abelian Lie algebra $\mathfrak{a}$ acting on $\mathfrak{f}$ by $D$ and $E$ has $\mathfrak{f}$ as derived subalgebra, which is not nilpotent. It is finitely generated as it is generated by the 4-dimensional subspace $\mathfrak{a}\oplus\mathfrak{f}_{(1,0)}\oplus\mathfrak{f}_{(0,1)}$.

Answer 3

There is a counterexample given in Jacobsons book, i.e., an example of a finite-dimensional solvable Lie algebra $L$ in characteristic $p>0$ such that $[L,L]$ is not nilpotent. More concretely, this Lie algebra $L$ has a basis $(x,y,e_1,\ldots ,e_p)$ with Lie brackets \begin{align*} [x,y] & = x, \\ [x,e_1] & = e_p, \\ [x,e_i] & = e_{i-1}, \; i\ge 2,\\ [y,e_i] & = (i-1)e_i, \; 1\le i\le p. \end{align*} Since $V=\operatorname{span}(e_1,\ldots ,e_p)$ is solvable (even abelian) and the quotient $L/V$ is solvable, so is $L$. But $[L,L]=kx\oplus V$ is not nilpotent. We have $[L,L]^1=[L,L]^2=\cdots =V$.