# Solvable equivalent to nilpotency of first derived Lie algebra?

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.

• This is not true in general, see here. A counterexample is given in Jacobson's book on Lie algebras. It is true, though, over algebraically closed fields of characteristic zero. – Dietrich Burde May 30 '15 at 14:24
• @DietrichBurde Many thanks. I don't have Jacobson's book so I'll need to look it up when I'm next in a university library (I'm not an academic). You seem to imply Quiochu Yuan's reasoning isn't quite right below: i.e. one thinks of the algebra as an algebra of $\mathcal{M}(\mathbb{K})$ matrices with elements in field $\mathbb{K}$ as special cases of $\mathcal{M}(\bar{\mathbb{K}})$ matrices with elements in the alg. closure $\bar{\mathbb{K}}$, find simulaneous upper triangularizing matrix in $T\in \mathcal{M}(\bar{\mathbb{K}})$ by Lie's theorem to argue that the transformed algebras ..... – Selene Routley May 31 '15 at 3:12
• @DietrichBurde .... $T\,\mathfrak{g}\,T^{-1}\subset \mathcal{M}(\bar{\mathbb{K}})$ and $T\,[\mathfrak{g},\,\mathfrak{g}]\,T^{-1}\subset\mathcal{M}(\bar{\mathbb{K}})$ have the claimed relationship, then transform back to the algebra of matrices in $\mathcal{M}(\mathbb{K})$ and the structure constants of the Lie algebra are unchanged by the inverse conjugation by $T^{-1}$, so the original algebras have the same properties and relationships. PS Of course I assume $\mathbb{K}$ has characteristic 0. Or am I missing something? I also see very clearly that it isn't true in general: I didn't think... – Selene Routley May 31 '15 at 3:15
• I should add that for every given dimension $n$ there exists $p_0$ such that every $n$-dimensional solvable Lie algebra of characteristic $p\ge p_0$ and dimension $n$ has nilpotent derived subalgebra. For if we have a sequence of counterexamples, they are all of bounded derived length $(\le n)$, hence taking an ultraproduct we have a Lie algebra in char. zero that is solvable and whose derived subalgebra is, say, not $n$-nilpotent, and hence is not nilpotent since the dimension is $\le n$, contradiction. Probably a careful explicit proof of the char zero case yields an explicit bound on $p_0$ – YCor May 31 '15 at 11:58
• I think it's suitable for here :) I'd first check if the char 0 proof yields a bound (working in a alg. closed field of char $p$). – YCor Jun 1 '15 at 6:11

## 3 Answers

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.

• As for a reference, this should be in any sensible textbook! – Mariano Suárez-Álvarez May 30 '15 at 4:56
• For example, it is Proposition 1.39 in Knapp's Lie Groups Beyond an Introduction (although I don't know what book of Knapp's the OP has in mind). – Qiaochu Yuan May 30 '15 at 4:58
• Indeed. Lie's theorem gives you a flag of ideals in the adjoint rep which the rep preserves, and elements in the derived subalgebra push things one level deeper. – Mariano Suárez-Álvarez May 30 '15 at 5:00
• @MarianoSuárez-Alvarez that's a nice statement of a proof sketch: instantly graspable. Commutators of upper triangulars are strictly upper triangular and the latter form a Lie subalgebra qed. I was missing something trivial, as I suspected. – Selene Routley May 30 '15 at 5:02
• @WetSavanna: second edition. – Qiaochu Yuan May 30 '15 at 5:25

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) 2-solvable Lie algebra has $c$-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, the product $\prod_n\mathfrak{g}_n$ (or the restricted product if you like) is 3-solvable and its derived subalgebra is non-nilpotent).

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 $\mathfrak{a}\oplus\mathfrak{f}_{(1,0)}\oplus\mathfrak{f}_{(0,1)}$.

• Not what I was thinking of (I was thinking mainly finite dimensional - but didn't make this clear, so my bad) but extremely interesting and enlightening: fantastic to see explicitly how countably infinite dimension exactly breaks the other arguments. Many thanks. – Selene Routley May 31 '15 at 10:42

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 $$[L,L]^1=[L,L]^2=\cdots =V$$.