= Solution
For a <Lie algebra> $\mathfrak g$, write $[A,B]$ for the linear span of brackets with one argument in each indicated subspace. The three definitions are
$$
\begin{aligned}
\text{abelian:}&\quad[\mathfrak g,\mathfrak g]=0,\\
\text{solvable:}&\quad\mathfrak g^{(r)}=0\text{ for some }r,\quad\mathfrak g^{(0)}=\mathfrak g,\quad\mathfrak g^{(j+1)}=[\mathfrak g^{(j)},\mathfrak g^{(j)}],\\
\text{nilpotent:}&\quad\gamma_s(\mathfrak g)=0\text{ for some }s,\quad\gamma_1=\mathfrak g,\quad\gamma_{j+1}=[\mathfrak g,\gamma_j].
\end{aligned}
$$
These are respectively an <Abelian Lie algebra>, a <Solvable Lie algebra> and a <Nilpotent Lie algebra>. The second and third sequences are the <derived series of a Lie algebra> and the <Lower central series of a Lie algebra>.
An <Abelian Lie algebra> has $\gamma_2=0$, and a <Nilpotent Lie algebra> is a <Solvable Lie algebra>: induction gives $\mathfrak g^{(j)}\subseteq\gamma_{j+1}$. Thus \b[all the implications are generated by]
$$
\boxed{\text{abelian}\ \Longrightarrow\ \text{nilpotent}\ \Longrightarrow\ \text{solvable}.}
$$
None of the reverse implications holds. The <Heisenberg Lie algebra> with basis $x,y,z$ and $[x,y]=z$, all other basic brackets zero, is nonabelian but has $\gamma_2=\mathbb Cz$, $\gamma_3=0$. The two-dimensional <affine Lie algebra of the line> with $[h,e]=e$ has $\mathfrak g^{(1)}=\mathbb Ce$ and $\mathfrak g^{(2)}=0$, but $\gamma_j=\mathbb Ce$ for every $j\ge2$. It is solvable and neither nilpotent nor abelian. These two examples answer all six ordered-pair comparisons.
The <Lie theorem> says that a finite-dimensional <Lie algebra representation> of a finite-dimensional <Solvable Lie algebra> over an <algebraically closed field> of <characteristic> zero has a common <eigenvector> whenever its representation space is nonzero. Equivalently it admits an invariant <complete flag>, or simultaneous upper triangularization. Here the field may be taken to be $\mathbb C$; these hypotheses are essential.
To prove the common-eigenvector assertion, induct on $\dim\mathfrak g$. The zero algebra is immediate. Since $\mathfrak g$ is nonzero and solvable, its <derived algebra> is proper. Choose a codimension-one <Lie algebra ideal> $\mathfrak h$ containing it, and write $\mathfrak g=\mathfrak h\oplus\mathbb Cx$. The <Lie algebra> $\mathfrak h$ is solvable, so induction supplies $v\ne0$ and a <linear functional> $\lambda:\mathfrak h\to\mathbb C$ with $hv=\lambda(h)v$.
Consider the finite-dimensional cyclic subspace $W=\operatorname{span}\{v,xv,x^2v,\ldots\}$. Until the first <linear dependence>, these powers form a basis. The identity
$$
hx^jv=x(hx^{j-1}v)+[h,x]x^{j-1}v
$$
and $[h,x]\in\mathfrak h$ show by induction, simultaneously for all $h\in\mathfrak h$, that $W$ is $\mathfrak h$-invariant and that $h$ is upper triangular on it with every diagonal entry $\lambda(h)$. It is also $x$-invariant by construction. Hence
$$
0=\operatorname{tr}_W[x,h]=(\dim W)\lambda([x,h]).
$$
The <trace> of a <commutator> is zero, and <characteristic> zero gives $\lambda([x,h])=0$.
The simultaneous <eigenspace> $E_\lambda=\{w:hw=\lambda(h)w\text{ for every }h\in\mathfrak h\}$ is nonzero and $x$-invariant, since
$$
h(xw)=x(hw)+[h,x]w=\lambda(h)xw.
$$
Over an <algebraically closed field>, $x|_{E_\lambda}$ has an <eigenvector>, which is therefore a common <eigenvector> for all of $\mathfrak g$. This finishes induction. Apply the same assertion to the <quotient representation> by its invariant line, and then to successive quotients. A <basis> adapted to the resulting <complete flag> gives the stated <simultaneous triangularization of a Lie algebra representation>, completing the proof of the <Lie theorem>.
A <Nilpotent Lie algebra> is a <Solvable Lie algebra>, so the inclusion $\mathfrak g\subseteq\mathfrak{gl}(V)$ satisfies the <Lie theorem> and is upper triangular in a suitable <basis>, for finite-dimensional complex $V$.
This is insufficient to prove the <Engel theorem>. Its matrix version starts with a <Lie subalgebra> of <nilpotent endomorphisms> and concludes that they are simultaneously strictly upper triangular; its abstract version concludes nilpotence from nilpotence of every <Adjoint representation> endomorphism. Merely <upper triangular matrices> can have nonzero diagonal entries: the one-dimensional algebra $\mathbb CI_V$ is an <Abelian Lie algebra> and a <Nilpotent Lie algebra>, but acts by a nonnilpotent identity matrix. This is <nilpotent Lie algebras need not act nilpotently>. Moreover, in the abstract <Engel theorem> nilpotence of the algebra is a conclusion, so assuming it first to invoke the <Lie theorem> would be circular. \b[Abstract nilpotence and nilpotence of each representing matrix are different conditions.]
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