Derived series of a Lie algebra

The derived series is ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(n+1)}=[{\mathfrak{g}}^{(n)},{\mathfrak{g}}^{(n)}]$.

The derived algebra is the linear span of all Lie brackets $[x,y]$ in a Lie algebra. The Jacobi identity makes it an ideal of a Lie algebra. Its quotient is the largest abelian Lie algebra quotient of $\mathfrak{g}$.

A finite-dimensional complex Lie algebra is a solvable Lie algebra exactly when its derived algebra is a nilpotent Lie algebra. The forward direction follows from the Lie theorem in the Adjoint representation and lifting nilpotence through its central kernel. The reverse direction follows because the derived series of a Lie algebra after its first term is the derived series of a Lie algebra of the derived algebra.

A Lie algebra is solvable when its derived series eventually becomes zero.

The four-dimensional diamond Lie algebra is the semidirect product of Lie algebras of the three-dimensional Heisenberg Lie algebra with a derivation having eigenvalues zero, one and minus one. With basis $c,d,p,q$, its nonzero brackets are $[d,p]=p$, $[d,q]=-q$, $[p,q]=c$. Its derived series of a Lie algebra has dimensions four, three, one and zero. The symmetric form with $B(c,d)=B(p,q)=1$ and all other basis pairings zero is a nondegenerate bilinear form and invariant. The alternating tensor $B([x,y],z)$ is the volume form on $⟨d,p,q⟩$, verifying invariance.

The two-dimensional Lie algebra of infinitesimal dilations and translations of a line has a basis $h,e$ with $[h,e]=e$. Its derived algebra is the abelian line spanned by $e$, but every nontrivial term of its lower central series of a Lie algebra is that same line. It is solvable and not nilpotent.

The solvable radical of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A Lie algebra is semisimple exactly when its solvable radical is zero.

Every finite-dimensional complex representation of a solvable Lie algebra has a common eigenvector. Equivalently, an irreducible finite-dimensional complex representation is one-dimensional; iterating gives simultaneous upper triangularization.

The solvable Lie algebra $\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, has an infinite-dimensional Irreducible Lie algebra representation on $\mathbb{C}[t]$: $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. Indeed $[x,y]f=yf$. An invariant subspace is an ideal $p(t)\mathbb{C}[t]$ because it is invariant under multiplication by $t$. Shift invariance gives $p(t)\mid p(t-1)$, hence $p(t-1)=p(t)$ and $p$ is constant. This shows why the Lie theorem needs finite-dimensional representations, even when the algebra itself is finite-dimensional.

A Lie algebra representation is simultaneously upper triangular precisely when it preserves a complete flag, with $\dim V_j=j$. A basis adapted to the invariant flag makes every representing matrix upper triangular. For a complex finite-dimensional representation of a solvable Lie algebra, the Lie theorem provides a common eigenvector; iteration on the quotient representations constructs the flag.

In a field of characteristic $p$, take a basis $e_0,\dots ,e_{p-1}$, and let $x{e}_j=e_{j-1}$ with indices modulo $p$, while $y{e}_j=j{e}_j$. Then $[x,y]=x$, so $kx+ky$ is a solvable Lie algebra, but the distinct eigenvalues of $y$ force any common eigenvector to be a coordinate vector, and $x$ cyclically permutes those vectors. There is consequently no common eigenvector, even over an algebraically closed field.

This is the solvability test in Cartan's criterion, distinct from its semisimplicity test.

A complex Lie subalgebra $\mathfrak{h}\subseteq \mathfrak{gl}(V)$ is solvable iffor every $x\in [\mathfrak{h},\mathfrak{h}]$ and $y\in \mathfrak{h}$. For an abstract Lie algebra, this is equivalent to $\kappa (\mathfrak{h},[\mathfrak{h},\mathfrak{h}])=0$.

For $x$ in the derived algebra of a complex matrix Lie algebra $L$, define $y$ by multiplication by $\overline{\lambda }$ on each generalized eigenspace $V_{\lambda }$ of $x$. Polynomial interpolation and adjoint compatibility of additive Jordan decomposition make $\mathrm{ad}y$ a polynomial in $\mathrm{ad}x$ with zero constant term, so $[y,L]\subseteq [L,L]$. Trace orthogonality of $[L,L]$ and $L$ forces the displayed sum to vanish. Thus $x$ is a nilpotent endomorphism, and the Engel theorem implies solvability. The auxiliary $y$ and the Jordan components are not required to lie in $L$.