Solvable Lie 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 $L$.