Finite Lie triangularization and rank-one complete reducibility

For solvable $L$, induct on $\dim L$. The zero algebra is immediate. A nonzero solvable algebra has $[L,L]\mathrm{\ne }L$, since otherwise its derived series never vanishes. Choose a codimension-one subspace $H$ containing $[L,L]$; it is an ideal and is solvable, because its derived series is contained in that of $L$. By induction choose $0\mathrm{\ne }v$ with $av=\lambda (a)v$ for $a\in H$. Take $x\in L\setminus H$ and let $W$ be the span of $v,xv,x^{2}v,\dots$. This span is finite-dimensional and $x$-invariant. The identity $a{x}^{j}v=x(a{x}^{j-1}v)+[a,x]x^{j-1}v$ and induction in $j$ show that $H$ preserves each initial span and acts on its new basis vector with diagonal coefficient $\lambda (a)$. Taking the first linearly dependent power therefore gives a basis of $W$ in which every $a\in H$ is upper triangular with constant diagonal $\lambda (a)$.

For the rank-one relations, direct induction gives $e{f}^{k}v=f^{k}ev+k{f}^{k-1}(h-k+1)v(k\ge 1)\mathrm{.}$ The commutation with $h$ similarly shifts generalized $h$-weights: $e$ sends the generalized weight-$\lambda$ space to weight $\lambda +2$, and $f$ to weight $\lambda -2$. F1 gives only finitely many such weights. Repeated application must eventually leave this finite set, so both $e$ and $f$ are nilpotent on any finite-dimensional representation. Choose an eigenvalue $\lambda$ with $\lambda +2$ absent and a nonzero $h$-eigenvector $v$ there. Then $ev=0$. Let $m\ge 0$ be the last index with $f^{m}v\mathrm{\ne }0$. The displayed identity at $k=m+1$ forces $\lambda =m$. The vectors $v,fv,\dots ,f^{m}v$ have distinct $h$-eigenvalues, are independent, and obey the action formulas in the Statement.

The operator $\mathrm{\Omega }=h^2+2h+4fe$ commutes with $h,e,f$. For example, $[e,h^2]=-2(eh+he)$ and $[e,fe]=he$, so $[e,\mathrm{\Omega }]=-2eh+2he-4e=0$; the computation with $f$ is identical after using $[f,h]=2f$ and $[f,e]=-h$. Thus its generalized eigenspaces are subrepresentations.

As $W$ is also $x$-invariant, the trace of $[a,x]$ on $W$ is zero. Step 1.1 identifies it as $(\dim W)\lambda ([a,x])$, so $\lambda ([a,x])=0$. Applying the recurrence there to every $a\in H$ now proves $a{x}^{j}v=\lambda (a)x^{j}v$ for all $j$: in the induction, the commutator term has scalar $\lambda ([a,x])=0$. Hence $H$ acts scalarly on $W$. F1 gives an eigenvector of $x{\mathrm{\mid }}_W$, which is a common eigenvector for $L$. This completes the dimension induction. Apply the result successively to the quotients by its invariant one-dimensional spaces and lift their finite bases. This yields a full invariant flag and upper triangularity.

Each string constructed in step 1.2 is simple. Any nonzero invariant subspace contains some nonzero individual $h$-weight component: for distinct eigenvalues, the elementary interpolation polynomials ${\prod }_{\nu \mathrm{\ne }\mu }(h-\nu )\mathrm{/}(\mu -\nu )$ project onto the weight-$\mu$ line. Raising from that line uses only the nonzero coefficients $k(m-k+1)$ until it reaches $v$, and lowering then spans the whole string. Thus every simple finite-dimensional rank-one module is one of these $V_m$, since it contains such a string. Conversely the displayed formulas directly satisfy the three bracket relations, including both endpoints, so each $V_m$ exists and is simple. Its highest vector has $\mathrm{\Omega }$-eigenvalue $m(m+2)$; centrality from step 1.3 makes that its scalar on the entire string. These scalars are distinct for nonnegative integers $m$, since $m(m+2)-n(n+2)=(m-n)(m+n+2)$.

Prove complete reducibility by induction on the dimension of a module $V$. A finite composition series exists by repeatedly taking a proper submodule of maximal dimension; each quotient is simple, and dimensions decrease. Step 2.2 identifies its factors as $V_m$. F1 decomposes $V$ into generalized $\mathrm{\Omega }$-eigenspaces. If there is more than one nonzero block, each has smaller dimension and the induction hypothesis proves the result on each. It remains to handle one block with eigenvalue $c$. Every simple factor of it has $c=m(m+2)$, because a power of $\mathrm{\Omega }-c$ annihilates that factor. Step 2.2 makes this the same $m$ for all factors. If $V$ itself is simple we are done. Otherwise choose a maximal proper submodule $U$. By induction it is a direct sum of copies of $V_m$, and $V\mathrm{/}U\cong V_m$.

In the single-block case of step 3.1, the generalized $h$-weights of $V$ are all among $m,m-2,\dots ,-m$: in a basis adapted to a composition series the characteristic polynomial is the product of those of the factors. Lift the highest vector of $V\mathrm{/}U$ to a vector $v$ in generalized $h$-weight $m$. Such a lift exists by decomposing any lift using F1; all other generalized-weight components have zero image in the weight-$m$ line of the quotient. There is no generalized weight $m+2$ or $-m-2$, so step 1.2 gives $ev=0$ and $f^{m+1}v=0$. Also $(h-m)v\in U$ lies in its weight-$m$ space, since $U$ is a direct sum of $V_m$ and $h$ is diagonalizable there. The identity from step 1.2 gives $0=e{f}^{m+1}v=(m+1)f^m(h-m)v\mathrm{.}$ On the top weight space of $U$, $f^m$ is injective, as it sends each highest basis vector of a summand $V_m$ to its nonzero bottom vector. Thus $(h-m)v=0$.

The string generated by this $v$ is now a copy of $V_m$ by steps 1.2 and 2.2 and maps nontrivially, hence isomorphically, to $V\mathrm{/}U$. Its intersection with $U$ is a proper submodule of that simple string, hence zero, and the surjection to $V\mathrm{/}U$ gives $V=U\oplus V_m$. This completes the induction. The formula for $fe$ and $ef$ follows at once by composing the displayed raising/lowering actions, giving the stated strict positivity. The case $m=0$ uses $f^0=\mathrm{id}$ in step 4.1 and is included; the zero module is the empty sum. Every series, interpolation, eigenvector choice and basis lift used was finite, so no AC occurs.