Solvable and Nilpotent: Engel and Lie

Two chains: derived and lower central

Define the derived series L^{(0)}=L, L^{(1)}=[L,L], L^{(k+1)}=[L^{(k)},L^{(k)}]. L is solvable if this reaches 0. Define the lower central series L^0=L, L^1=[L,L], L^{k+1}=[L,L^k]. L is nilpotent if *this* reaches 0. Since L^{(k)}⊆L^k, nilpotent ⇒ solvable, never the reverse. These are the exact analogues of solvable and nilpotent groups, with the bracket replacing the group commutator.

b = upper-triangular 2x2,  basis: h=[1,0;0,0], h'=[0,0;0,1], e=[0,1;0,0]
(equivalently span of diagonal d=h, d'=h', and e).

Derived series of b:
  [b,b] = span{ [d, e] } : [diag, e] is a multiple of e, [d,d']=0
     so  b^(1) = [b,b] = span{e}     (1-dimensional, abelian)
  b^(2) = [b^(1), b^(1)] = [span e, span e] = 0
  --> derived series: b ⊃ <e> ⊃ 0.  b is SOLVABLE.

Lower central series of b:
  b^1 = [b,b] = span{e}
  b^2 = [b, span e] = span{ [d,e], [d',e], [e,e] } = span{e}  (NOT smaller!)
  b^3 = [b, span e] = span{e} ...  stabilizes at <e>, never 0.
  --> b is NOT nilpotent.

Now n = strictly upper-triangular (just span{e}): [n,n]=0, abelian,
  hence n is nilpotent (and solvable).  Contrast: n nilpotent, b only solvable.

Engel's theorem

Call x ∈ L ad-nilpotent if ad(x) is a nilpotent operator. Engel's theorem says: a finite-dimensional Lie algebra L is nilpotent if and only if every element is ad-nilpotent. One direction is easy; the substance is the converse. The sharp form, from which the rest follows, is the statement about linear Lie algebras: if L⊆gl(V) consists of nilpotent operators and V≠0, then there is a nonzero v ∈ V killed by all of L (Lv=0). Iterating produces a full flag in which every element of L is strictly upper-triangular.

Lie's theorem and the common flag idea

Lie's theorem is the solvable counterpart, and it needs an algebraically closed field of characteristic 0 (take ℂ). It says: if L⊆gl(V) is solvable and V≠0, then L has a common eigenvector — a single v with x·v=λ(x)v for all x ∈ L, where λ:L→ℂ is linear (a *weight*). Iterating up a flag, every x ∈ L is simultaneously upper-triangular. So over ℂ, solvable = simultaneously triangularizable, exactly as Engel gave nilpotent = simultaneously strictly-triangularizable.

1. Both proofs run by induction on dim L: peel off a codimension-1 ideal M (solvability/nilpotency hands you one).
2. Find the common eigenvector (or annihilated vector) for M by induction; collect the eigenspace V_λ.
3. Show V_λ is invariant under the remaining direction x∉M (Lie's key lemma: λ([x,m])=0); diagonalize x on V_λ to extend the eigenvector.