# The Unapologetic Mathematician

## Lie’s Theorem

The lemma leading to Engel’s theorem boils down to the assertion that there is some common eigenvector for all the endomorphisms in a nilpotent linear Lie algebra $L\subseteq\mathfrak{gl}(V)$ on a finite-dimensional nonzero vector space $V$. Lie’s theorem says that the same is true of solvable linear Lie algebras. Of course, in the nilpotent case the only possible eigenvalue was zero, so we may find things a little more complicated now. We will, however, have to assume that $\mathbb{F}$ is algebraically closed and that no multiple of the unit in $\mathbb{F}$ is zero.

We will proceed by induction on the dimension of $L$ using the same four basic steps as in the lemma: find an ideal $K\subseteq L$ of codimension one, so we can write $L=K+\mathbb{F}z$ for some $z\in K\setminus L$; find common eigenvectors for $K$; find a subspace of such common eigenvectors stabilized by $L$; find in that space an eigenvector for $z$.

First, solvability says that $L$ properly includes $[L,L]$, or else the derived series wouldn’t be able to even start heading towards $0$. The quotient $L/[L,L]$ must be abelian, with all brackets zero, so we can pick any subspace of this quotient with codimension one and it will be an ideal. The preimage of this subspace under the quotient projection will then be an ideal $K\subseteq L$ of codimension one.

Now, $K$ is a subalgebra of $L$, so we know it’s also solvable, so induction tells us that there’s a common eigenvector $v\in V$ for the action of $K$. If $K$ is zero, then $L$ must be one-dimensional abelian, in which case the proof is obvious. Otherwise there is some linear functional $\lambda\in K^*$ defined by

$\displaystyle k(v)=\lambda(k)v$

Of course, $v$ is not the only such eigenvector; we define the (nonzero) subspace $W$ by

$\displaystyle W=\{w\in V\vert\forall k\in K, k(w)=\lambda(k)w\}$

Next we must show that $L$ sends $W$ back into itself. To see this, pick $l\in L$ and $k\in K$ and check that

\displaystyle\begin{aligned}k(l(w))&=l(k(w))-[l,k](w)\\&=l(\lambda(k)w)-\lambda([l,k])w\\&=\lambda(k)l(w)-\lambda([l,k])w\end{aligned}

But if $l(w)\in W$, then we’d have $k(l(w))=\lambda(k)l(w)$; we need to verify that $\lambda([l,k])=0$. In the nilpotent case — Engel’s theorem — the functional $\lambda$ was constantly zero, so this was easy, but it’s a bit harder here.

Fixing $w\in W$ and $l\in L$ we pick $n$ to be the first index where the collection $\{l^i(w)\}_{i=0}^n$ is linearly independent — the first one where we can express $l^n(w)$ as the linear combination of all the previous $l^i(w)$. If we write $W_i$ for the subspace spanned by the first $i$ of these vectors, then the dimension of $W_i$ grows one-by-one until we get to $\dim(W_n)=n$, and $W_{n+i}=W_n$ from then on.

I say that each of the $W_i$ are invariant under each $k\in K$. Indeed, we can prove the congruence

$\displaystyle k(l^i(w))\equiv\lambda(k)l^i(w)\quad\mod W_i$

that is, $k$ acts on $l^i(w)$ by multiplication by $\lambda(k)$, plus some “lower-order terms”. For $i=0$ this is the definition of $\lambda$; in general we have

\displaystyle\begin{aligned}k(l^i(w))&=k(l(l^{i-1}(w)))\\&=l(k(l^{i-1}(w)))-[l,k](l^{i-1}(w))\\&=\lambda(k)l^i(w)+l(w')-\lambda([l,k])l^{i-1}(w)-w''\end{aligned}

for some $w',w''\in W_{i-1}$.

And so we conclude that, using the obvious basis of $W_n$ the action of $k$ on this subspace is in the form of an upper-triangular matrix with $\lambda(k)$ down the diagonal. The trace of this matrix is $n\lambda(k)$. And in particular, the trace of the action of $[l,k]$ on $W_n$ is $n\lambda([l,k])$. But $l$ and $k$ both act as endomorphisms of $W_n$ — the one by design and the other by the above proof — and the trace of any commutator is zero! Since $n$ must have an inverse we conclude that $\lambda([l,k])=0$.

Okay so that checks out that the action of $L$ sends $W$ back into itself. We finish up by picking some eigenvector $v_0\in W$ of $z$, which we know must exist because we’re working over an algebraically closed field. Incidentally, we can then extend $\lambda$ to all of $L$ by using $z(v_0)=\lambda(z)v_0$.