Yesterday I was talking about Lie’s theorem for solvable Lie algebras. I went through most of the proof, but didn’t finish the last step. We had a solvable Lie algebra ${L}$ and an ideal ${I \subset L}$ such that ${I}$ was of codimension one.

There was a finite-dimensional representation ${V}$ of ${L}$. For ${\lambda \in I^*}$, we set $\displaystyle V_\lambda := \{ v \in V: Yv = \lambda(Y) v, \ \mathrm{all} \ Y \in I \}.$

We assumed ${V_\lambda \neq 0}$ for some ${\lambda}$ by the induction hypothesis. Then the following then completes the proof of Lie’s theorem, by the “fundamental calculation:”

Lemma 1 If ${V_\lambda \neq 0}$, then ${\lambda([L,I])=0}$.

Proof: Let ${X \in L}$ have an image in ${L/I}$ that generates ${L/I}$. We have to show ${\lambda([X,Y])=0}$ for any ${Y \in I}$.

So, fix ${v \in v_\lambda}$. Consider the subspaces $\displaystyle W_i := \{ \mathrm{ \ span \ of \ } v, Xv, \dots, X^i v \} \subset V.$

Eventually, the ${W_i}$ have to stabliize, say at ${W := W_M}$. Computing the action of elements of ${I}$ on ${W}$ and taking the trace will give the lemma.

I claim, first of all, that any ${Y \in L}$ stabilizes ${W_i}$. This follows by induction. When ${i=0}$, it’s the meaning of being an eigenvector. If it’s true for ${i-1}$, then by the inductive hypothesis and ${[Y,X] \in I}$: $\displaystyle Y(X^iv) = XYX^{i-1}v + [Y,X] X^{i-1}v \in X(W_{i-1}) + W_{i-1} \subset W_i.\ \ \ \ \ (1)$

Thus, via the flag ${W_0 \subset W_1 \subset \dots \subset W_M}$, we can represent the action of ${I}$ as upper-triangular matrices. But we now want to know what the diagonal elements are.

I claim next these are just all the diagonal elements are just ${\lambda(Y)}$. More precisely, we have $\displaystyle Y X^i v \equiv \lambda(Y) X^i v \quad mod \ W_{i-1}.$

This is checked by induction on ${i}$, as before; it’s true for ${i=0}$, and we use the same type of argument as in (1) to induct on ${i}$. So, we now know how elements of ${I}$ act on the filtration. In particular, $\displaystyle Tr(Y_W) = (M+1) \lambda(Y).$

(Here ${Y_W}$ denotes the operator from the action of ${Y}$ on ${W}$.) If we replace ${Y}$ with ${[Y,X]}$, we get ${\lambda([Y,X]_W) = \frac{1}{M+1} Tr( [Y,X]) = 0}$ since the trace of a commutator is always zero.

Note that in the last step only, we used the hypothesis of characteristic zero. $\Box$

The proof of Lie’s theorem now is basically complete. Since the element ${X}$ generating ${L/I}$ stabilizes ${V_\lambda}$, we can find an eigenvector for ${X}$ since the field is algebraically closed, and this is what we wanted.

So we now have, incidentally, a classification of simple representations of a solvable Lie algebra- they’re just one-dimensional.  The case for semisimple Lie algebras is much more complicated and interesting.  That is the ultimate aim of this series of posts.