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.

Advertisements