Lie’s theorem | Climbing Mount Bourbaki

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 ${\lambda([L,I])=0}$ .

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I talked a bit earlier about nilpotent Lie algebras and Engel’s theorem. There is an analog for solvable Lie algebras, and the corresponding Lie’s theorem.

So, first the definitions. Solvability is similar to nilpotence in that one takes repeated commutators, except one uses the derived series instead of the lower central series.

In the future, fix a Lie algebra ${L}$ over an algebraically closed field ${k}$ of characteristic zero.

Definition 1 The derived series of ${L}$ is the descending filtration ${D_n}$ defined by ${D_0 := L, D_n := [D_{n-1}, D_{n-1}]}$ . The Lie algebra ${L}$ is solvable if ${D_M=0}$ for some ${M}$ .

For instance, a nilpotent Lie algebra is solvable, since if ${\{C_n\}}$ is the lower central series, then ${D_n \subset C_n}$ for each ${n}$ .

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Climbing Mount Bourbaki

Lie's Theorem II

Lie's Theorem I

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