# How Do Engel and Lie Arise Naturally?

I am wondering how Lie's theorem and Engel's theorem fit into the theory of Lie algebras naturally, perhaps they are motivated by the Levi decomposition and the JNF decomposition?

I find it jarring to prove Engel for nilpotent Lie algebras for no real reason, but nilpotent operators arise in the JNF naturally, so perhaps this motivates Engel's theorem? It would be great to run into the necessity of needing these theorems without realizing it, as can be done for the Levi decomposition!

Overall, my best attempt so far is:

In a general real/complex Lie algebra $L$, if you write a JNF decomposition $$x = n + s,$$ with $s$ diagonalizable, $n$ nilpotent, what happens to $x$ when you do a Levi decomposition $$L = N \rtimes S,$$ $N$ solvable, $S$ semi-simple? On the one hand it kind of looks like a Levi decomposition is motivated by the JNF decomposition, but Erdmann's Lie Algebras book seems to only apply the JNF to the semi-simple part $S$ of $J$, so it seems like they are different things?

If the JNF applies only to the semi-simple part, then am I right in saying, given $L$, you first do a Levi decomposition $$L = N \rtimes S,$$ apply Lie's theorem on solvable Lie algebras to $N$ to decompose some of $x$ into upper triangular form $u$ and the rest, $x'$, then just needs to be dealt with, $$x = u + x',$$ so for $x'$ we can decompose it using the JNF into $$x' = n + s,$$ so that $$x = u + n + s,$$ and then on the nilpotent part $n$ we apply Engel's theorem to bring $n$ into strictly-upper-triangular form?

• You may already know this, but some structure theory for lie algebras, $\frak g$, uses the adjoint action of a cartan sub algebra $\frak h$ on $\frak g$. $\frak h$ is, in particular, nilpotent by definition, so Engel's theorem is useful. When $\frak g$ is semi simple, it turns out $\frak h$ is abelian, but you may actually need Engel's theorem to prove that- I can't remember for sure though. – Tim kinsella Oct 17 '16 at 0:03
• JNF = "Jordan Normal Form" (you could have said it). What do you can the JNF decomposition in a Lie algebra? it sounds senseless. First of all, JNF is not really a decomposition but a choice of matrix form. So you mean the Jordan (or Jordan-Chevalley) decomposition, but this is for matrices/endomorphisms. So for an element of a Lie algebra, this is senseless, and even if the Lie algebra is embedded in a matrix algebra, it is not stable under taking diagonalizable/nilpotent parts of the decomposition. – YCor Oct 17 '16 at 2:14

The importance of Lie's Theorem is immense. It holds for algebraic groups as well (Lie-Kolchin). Also here I want to give only one example. By Lie's Theorem every irreducible representation of a complex solvable Lie algebra is $1$-dimensional. This shows how different the representation theory of solvable Lie algebras is from the representation theory of, say, semisimple Lie algebras.